Ch 08 Hw 1a1z16
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- Pages: 34
4/2/2016
Ch 08 HW
Ch 08 HW Due: 12:59pm on Saturday, April 2, 2016 You will receive no credit for items you complete after the assignment is due. Grading Policy
Center of Mass and External Forces Learning Goal: Understand that, for many purposes, a system can be treated as a pointlike particle with its mass concentrated at the center of mass. A complex system of objects, both pointlike and extended ones, can often be treated as a point particle, located at the system's center of mass. Such an approach can greatly simplify problem solving. Before you use the center of mass approach, you should first understand the following : System: Any collection of objects that are of interest to you in a particular situation. In many problems, you have a certain freedom in choosing your system. Making a wise choice for the system is often the first step in solving the problem efficiently. Center of mass: The point that represents the "average" position of the entire mass of a system. The ⃗ postion of the center of mass r cm can be expressed in of the position vectors r i⃗ of the particles as ⃗ r cm =
∑ m i r i⃗ ∑ mi
.
The x coordinate of the center of mass xcm can be expressed in of the x coordinates (r x ) i of the particles as x cm =
∑ m i (rx ) i ∑ mi
.
Similarly, the y coordinate of the center of mass can be expressed. Internal force: Any force that results from an interaction between the objects inside your system. As we will show, the internal forces do not affect the motion of the system's center of mass. External force: Any force acting on an object inside your system that results from an interaction with an object outside your system. Consider a system of two blocks that have masses m1 and m2 . Assume that the blocks are pointlike particles and are located along the x axis at the coordinates x1 and x2 as shown . In this problem, the blocks can only move along the x axis.
Typesetting math: 45% https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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Part A Find the x coordinate xcm of the center of mass of the system. Express your answer in of m1 , m2 , x1 , and x2 . ANSWER:
xcm
=
m 1 x 1 +m 2 x 2 m 1 +m 2
Correct
Part B If m2
≫ m1
, then the center of mass is located:
ANSWER: to the left of m1 at a distance much greater than x2 to the left of m1 at a distance much less than x2
− x1
− x1
to the right of m1 at a distance much less than x2
− x1
to the right of m2 at a distance much greater than x2 to the right of m2 at a distance much less than x2 to the left of m2 at a distance much less than x2
− x1
− x1
− x1
Correct
Part C If m2
= m1
, then the center of mass is located:
ANSWER: at m1 at m2 halfway between m1 and m2 the answer depends on x1 and x2
Correct
Part D Recall that the blocks can only move along the x axis. The x components of their velocities at a certain moment are and . Find the component of the velocity of the center of mass ( https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
)
at that moment. Keep in
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are v 1x and v 2x . Find the x component of the velocity of the center of mass (v cm ) x at that moment. Keep in mind that, in general: v x = dx/dt. Express your answer in of m1 , m2 , v 1x , and v 2x . ANSWER: (v cm )
x
=
m 1 v 1x +m 2 v 2x m 1 +m 2
Correct Because v 1x and v 2x are the x components of the velocities of m1 and m2 their values can be positive or negative or equal to zero.
Part E →
→
Suppose that v 1x and v 2x have equal magnitudes. Also, v 1 is directed to the right and v 2 is directed to the left. The velocity of the center of mass is then: ANSWER: directed to the left directed to the right zero m1
the answer depends on the ratio m
2
Correct
Part F Assume that the x components of the blocks' momenta at a certain moment are p1x and p2x . Find the x component of the velocity of the center of mass (v cm ) x at that moment. Express your answer in of m1 , m2 , p1x , and p2x . ANSWER:
(v cm )
x
=
p 1x +p 2x m 1 +m 2
Correct
Part G ⃗ Suppose that v cm
= 0
. Which of the following must be true?
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ANSWER: |p
1x
| = |p
2x
|
|v 1x | = |v 2x | m1 = m2
none of the above
Correct
Part H Assume that the blocks are accelerating, and the x components of their accelerations at a certain moment are a 1x and a 2x . Find the x component of the acceleration of the center of mass (a cm ) x at that moment. Keep in mind that, in general, a x = dv x /dt . Express your answer in of m1 , m2 , a 1x , and a 2x . ANSWER: (a cm )
x
=
m 1 a1x +m 2 a2x m 1 +m 2
Correct Because a 1x and a 2x are the x components of the velocities of m1 and m2 their values can be positive or negative or equal to zero.
We will now consider the effect of external and internal forces on the acceleration of the center of mass.
Part I Consider the same system of two blocks. An external force F ⃗ is now acting on block m1 . No forces are applied to block m2 as shown . Find the x component of the acceleration of the center of mass (a cm ) x of the system. Express your answer in of the x component Fx of the force, m1 ,and m2 .
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Hint 1. Using Newton's laws Find the acceleration of each block from Newton's second law and then apply the formula for (a cm ) x found earlier. ANSWER:
(a cm )
x
=
Fx m 1 +m 2
Correct
Part J →
Consider the same system of two blocks. Now, there are two forces involved. An external force F1 is acting on →
block m1 and another external force F2 is acting on block m2 . Find the x component of the acceleration of the center of mass (a cm ) x of the system. Express your answer in of the x components F 1x and F 2x of the forces, m 1 and m 2 .
ANSWER: (a cm )
x
=
F1x +F2x m 1 +m 2
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Correct Note that, in both cases, the acceleration of the center of mass can be found as (acm )
x
=
(Fnet )
x
Mtotal
where Fnet is the net external force applied to the system, and M total is the total mass of the system. Even though each force is only applied to one object, it affects the acceleration of the center of mass of the entire system. This result is especially useful since it can be applied to a general case, involving any number of objects moving in all directions and being acted upon by any number of external forces.
Part K Consider the previous situation. Under what condition would the acceleration of the center of mass be zero? Keep in mind that F1x and F2x represent the components, of the corresponding forces. ANSWER: F 1x = −F 2x F 1x = F 2x m1 = m2 m1 ≪ m2
Correct
Part L ⃗ Consider the same system of two blocks. Now, there are two internal forces involved. An internal force F 12 is ⃗ applied to block m1 by block m2 and another internal force F 21 is applied to block m2 by block m1 . Find the x component of the acceleration of the center of mass (a cm ) of the system. x
Express your answer in of the x components F 12x and F 21x of the forces, m 1 and m 2 .
ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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\texttip{(a_{\rm cm})_x}{a_cm_x} = \large{\frac{F_{\rm{12x}}+F_{\rm{21x}}}{m_{1}+m_{2}}}
Correct Newton's 3rd law tells you that |F_{12x}|=|F_{21x}|. From your answers above, you can conclude that (a_{\rm cm})_x=0.The internal forces do not change the velocity of the center of mass of the system. In the absence of any external forces, (a_{\rm cm})_x=0 and \texttip{(v_{\rm cm})_x}{v_cm_x} is constant. You just demonstrated this to be the case for the twobody situation moving along the x axis; however, it is true in more general cases as well.
A Bullet Is Fired into a Wooden Block A bullet of mass \texttip{m_{\rm b}}{m_b} is fired horizontally with speed \texttip{v_{\rm i}}{v_i} at a wooden block of mass \texttip{m_{\rm w}}{m_w} resting on a frictionless table. The bullet hits the block and becomes completely embedded within it. After the bullet has come to rest within the block, the block, with the bullet in it, is traveling at speed \texttip{v_{\rm f \hspace{1 pt}}}{v_f}.
Part A Which of the following best describes this collision?
Hint 1. Types of collisions An inelastic collision is a collision in which kinetic energy is not conserved. In a partially inelastic collision, kinetic energy is lost, but the objects colliding do not stick together. From this information, you can infer what completely inelastic and elastic collisions are. ANSWER:
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perfectly elastic partially inelastic perfectly inelastic
Correct
Part B Which of the following quantities, if any, are conserved during this collision?
Hint 1. When is kinetic energy conserved? Kinetic energy is conserved only in perfectly elastic collisions. ANSWER: kinetic energy only momentum only kinetic energy and momentum neither momentum nor kinetic energy
Correct
Part C What is the speed of the block/bullet system after the collision? Express your answer in of \texttip{v_{\rm i}}{v_i}, \texttip{m_{\rm w}}{m_w}, and \texttip{m_{\rm b}} {m_b}.
Hint 1. Find the momentum after the collision What is the total momentum \texttip{p_{\rm total}}{p_total} of the block/bullet system after the collision? Express your answer in of \texttip{v_{\rm f \hspace{1 pt}}}{v_f} and other given quantities. ANSWER: \texttip{p_{\rm total}}{p_total} = \left(m_{w}+m_{b}\right) v_{f}
Hint 2. Use conservation of momentum The momentum of the block/bullet system is conserved. Therefore, the momentum before the collision is the same as the momentum after the collision. Find a second expression for \texttip{p_{\rm total}}{p_total}, https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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this time expressed as the total momentum of the system before the collision. Express your answer in of \texttip{v_{\rm i}}{v_i} and other given quantities. ANSWER: \texttip{p_{\rm total}}{p_total} = m_{b} v_{i}
ANSWER: \texttip{v_{\rm f \hspace{1 pt}}}{v_f} = \large{\frac{m_{b} v_{i}}{m_{b}+m_{w}}}
Correct
A Ball Hits a Wall Elastically A ball of mass \texttip{m}{m} moving with velocity \texttip{\vec{v}_{\rm i}}{v_i_vec} strikes a vertical wall as shown in . The angle between the ball's initial velocity vector and the wall is \texttip{\theta _{\rm i}}{theta_i} as shown on the diagram, which depicts the situation as seen from above. The duration of the collision between the ball and the wall is \texttip{\Delta t}{Deltat}, and this collision is completely elastic. Friction is negligible, so the ball does not start spinning. In this idealized collision, the force exerted on the ball by the wall is parallel to the x axis.
Part A What is the final angle \texttip{\theta _{\rm f \hspace{1 pt}}}{theta_f} that the ball's velocity vector makes with the negative y axis? Express your answer in of quantities given in the problem introduction.
Hint 1. How to approach the problem Relate the vector components of the ball's initial and final velocities. This will allow you to determine \texttip{\theta _{\rm f \hspace{1 pt}}}{theta_f} in of \texttip{\theta _{\rm i}}{theta_i}.
Hint 2. Find the y component of the ball's final velocity https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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What is \texttip{v_{\rm fy}}{v_fy}, the \texttip{y}{y} component of the final velocity of the ball? Express your answer in of quantities given in the problem introduction and/or \texttip{v_{\rm ix}}{v_ix} and \texttip{v_{\rm iy}}{v_iy}, the \texttip{x}{x} and \texttip{y}{y} components of the ball's initial velocity.
Hint 1. How to approach this part There is no force on the ball in the y direction. From the impulsemomentum theorem, this means that the change in the y component of the ball's momentum must be zero. ANSWER: \texttip{v_{\rm fy}}{v_fy} = v_{i} {\cos}\left({\theta}_{i}\right)
Hint 3. Find the \texttip{x}{x} component of the ball's final velocity What is \texttip{v_{\rm fx}}{v_fx}, the \texttip{x}{x} component of the ball' final velocity? Express your answer in of quantities given in the problem introduction and/or \texttip{v_{\rm ix}}{v_ix} and \texttip{v_{\rm iy}}{v_iy}, the \texttip{x}{x} and \texttip{y}{y} components of the ball's initial velocity.
Hint 1. How to approach this problem Since energy is conserved in this collision, the final speed of the ball must be equal to its initial speed. ANSWER: \texttip{v_{\rm fx}}{v_fx} = v_{i} {\sin}\left({\theta}_{i}\right)
Hint 4. Putting it together Once you find the vector components of the final velocity in of the initial velocity, use the geometry of similar triangles to determine \texttip{\theta _{\rm f \hspace{1 pt}}}{theta_f} in of \texttip{\theta _{\rm i}}{theta_i}. ANSWER: \texttip{\theta _{\rm f \hspace{1 pt}}}{theta_f} = {\theta}_{i}
Correct
Part B What is the magnitude \texttip{F}{F} of the average force exerted on the ball by the wall? https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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Express your answer in of variables given in the problem introduction and/or \texttip{v_{\rm ix}} {v_ix}.
Hint 1. What physical principle to use Use the impulsemomentum theorem, \vec{J}=\vec{p}_{\rm f \hspace{1 pt}}\vec{p}_{\rm i}, along with the definition of impulse, \vec{J}=\sum\vec{F}\Delta t. In this case, only one force is acting, so \left|\vec{J}\right|=F\Delta t. Putting everything together, \large{F = \frac{\vec{p}_{\rm f \hspace{1 pt}} \vec{p}_{\rm i}}{\Delta t}}.
Hint 2. Change in momentum of the ball The fact that \theta_{\rm f \hspace{1 pt}}=\theta_{\rm i} implies that the \texttip{y}{y} component of the ball's momentum does not change during the collision. What is \texttip{\Delta p_{\mit x}}{Deltap_x}, the magnitude of the change in the ball's \texttip{x}{x} momentum? Express your answer in of quantities given in the problem introduction and/or \texttip{v_{\rm ix}}{v_ix}. ANSWER: \texttip{\Delta p_{\mit x}}{Deltap_x} = 2 m v_{i} {\sin}\left({\theta}_{i}\right)
ANSWER: \texttip{F}{F} = \large{\frac{2mv_{i} {\sin}\left({\theta}_{f}\right)}{{\Delta}t}}
Correct
Video Tutor: Water Rocket First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question at right. You can watch the video again at any point.
Part A https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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Suppose we repeat the experiment from the video, but this time we use a rocket three times as massive as the one in the video, and in place of water we use a fluid that is twice as massive (dense) as water. If the new fluid leaves the rocket at the same speed as the water in the video, what will be the ratio of the horizontal speed of our rocket to the horizontal speed of the rocket in the video after all the fluid has left the rocket? (Ignore air resistance.)
Hint 1. How to approach the problem First, can we use momentum conservation? Let's take the rocket and fluid as the system. In the vertical direction, momentum is not conserved because gravity exerts an external force on the system. But in the horizontal direction, momentum is conserved (since we ignore air resistance). Next, we draw beforeandafter diagrams, labeling mass and velocity in each case. (For convenience, we use “exhaust” to label the fluid both before and after it is expelled.)
Momentum conservation enables you to relate the final horizontal momenta of the rocket and the exhaust. The problem says that the fluid leaves the rocket “at the same speed” as the water in the video. What does that tell you about \vec v_{e.f}? \vec p_{e,f}? Now, write an equation for the rocket’s velocity in of known quantities, including m_e and m_r. How will doubling m_e and tripling m_r change the magnitude of \vec v_{r.f}? ANSWER: 1/2 1 2/3 1/3 3/2
Correct The rocket's speed is proportional to the ratio of the fluid's mass to the rocket's mass.
Exercise 8.12 A bat strikes a 0.145{\rm kg} baseball. Just before impact, the ball is traveling horizontally to the right at 60.0 {\rm m/s} , and it leaves the bat traveling to the left at an angle of 35 {\rm ^\circ} above horizontal with a speed of 55.0 {\rm m/s} . https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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The ball and bat are in for 1.85 {\rm ms} .
Part A Find the horizontal component of the average force on the ball. Take the xdirection to be positive to the right Express your answer using two significant figures. ANSWER: F_{\rm x} = 8200 {\rm N}
Answer Requested
Part B Find the vertical component of the average force on the ball. Express your answer using two significant figures. ANSWER: F_{\rm y} = 2500 {\rm N}
Correct
Exercise 8.17 The expanding gases that leave the muzzle of a rifle also contribute to the recoil. A .30 caliber bullet has mass 7.20×10−3 {\rm kg} and a speed of 601 {\rm m/s} relative to the muzzle when fired from a rifle that has mass 2.90 {\rm kg} . The loosely held rifle recoils at a speed of 2.05 {\rm m/s} relative to the earth.
Part A Find the momentum of the propellant gases in a coordinate system attached to the earth as they leave the muzzle of the rifle. ANSWER: p = 1.63 {\rm kg \cdot m/s}
Correct
A Girl on a Trampoline A girl of mass m_1 = 60.0 kilograms springs from a trampoline with an initial upward velocity of v_i = 8.00 meters per https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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second. At height h = 2.00 meters above the trampoline, the girl grabs a box of mass m_2 = 15.0 kilograms. For this problem, use g = 9.80 meters per second per second for the magnitude of the acceleration due to gravity.
Part A What is the speed \texttip{v_{\rm before}}{v_before} of the girl immediately before she grabs the box? Express your answer numerically in meters per second.
Hint 1. How to approach the problem Use conservation of energy. Find the initial kinetic energy \texttip{K_{\rm i}}{K_i} of the girl as she leaves the trampoline. Then find her gravitational potential energy \texttip{U_{\rm before}}{U_before} just before she grabs the box (define her initial potential energy to be zero). According to the principle of conservation of energy, K_i = U_{\rm before} + K_{\rm before}. Once you have \texttip{K_{\rm before}}{K_before}, use the definition of translational kinetic energy to find the girl's speed \texttip{v_{\rm before}}{v_before}.
Hint 2. Initial kinetic energy What is the girl's initial kinetic energy \texttip{K_{\rm i}}{K_i} as she leaves the trampoline? Express your answer numerically in joules. ANSWER: \texttip{K_{\rm i}}{K_i} = 1920 \rm{J}
Hint 3. Potential energy at height \texttip{h}{h} What is the girl's gravitational potential energy \texttip{U_{\rm before}}{U_before} immediately before she grabs the box? Express your answer numerically in joules. ANSWER: \texttip{U_{\rm before}}{U_before} = 1180 \rm{J}
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ANSWER: \texttip{v_{\rm before}}{v_before} = 4.98 \rm{m/s}
Correct
Part B What is the speed \texttip{v_{\rm after}}{v_after} of the girl immediately after she grabs the box? Express your answer numerically in meters per second.
Hint 1. How to approach the problem Think of the process of grabbing the box as a collision. Though the girl and the box don't collide as such, any interaction between two objects that takes place extremely fast can be thought of as a collision. To find the velocity at a later time, which of the following principles could you use? ANSWER: conservation of momentum alone conservation of energy alone both conservation of momentum and conservation of energy Newton's second law
Hint 2. Total initial momentum What is the total momentum before the collision? Answer in kilogram meters per second. ANSWER: \texttip{p_{\rm before}}{p_before} = 299 \rm kg \, m/s
ANSWER: \texttip{v_{\rm after}}{v_after} = 3.98 \rm m/s
Correct
Part C Is this "collision" elastic or inelastic? https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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Hint 1. Definition of an inelastic collision If two objects move together with the same velocity after a collision, the collision is said to be inelastic. ANSWER: elastic inelastic
Correct In inelastic collisions, some of the system's kinetic energy is lost. In this case the kinetic energy lost is converted to heat energy in the girl's muscles as she grabs the box, and sound energy.
Part D What is the maximum height \texttip{h_{\rm max}}{h_max} that the girl (with box) reaches? Measure \texttip{h_{\rm max}}{h_max} with respect to the top of the trampoline. Express your answer numerically in meters.
Hint 1. How to approach the problem Use conservation of energy. From Part B you know the velocity of the girl/box system just after the girl grabs the box. Therefore, you can compute the kinetic energy \texttip{K_{\rm after}}{K_after} of the girl/box system just after the collision. You can also compute the gravitational potential energy \texttip{U_{\rm after}} {U_after} of the girl/box system at this point. The sum of these two quantities must equal the gravitational potential energy of the girl/box system at the height \texttip{h_{\rm max}}{h_max} (where their velocity, and therefore kinetic energy, will be zero).
Hint 2. Finding \texttip{U_{\rm after}}{U_after} What is the girl/box system's gravitational potential energy \texttip{U_{\rm after}}{U_after} immediately after she grabs the box? Express your answer numerically in joules. ANSWER: \texttip{U_{\rm after}}{U_after} = 1470 \rm J
Hint 3. Finding \texttip{K_{\rm after}}{K_after} What is the girl/box system's kinetic energy \texttip{K_{\rm after}}{K_after} immediately after she grabs the box? Express your answer numerically in joules. ANSWER:
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\texttip{K_{\rm after}}{K_after} = 594 \rm J
ANSWER: \texttip{h_{\rm max}}{h_max} = 2.81 \rm m
Correct
Conservation of Momentum in Inelastic Collisions Learning Goal: To understand the vector nature of momentum in the case in which two objects collide and stick together. In this problem we will consider a collision of two moving objects such that after the collision, the objects stick together and travel off as a single unit. The collision is therefore completely inelastic. You have probably learned that "momentum is conserved" in an inelastic collision. But how does this fact help you to solve collision problems"
Part A What physical quantities are conserved in this collision? ANSWER: the magnitude of the momentum only the net momentum (considered as a vector) only the momentum of each object considered individually
Correct
Part B Two cars of equal mass collide inelastically and stick together after the collision. Before the collision, their speeds are \texttip{v_{\rm 1}}{v_1} and \texttip{v_{\rm 2}}{v_2}. What is the speed of the twocar system after the collision?
Hint 1. How to approach the problem Think about how you would calculate the final speed of the two cars with the information provided and using the idea of conservation of momentum. Better yet, try the calculation out. What do you get? ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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v_1+v_2 v_1v_2 v_2v_1 \sqrt{v_1v_2} \large{\frac{v_1+v_2}{2}} \sqrt{{v_1}^2+{v_2}^2} The answer depends on the directions in which the cars were moving before the collision.
Correct
Part C Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are \texttip{p_{\rm 1}}{p_1} and \texttip{p_{\rm 2}}{p_2}. After the collision, what is the magnitude of their combined momentum?
Hint 1. How to approach the problem Think about how you would calculate the final momentum of the two cars using the information provided and the idea of conservation of momentum. Better yet, try the calculation out. What do you get? Keep in mind that momentum is a vector, but you are asked about the magnitude of the momentum, which is a scalar. ANSWER: p_1+p_2 p_1p_2 p_2p_1 \sqrt{p_1p_2} \large{\frac{p_1+p_2}{2}} \sqrt{{p_1}^2+{p_2}^2} The answer depends on the directions in which the cars were moving before the collision.
Correct
Part D Two cars collide inelastically and stick together after the collision. Before the collision, their momenta are \texttip{\vec{p}_{\rm 1}}{p_1_vec} and \texttip{\vec{p}_{\rm 2}}{p_2_vec}. After the collision, their combined momentum is \texttip{\vec{p}}{p_vec}. Of what can one be certain?
Hint 1. Momentum is a vector https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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Momentum is a vector quantity, and conservation of momentum holds for twodimensional and three dimensional collisions as well as for onedimensional collisions. ANSWER: \vec{p}=\vec{p_1}+\vec{p_2} \vec{p}=\vec{p_1}\vec{p_2} \vec{p}=\vec{p_2}\vec{p_1}
Correct You can decompose the vector equation that states the conservation of momentum into individual equations for each of the orthogonal components of the vectors.
Part E Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are \texttip{p_{\rm 1}}{p_1} and \texttip{p_{\rm 2}}{p_2}. After the collision, the magnitude of their combined momentum is \texttip{p}{p}. Of what can one be certain?
Hint 1. How to approach the problem mathematically Momentum is a vector quantity. It is impossible to make exact predictions about the direction of motion after a collision if nothing is known about the direction of motion before the collision. However, one can put some bounds on the values of the final momentum. Start with the expression for \texttip{\vec{p}}{p_vec} from Part D: \vec{p} = \vec{p_1} + \vec{p_2}. Therefore, |p| = \left|\vec{p_1} + \vec{p_2}\right| = \sqrt{|p_1|^2 + |p_2|^2 + 2 \vec{p_1} \cdot \vec{p_2}} = \sqrt{|p_1|^2 + |p_2|^2 + 2|p_1||p_2|\cos{\theta}}, where \texttip{\theta }{theta} is the angle between \texttip{\vec{p}_{\rm 1}}{p_1_vec} and \texttip{\vec{p}_{\rm 2}}{p_2_vec}. (To derive the above, you would have to break each vector into components.) So the value of |p| is controlled by \texttip{\theta }{theta}.
Hint 2. How to approach the problem empirically Consider the directions for the initial momenta that will give the largest and smallest final momentum. ANSWER: p_1+p_2\ge p\ge \sqrt{p_1 p_2} \large{p_1+p_2\ge p\ge \frac{p_1+p_2}{2}} p_1+p_2\ge p\ge |p_1p_2| p_1+p_2\ge p\ge \sqrt{{p_1}^2+{p_2}^2}
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Correct When the two cars collide, the magnitude of the final momentum will always be at most p_1+p_2 (a value attained if the cars were moving in the same direction before the collision) and at least |p_1p_2| (a value attained if the cars were moving in opposite directions before the collision).
Colliding Cars In this problem we will consider the collision of two cars initially moving at right angles. We assume that after the collision the cars stick together and travel off as a single unit. The collision is therefore completely inelastic. Two cars of masses \texttip{m_{\rm 1}}{m_1} and \texttip{m_{\rm 2}}{m_2} collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of \texttip{v_{\rm 1}}{v_1}, and car 2 was traveling northward at a speed of \texttip{v_{\rm 2}}{v_2}. After the collision, the two cars stick together and travel off in the direction shown.
Part A First, find the magnitude of \texttip{\vec{v}}{v_vec}, that is, the speed \texttip{v}{v} of the twocar unit after the collision. Express \texttip{v}{v} in of \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, and the cars' initial speeds \texttip{v_{\rm 1}}{v_1} and \texttip{v_{\rm 2}}{v_2}.
Hint 1. Conservation of momentum Recall that conservation of linear momentum may be expressed as a vector equation, \vec{p}_{\rm initial}=\vec{p}_{\rm final}. Each vector component of linear momentum is conserved separately.
Hint 2. x and y components of momentum The momentum of the twocar system immediately after the collision may be written as \vec{p}=p_x\hat{i}+p_y\hat{j}, where the x and y directions are the eastward and northward directions, respectively. Find \texttip{p_{\mit x}}{p_x} and \texttip{p_{\mit y}}{p_y} https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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Express the two components, separated by a comma, in of \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, \texttip{v_{\rm 1}}{v_1} and \texttip{v_{\rm 2}}{v_2}. ANSWER: \texttip{p_{\mit x}}{p_x}, \texttip{p_{\mit y}}{p_y} = m_{1} v_{1},m_{2} v_{2}
Hint 3. A vector and its components Recall that the square of the magnitude of a vector is given by the Pythagorean formula: p^2=p_x^2+p_y^2.
Hint 4. Velocity and momentum Find \texttip{v}{v}, the magnitude of the final velocity. Express \texttip{v}{v} in of the magnitude of the final momentum \texttip{p}{p} and the masses \texttip{m_{\rm 1}}{m_1} and \texttip{m_{\rm 2}}{m_2}. ANSWER: \texttip{v}{v} = \large{\frac{p}{m_{1} + m_{2}}}
ANSWER: \texttip{v}{v} = \large{{\frac{1}{m_{1}+m_{2}}} \sqrt{\left(m_{1}{^2}v_{1}{^2}\right)+\left(m_{2}{^2}v_{2}{^2}\right)}}
Correct
Part B Find the tangent of the angle \texttip{\theta }{theta}. Express your answer in of the momenta of the two cars, \texttip{p_{\rm 1}}{p_1} and \texttip{p_{\rm 2}}{p_2}. ANSWER: \tan(\theta) = \large{\frac{p_{2}}{p_{1}}}
Correct
Part C Suppose that after the collision, \tan \theta=1; in other words, \texttip{\theta }{theta} is 45^\circ. This means that before the collision: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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ANSWER: The magnitudes of the momenta of the cars were equal. The masses of the cars were equal. The velocities of the cars were equal.
Correct
Exercise 8.42 A bullet of mass 4.00 {\rm g} is fired horizontally into a wooden block of mass 1.13 {\rm kg} resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.150. The bullet remains embedded in the block, which is observed to slide a distance 0.210 {\rm m} along the surface before stopping.
Part A What was the initial speed of the bullet? ANSWER: v = 223 {\rm m/s}
Correct
An Exciting Encounter An atom of mass \texttip{M}{M} is initially at rest, in its ground state. A moving (nonrelativistic) electron of mass \texttip{m_{\rm e}}{m_e} collides with the atom. The atom+electron system can exist in an excited state in which the electron is absorbed into the atom. The excited state has an extra, "internal," energy \texttip{E}{E} relative to the atom's ground state.
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Part A Find the kinetic energy \texttip{K_{\rm e}}{K_e} that the electron must have in order to excite the atom. Express your answer in of \texttip{E}{E}, \texttip{m_{\rm e}}{m_e}, and \texttip{M}{M}.
Hint 1. How to approach the problem Depending on the initial kinetic energy of the electron, there are two possible outcomes to the collision between the electron and the atom. If the electron's kinetic energy is too low or too high, the collision will be elastic. If the electron's kinetic energy is just right, the collision will be perfectly inelastic: The electron will be absorbed into the atom, and the two will stick together. In both cases, total momentum must be conserved. In the case of an elastic collision, total kinetic energy is conserved. However, in the case of a perfectly inelastic collision, the final kinetic energy of the system is less than the initial kinetic energy. The loss of kinetic energy goes into exciting the atom, so total energy is conserved. From the perspective of Newtonian mechanics, you can think of the excited atom as having an internal potential energy. For example, it is as if the electron has compressed a spring in the atom and is held there by a catch mechanism. Thus, you can think of the system as having acquired a final potential energy \texttip{E}{E}. The usual expression for energy conservation then becomes K_{\rm final} + E = K_{\rm initial}.
Hint 2. Find the final kinetic energy in of the initial kinetic energy of the electron Find the kinetic energy \texttip{K_{\rm final}}{K_final} that the atom+electron system would have after a perfectly inelastic collision. Express your answer in of the initial kinetic energy of the electron \texttip{K_{\rm initial}} {K_initial}, \texttip{m_{\rm e}}{m_e}, and \texttip{M}{M}. The initial velocity of the electron, \texttip{v_{\rm 0}}{v_0}, should not appear in your answer.
Hint 1. Find the final velocity of the atom Suppose the initial velocity of the electron is \texttip{v_{\rm 0}}{v_0}. Use conservation of momentum to find the velocity \texttip{v_{\rm final}}{v_final} of the atom after it has absorbed the electron. Express your answer in of \texttip{v_{\rm 0}}{v_0}, \texttip{m_{\rm e}}{m_e}, and \texttip{M}{M}. ANSWER: \texttip{v_{\rm final}}{v_final} = \large{\frac{m_{e} v_{0}}{M+m_{e}}}
Hint 2. Find the final kinetic energy in of initial velocity In of \texttip{v_{\rm 0}}{v_0}, what is \texttip{K_{\rm final}}{K_final}, the kinetic energy of the atom after it has absorbed the electron? Express \texttip{K_{\rm final}}{K_final} in of \texttip{m_{\rm e}}{m_e}, \texttip{M}{M}, and \texttip{v_{\rm 0}}{v_0} (but without \texttip{K_{\rm initial}}{K_initial}). ANSWER: \texttip{K_{\rm final}}{K_final} = \large{\frac{1}{2} {v_{0}}^{2} \frac{{m_{e}}^{2}}{M+m_{e}}}
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ANSWER: \texttip{K_{\rm final}}{K_final} = \large{\frac{m_{e}}{M+m_{e}} K_{\rm{initial}}}
ANSWER: \texttip{K_{\rm e}}{K_e} = \large{{\frac{M+m_{e}}{M}}E}
Correct
Part B We can use the result from Part A to study a process of interest in atomic physics: a collision of two atoms that causes one of the atoms to ionize (lose an electron). In this case, \texttip{E}{E} is the energy needed to ionize one of the atoms, called the ionization energy. The most efficient way to ionize an atom in a collision with another atom is for the collision to be completely inelastic (atoms stick together after the collision). If the collision were perfectly elastic, then translational kinetic energy would be conserved, and there would be no energy left over for exciting the atom. If the collision were partially elastic, then some of the initial kinetic energy would be converted into internal energy, but not as much as in a perfectly inelastic collision. In practice, interatomic collisions are never perfectly inelastic, but analyzing this case can give a lower bound on the amount of kinetic energy needed for ionization. Is it possible to ionize an atom of {}^{133}\rm Cs, initially at rest, by a collision with an atom of {}^{16}\rm O that has kinetic energy \texttip{K_{\rm initial}}{K_initial} of 4.0 electron volts? The ionization energy of the cesium atom is 3.9 electron volts. It doesn't matter what energy units you choose, as long as you are consistent. For this question, it is most convenient to use electron volts (\rm {eV}) throughout. Note that 1\ \rm {eV} = 1.60 \times 10^{19} \rm {J}, which you maybe more familiar with. You can take the mass of the oxygen atom to be 16 atomic mass units and that of the cesium atom to be 133 atomic mass units. It doesn't matter what mass units you choose, as long as you are consistent. For this question, it is most convenient to use atomic mass units, since these are the numbers you are provided with.
Hint 1. How to approach the problem The cesium atom can be ionized only if a sufficient amount of kinetic energy is available. That is, the initial kinetic energy of the oxygen minus the final kinetic energy of the oxygen and cesium must be greater than the ionization energy of the cesium atom. The final kinetic energy of the oxygen and cesium will be minimal in a perfectly inelastic collision. Think about how this situation corresponds to the one you analyzed in Part A.
Hint 2. What is the correspondence with Part A? The oxygen atom in this collision corresponds to which of the two particles in the collision in Part A? ANSWER:
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the electron the atom
Hint 3. The masses of cesium and oxygen For this part of the problem, all you need to know is that the cesium atom is less than ten times as massive as the oxygen atom. ANSWER: yes no
Correct
Part C What is the least possible initial kinetic energy \texttip{K_{\rm min}}{K_min} the oxygen atom could have and still excite the cesium atom? Express your answer in electron volts, to one decimal place.
Hint 1. How to approach the problem Use the equation you found in Part A. Here, you can take the mass of the (initially moving) oxygen atom to be 16 atomic mass units and that of the (initially stationary) cesium atom to be 133 atomic mass units. It doesn't matter what mass units you choose, as long as you are consistent. For this question, it is most convenient to use atomic mass units, since these are the numbers you are provided with. ANSWER: \texttip{K_{\rm min}}{K_min} = 4.4 \rm eV
Correct
Trading Momenta in a Collision Two particles move perpendicular to each other until they collide. Particle 1 has mass \texttip{m}{m} and momentum of magnitude 2p, and particle 2 has mass 2m and momentum of magnitude \texttip{p}{p}. Note: Magnitudes are not drawn to scale in any of the figures.
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Part A Suppose that after the collision, the particles "trade" their momenta, as shown in the figure. That is, particle 1 now has magnitude of momentum \texttip{p}{p}, and particle 2 has magnitude of momentum 2p; furthermore, each particle is now moving in the direction in which the other had been moving. How much kinetic energy, \texttip{K_{\rm lost}}{K_lost}, is lost in the collision? Express your answer in of \texttip{m}{m} and \texttip{p}{p}.
Hint 1. How to approach the problem To find the kinetic energy lost in the collision, compute the initial kinetic energy \texttip{K_{\rm initial}} {K_initial} (before the collision) and the final kinetic energy \texttip{K_{\rm final}}{K_final} (after the collision). Then take the difference: K_{\rm lost} = K_{\rm initial} K_{\rm final}.
Hint 2. Find the relationship between energy and momentum What is the kinetic energy \texttip{K}{K} of a particle with mass \texttip{m}{m} and magnitude of momentum \texttip{p}{p}? ANSWER: \texttip{K}{K} = \large{\frac{p^{2}}{2 m}}
Hint 3. Find the initial kinetic energy What is the initial kinetic energy \texttip{K_{\rm initial}}{K_initial} of the twoparticle system? Express your answer in of \texttip{p}{p} and \texttip{m}{m}.
Hint 1. Find the kinetic energy of particle 1 Find the kinetic energy \texttip{K_{\rm 1.initial}}{K_1.initial} of particle 1 before the collision. Express your answer in of \texttip{p}{p} and \texttip{m}{m}. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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ANSWER: \texttip{K_{\rm 1.initial}}{K_1.initial} = \large{\frac{\left(2 p\right)^{2}}{2 m}}
Hint 2. Find the kinetic energy of particle 2 Find the kinetic energy \texttip{K_{\rm 2.initial}}{K_2.initial} of particle 2 before the collision. Express your answer in of \texttip{p}{p} and \texttip{m}{m}. ANSWER: \texttip{K_{\rm 2.initial}}{K_2.initial} = \large{\frac{p^{2}}{2 \left(2m\right)}}
ANSWER: \texttip{K_{\rm initial}}{K_initial} = \large{\frac{9 p^{2}}{4 m}}
Hint 4. Find the final kinetic energy What is the final kinetic energy \texttip{K_{\rm final}}{K_final} of the twoparticle system? Express your answer in of \texttip{p}{p} and \texttip{m}{m}. ANSWER: \texttip{K_{\rm final}}{K_final} = \large{\frac{3 p^{2}}{2 m}}
ANSWER: \texttip{K_{\rm lost}}{K_lost} = \large{\frac{3p^{2}}{4m}}
Correct
Part B Consider an alternative situation: This time the particles collide completely inelastically. How much kinetic energy \texttip{K_{\rm lost}}{K_lost} is lost in this case? Express your answer in of \texttip{m}{m} and \texttip{p}{p}.
Hint 1. How to approach the problem As in Part A, to find the kinetic energy lost in the collision, compute the initial kinetic energy \texttip{K_{\rm initial}}{K_initial} (before the collision) and the final kinetic energy \texttip{K_{\rm final}}{K_final} (after the collision). Then take the difference: K_{\rm lost} = K_{\rm initial} K_{\rm final}. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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Hint 2. Definition of completely inelastic A completely inelastic collision is one in which the particles stick together after the collision.
Hint 3. Initial kinetic energy The initial kinetic energy in this case is the same as it was in the partially elastic collision.
Hint 4. Find the final kinetic energy What is the combined kinetic energy \texttip{K_{\rm final}}{K_final} of the two particles after the perfectly inelestic collision? If you use the formula \large{K=\frac{p^2}{2m}}, that \texttip{p}{p} is the magnitude of the momentum vector. Express your answer in of \texttip{m}{m} and \texttip{p}{p}.
Hint 1. Find the magnitude of the final momentum What is the magnitude of \texttip{\vec{p}_{\rm final}}{p_final_vec}, the total momentum of the two particles after the collision? Express \texttip{p_{\rm final}}{p_final} in of \texttip{p}{p}.
Hint 1. Find the final momentum vector What is the total momentum \texttip{\vec{p}_{\rm final}}{p_final_vec} of the two particles after the collision? Take the positive x direction to be to the right in the figures and the positive y direction to be upward. Express your answer as a vector in of \texttip{p}{p} and the unit vectors \texttip{\hat{i}}{i_unit} and \texttip{\hat{j}}{j_unit}. ANSWER: \texttip{\vec{p}_{\rm final}}{p_final_vec} = 2 p \hat{i}+p \hat{j}
ANSWER: \texttip{p_{\rm final}}{p_final} = \sqrt{5} p
ANSWER: \texttip{K_{\rm final}}{K_final} = \large{\frac{5 p^{2}}{6 m}}
ANSWER: \texttip{K_{\rm lost}}{K_lost} = \large{\frac{17p^{2}}{12m}}
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Correct
Problem 8.101 A neutron at rest decays (breaks up) to a proton and an electron. Energy is released in the decay and appears as kinetic energy of the proton and electron. The mass of a proton is 1836 times the mass of an electron.
Part A What fraction of the total energy released goes into the kinetic energy of the proton? Express your answer as a percentage. ANSWER: \large{\frac{K_{\rm p}}{K_{\rm tot}}\,} = 5.44×10−2 \%
Correct
Problem 8.73 Spheres A (mass 0.020 {\rm kg}), B (mass 0.030 {\rm kg}), and C (mass 0.050 {\rm kg}), are each approaching the origin as they slide on a frictionless air table . The initial velocities of A and B are 1.50 {\rm m/s} and 0.50 .{\rm m/s} All three spheres arrive at the origin at the same time and stick together.
Part A What must the xcomponent of the initial velocity of C be if all three objects are to end up moving at 0.50 {\rm m/s} in the +xdirection after the collision? ANSWER:
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v_{Cx} = 1.75 {\rm m/s}
Correct
Part B What must the ycomponent of the initial velocity of C be if all three objects are to end up moving at 0.50 {\rm m/s} in the +xdirection after the collision? ANSWER: v_{Cy} = 0.260 {\rm m/s}
Correct
Part C If C has the velocity found in parts (A) and (B), what is the change in the kinetic energy of the system of three spheres as a result of the collision? Express your answer using two significant figures. ANSWER: \Delta K = −9.2×10−2 {\rm J}
Correct
Problem 8.109 A fireworks rocket is fired vertically upward. At its maximum height of 80.0 {\rm m} , it explodes and breaks into two pieces, one with mass \texttip{m_{\rm A}}{m_A} = 1.45 {\rm kg} and the other with mass \texttip{m_{\rm B}}{m_B} = 0.300 {\rm kg} . In the explosion, 900 {\rm J} of chemical energy is converted to kinetic energy of the two fragments.
Part A What is the speed of each fragment just after the explosion? Enter your answers numerically separated by a comma. ANSWER: v_{\rm A}, v_{\rm B} = 14.6,70.5 {\rm m/s}
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Correct
Part B It is observed that the two fragments hit the ground at the same time. What is the distance between the points on the ground where they land? Assume that the ground is level and air resistance can be ignored. ANSWER: \left|x_{\rm A}\;\;x_{\rm B}\right| = 344 {\rm m}
Correct
Exercise 8.49 Canadian nuclear reactors use heavy water s in which elastic collisions occur between the neutrons and deuterons of mass 2.0 u.
Part A What is the speed of a neutron, expressed as a fraction of its original speed, after a headon, elastic collision with a deuteron which is initially at rest? Express your answer using three decimal places. ANSWER: 0.333
Correct
Part B What is its kinetic energy, expressed as a fraction of its original kinetic energy? Express your answer using three decimal places. ANSWER: 0.111
Correct
Part C https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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How many such successive collisions will reduce the speed of a neutron to 1/2190 of its original value? Express your answer as a whole number. ANSWER: 7 collisions
Correct
Problem 8.76 At a classic auto show, a 810 {\rm kg} 1955 Nash Metropolitan motors by at 8.4 {\rm m/s} , followed by a 1630 {\rm kg} 1957 Packard Clipper purring past at 4.4 {\rm m/s} .
Part A What is the ratio of the kinetic energy of the Nash to that of the Packard? ANSWER: \large{\frac{K_{\rm N}}{K_{\rm P}}} = 1.81
Correct
Part B What is the ratio of the magnitude of momentum of the Nash to that of the Packard? ANSWER: \large{\frac{p_{\rm N}}{p_{\rm P}}} = 0.949
Correct
Part C Let F_{\rm{N}} be the net force required to stop the Nash in time \texttip{t}{t}, and let F_{\rm{P}} be the net force required to stop the Packard in the same time. What is the ratio F_{\rm{N}}/F_{\rm{P}} of these two forces? ANSWER: \large{\frac{F_{\rm{N}}}{F_{\rm{P}}}} = 0.949
Correct https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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Part D Now let F_{\rm{N}} be the net force required to stop the Nash in a distance d, and let F_{\rm{P}} be the net force required to stop the Packard in the same distance. What is the ratio F_{\rm{N}}/F_{\rm{P}} {\kern 1pt}? ANSWER: \large{\frac{F_{\rm{N}}}{F_{\rm{P}}} {\kern 1pt}} = 1.81
Correct
Problem 8.112 In Section 8.6, we considered a rocket fired in outer space where there is no air resistance and where gravity is negligible. Suppose instead that the rocket is accelerating vertically upward from rest on the earth's surface. Continue to ignore air resistance and consider only that part of the motion where the altitude of the rocket is small so that g may be assumed to be constant.
Part A How is equation \large{m\frac{{dv}}{{dt}} = v_{ex} \frac{{dm}}{{dt}}} modified by the presence of the gravity force? ANSWER: Essay answers are limited to about 500 words (3800 characters maximum, including spaces). 3785 Character(s) remaining (none provided)
Part B Derive an expression for the acceleration a of the rocket, analogous to equation \large{a = \frac{{dv}}{{dt}} = \frac{{v_{ex} }}{m}\frac{{dm}}{{dt}}}. ANSWER: Essay answers are limited to about 500 words (3800 characters maximum, including spaces). 3785 Character(s) remaining (none provided)
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Part C What is the acceleration of the rocket if it is near the earth's surface? In the first second of firing, the rocket ejects \large{\frac{1}{120}} of its mass with a relative speed of 2300 {\rm {\rm m/s}} . You can ignore air resistance. ANSWER: a_{\rm earth} = 10 {\rm m/s^2}
Incorrect; Try Again; 5 attempts remaining
Part D Now suppose that \large{\frac{3}{4}} of the initial mass m_0 of the rocket is fuel, so the final mass m=m_0/4 is and that the fuel is completely consumed at a constant rate in a total time 90 {\rm {\rm s}} . Find the speed of the rocket after 90 {\rm {\rm s}} if the rocket is fired from the earth's surface rather than in outer space. You can ignore air resistance. ANSWER: v_{\rm earth} =
{\rm m/s}
Part E How does your answer in previous part compare with the rocket speed calculated in outer space at the same conditions? ANSWER: \large{\frac{v_{\rm space}}{v_{\rm earth}}} =
Score Summary: Your score on this assignment is 83.0%. You received 34.03 out of a possible total of 41 points.
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Ch 08 HW Due: 12:59pm on Saturday, April 2, 2016 You will receive no credit for items you complete after the assignment is due. Grading Policy
Center of Mass and External Forces Learning Goal: Understand that, for many purposes, a system can be treated as a pointlike particle with its mass concentrated at the center of mass. A complex system of objects, both pointlike and extended ones, can often be treated as a point particle, located at the system's center of mass. Such an approach can greatly simplify problem solving. Before you use the center of mass approach, you should first understand the following : System: Any collection of objects that are of interest to you in a particular situation. In many problems, you have a certain freedom in choosing your system. Making a wise choice for the system is often the first step in solving the problem efficiently. Center of mass: The point that represents the "average" position of the entire mass of a system. The ⃗ postion of the center of mass r cm can be expressed in of the position vectors r i⃗ of the particles as ⃗ r cm =
∑ m i r i⃗ ∑ mi
.
The x coordinate of the center of mass xcm can be expressed in of the x coordinates (r x ) i of the particles as x cm =
∑ m i (rx ) i ∑ mi
.
Similarly, the y coordinate of the center of mass can be expressed. Internal force: Any force that results from an interaction between the objects inside your system. As we will show, the internal forces do not affect the motion of the system's center of mass. External force: Any force acting on an object inside your system that results from an interaction with an object outside your system. Consider a system of two blocks that have masses m1 and m2 . Assume that the blocks are pointlike particles and are located along the x axis at the coordinates x1 and x2 as shown . In this problem, the blocks can only move along the x axis.
Typesetting math: 45% https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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Part A Find the x coordinate xcm of the center of mass of the system. Express your answer in of m1 , m2 , x1 , and x2 . ANSWER:
xcm
=
m 1 x 1 +m 2 x 2 m 1 +m 2
Correct
Part B If m2
≫ m1
, then the center of mass is located:
ANSWER: to the left of m1 at a distance much greater than x2 to the left of m1 at a distance much less than x2
− x1
− x1
to the right of m1 at a distance much less than x2
− x1
to the right of m2 at a distance much greater than x2 to the right of m2 at a distance much less than x2 to the left of m2 at a distance much less than x2
− x1
− x1
− x1
Correct
Part C If m2
= m1
, then the center of mass is located:
ANSWER: at m1 at m2 halfway between m1 and m2 the answer depends on x1 and x2
Correct
Part D Recall that the blocks can only move along the x axis. The x components of their velocities at a certain moment are and . Find the component of the velocity of the center of mass ( https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
)
at that moment. Keep in
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are v 1x and v 2x . Find the x component of the velocity of the center of mass (v cm ) x at that moment. Keep in mind that, in general: v x = dx/dt. Express your answer in of m1 , m2 , v 1x , and v 2x . ANSWER: (v cm )
x
=
m 1 v 1x +m 2 v 2x m 1 +m 2
Correct Because v 1x and v 2x are the x components of the velocities of m1 and m2 their values can be positive or negative or equal to zero.
Part E →
→
Suppose that v 1x and v 2x have equal magnitudes. Also, v 1 is directed to the right and v 2 is directed to the left. The velocity of the center of mass is then: ANSWER: directed to the left directed to the right zero m1
the answer depends on the ratio m
2
Correct
Part F Assume that the x components of the blocks' momenta at a certain moment are p1x and p2x . Find the x component of the velocity of the center of mass (v cm ) x at that moment. Express your answer in of m1 , m2 , p1x , and p2x . ANSWER:
(v cm )
x
=
p 1x +p 2x m 1 +m 2
Correct
Part G ⃗ Suppose that v cm
= 0
. Which of the following must be true?
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ANSWER: |p
1x
| = |p
2x
|
|v 1x | = |v 2x | m1 = m2
none of the above
Correct
Part H Assume that the blocks are accelerating, and the x components of their accelerations at a certain moment are a 1x and a 2x . Find the x component of the acceleration of the center of mass (a cm ) x at that moment. Keep in mind that, in general, a x = dv x /dt . Express your answer in of m1 , m2 , a 1x , and a 2x . ANSWER: (a cm )
x
=
m 1 a1x +m 2 a2x m 1 +m 2
Correct Because a 1x and a 2x are the x components of the velocities of m1 and m2 their values can be positive or negative or equal to zero.
We will now consider the effect of external and internal forces on the acceleration of the center of mass.
Part I Consider the same system of two blocks. An external force F ⃗ is now acting on block m1 . No forces are applied to block m2 as shown . Find the x component of the acceleration of the center of mass (a cm ) x of the system. Express your answer in of the x component Fx of the force, m1 ,and m2 .
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Hint 1. Using Newton's laws Find the acceleration of each block from Newton's second law and then apply the formula for (a cm ) x found earlier. ANSWER:
(a cm )
x
=
Fx m 1 +m 2
Correct
Part J →
Consider the same system of two blocks. Now, there are two forces involved. An external force F1 is acting on →
block m1 and another external force F2 is acting on block m2 . Find the x component of the acceleration of the center of mass (a cm ) x of the system. Express your answer in of the x components F 1x and F 2x of the forces, m 1 and m 2 .
ANSWER: (a cm )
x
=
F1x +F2x m 1 +m 2
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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Correct Note that, in both cases, the acceleration of the center of mass can be found as (acm )
x
=
(Fnet )
x
Mtotal
where Fnet is the net external force applied to the system, and M total is the total mass of the system. Even though each force is only applied to one object, it affects the acceleration of the center of mass of the entire system. This result is especially useful since it can be applied to a general case, involving any number of objects moving in all directions and being acted upon by any number of external forces.
Part K Consider the previous situation. Under what condition would the acceleration of the center of mass be zero? Keep in mind that F1x and F2x represent the components, of the corresponding forces. ANSWER: F 1x = −F 2x F 1x = F 2x m1 = m2 m1 ≪ m2
Correct
Part L ⃗ Consider the same system of two blocks. Now, there are two internal forces involved. An internal force F 12 is ⃗ applied to block m1 by block m2 and another internal force F 21 is applied to block m2 by block m1 . Find the x component of the acceleration of the center of mass (a cm ) of the system. x
Express your answer in of the x components F 12x and F 21x of the forces, m 1 and m 2 .
ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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\texttip{(a_{\rm cm})_x}{a_cm_x} = \large{\frac{F_{\rm{12x}}+F_{\rm{21x}}}{m_{1}+m_{2}}}
Correct Newton's 3rd law tells you that |F_{12x}|=|F_{21x}|. From your answers above, you can conclude that (a_{\rm cm})_x=0.The internal forces do not change the velocity of the center of mass of the system. In the absence of any external forces, (a_{\rm cm})_x=0 and \texttip{(v_{\rm cm})_x}{v_cm_x} is constant. You just demonstrated this to be the case for the twobody situation moving along the x axis; however, it is true in more general cases as well.
A Bullet Is Fired into a Wooden Block A bullet of mass \texttip{m_{\rm b}}{m_b} is fired horizontally with speed \texttip{v_{\rm i}}{v_i} at a wooden block of mass \texttip{m_{\rm w}}{m_w} resting on a frictionless table. The bullet hits the block and becomes completely embedded within it. After the bullet has come to rest within the block, the block, with the bullet in it, is traveling at speed \texttip{v_{\rm f \hspace{1 pt}}}{v_f}.
Part A Which of the following best describes this collision?
Hint 1. Types of collisions An inelastic collision is a collision in which kinetic energy is not conserved. In a partially inelastic collision, kinetic energy is lost, but the objects colliding do not stick together. From this information, you can infer what completely inelastic and elastic collisions are. ANSWER:
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perfectly elastic partially inelastic perfectly inelastic
Correct
Part B Which of the following quantities, if any, are conserved during this collision?
Hint 1. When is kinetic energy conserved? Kinetic energy is conserved only in perfectly elastic collisions. ANSWER: kinetic energy only momentum only kinetic energy and momentum neither momentum nor kinetic energy
Correct
Part C What is the speed of the block/bullet system after the collision? Express your answer in of \texttip{v_{\rm i}}{v_i}, \texttip{m_{\rm w}}{m_w}, and \texttip{m_{\rm b}} {m_b}.
Hint 1. Find the momentum after the collision What is the total momentum \texttip{p_{\rm total}}{p_total} of the block/bullet system after the collision? Express your answer in of \texttip{v_{\rm f \hspace{1 pt}}}{v_f} and other given quantities. ANSWER: \texttip{p_{\rm total}}{p_total} = \left(m_{w}+m_{b}\right) v_{f}
Hint 2. Use conservation of momentum The momentum of the block/bullet system is conserved. Therefore, the momentum before the collision is the same as the momentum after the collision. Find a second expression for \texttip{p_{\rm total}}{p_total}, https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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this time expressed as the total momentum of the system before the collision. Express your answer in of \texttip{v_{\rm i}}{v_i} and other given quantities. ANSWER: \texttip{p_{\rm total}}{p_total} = m_{b} v_{i}
ANSWER: \texttip{v_{\rm f \hspace{1 pt}}}{v_f} = \large{\frac{m_{b} v_{i}}{m_{b}+m_{w}}}
Correct
A Ball Hits a Wall Elastically A ball of mass \texttip{m}{m} moving with velocity \texttip{\vec{v}_{\rm i}}{v_i_vec} strikes a vertical wall as shown in . The angle between the ball's initial velocity vector and the wall is \texttip{\theta _{\rm i}}{theta_i} as shown on the diagram, which depicts the situation as seen from above. The duration of the collision between the ball and the wall is \texttip{\Delta t}{Deltat}, and this collision is completely elastic. Friction is negligible, so the ball does not start spinning. In this idealized collision, the force exerted on the ball by the wall is parallel to the x axis.
Part A What is the final angle \texttip{\theta _{\rm f \hspace{1 pt}}}{theta_f} that the ball's velocity vector makes with the negative y axis? Express your answer in of quantities given in the problem introduction.
Hint 1. How to approach the problem Relate the vector components of the ball's initial and final velocities. This will allow you to determine \texttip{\theta _{\rm f \hspace{1 pt}}}{theta_f} in of \texttip{\theta _{\rm i}}{theta_i}.
Hint 2. Find the y component of the ball's final velocity https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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What is \texttip{v_{\rm fy}}{v_fy}, the \texttip{y}{y} component of the final velocity of the ball? Express your answer in of quantities given in the problem introduction and/or \texttip{v_{\rm ix}}{v_ix} and \texttip{v_{\rm iy}}{v_iy}, the \texttip{x}{x} and \texttip{y}{y} components of the ball's initial velocity.
Hint 1. How to approach this part There is no force on the ball in the y direction. From the impulsemomentum theorem, this means that the change in the y component of the ball's momentum must be zero. ANSWER: \texttip{v_{\rm fy}}{v_fy} = v_{i} {\cos}\left({\theta}_{i}\right)
Hint 3. Find the \texttip{x}{x} component of the ball's final velocity What is \texttip{v_{\rm fx}}{v_fx}, the \texttip{x}{x} component of the ball' final velocity? Express your answer in of quantities given in the problem introduction and/or \texttip{v_{\rm ix}}{v_ix} and \texttip{v_{\rm iy}}{v_iy}, the \texttip{x}{x} and \texttip{y}{y} components of the ball's initial velocity.
Hint 1. How to approach this problem Since energy is conserved in this collision, the final speed of the ball must be equal to its initial speed. ANSWER: \texttip{v_{\rm fx}}{v_fx} = v_{i} {\sin}\left({\theta}_{i}\right)
Hint 4. Putting it together Once you find the vector components of the final velocity in of the initial velocity, use the geometry of similar triangles to determine \texttip{\theta _{\rm f \hspace{1 pt}}}{theta_f} in of \texttip{\theta _{\rm i}}{theta_i}. ANSWER: \texttip{\theta _{\rm f \hspace{1 pt}}}{theta_f} = {\theta}_{i}
Correct
Part B What is the magnitude \texttip{F}{F} of the average force exerted on the ball by the wall? https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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Express your answer in of variables given in the problem introduction and/or \texttip{v_{\rm ix}} {v_ix}.
Hint 1. What physical principle to use Use the impulsemomentum theorem, \vec{J}=\vec{p}_{\rm f \hspace{1 pt}}\vec{p}_{\rm i}, along with the definition of impulse, \vec{J}=\sum\vec{F}\Delta t. In this case, only one force is acting, so \left|\vec{J}\right|=F\Delta t. Putting everything together, \large{F = \frac{\vec{p}_{\rm f \hspace{1 pt}} \vec{p}_{\rm i}}{\Delta t}}.
Hint 2. Change in momentum of the ball The fact that \theta_{\rm f \hspace{1 pt}}=\theta_{\rm i} implies that the \texttip{y}{y} component of the ball's momentum does not change during the collision. What is \texttip{\Delta p_{\mit x}}{Deltap_x}, the magnitude of the change in the ball's \texttip{x}{x} momentum? Express your answer in of quantities given in the problem introduction and/or \texttip{v_{\rm ix}}{v_ix}. ANSWER: \texttip{\Delta p_{\mit x}}{Deltap_x} = 2 m v_{i} {\sin}\left({\theta}_{i}\right)
ANSWER: \texttip{F}{F} = \large{\frac{2mv_{i} {\sin}\left({\theta}_{f}\right)}{{\Delta}t}}
Correct
Video Tutor: Water Rocket First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question at right. You can watch the video again at any point.
Part A https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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Suppose we repeat the experiment from the video, but this time we use a rocket three times as massive as the one in the video, and in place of water we use a fluid that is twice as massive (dense) as water. If the new fluid leaves the rocket at the same speed as the water in the video, what will be the ratio of the horizontal speed of our rocket to the horizontal speed of the rocket in the video after all the fluid has left the rocket? (Ignore air resistance.)
Hint 1. How to approach the problem First, can we use momentum conservation? Let's take the rocket and fluid as the system. In the vertical direction, momentum is not conserved because gravity exerts an external force on the system. But in the horizontal direction, momentum is conserved (since we ignore air resistance). Next, we draw beforeandafter diagrams, labeling mass and velocity in each case. (For convenience, we use “exhaust” to label the fluid both before and after it is expelled.)
Momentum conservation enables you to relate the final horizontal momenta of the rocket and the exhaust. The problem says that the fluid leaves the rocket “at the same speed” as the water in the video. What does that tell you about \vec v_{e.f}? \vec p_{e,f}? Now, write an equation for the rocket’s velocity in of known quantities, including m_e and m_r. How will doubling m_e and tripling m_r change the magnitude of \vec v_{r.f}? ANSWER: 1/2 1 2/3 1/3 3/2
Correct The rocket's speed is proportional to the ratio of the fluid's mass to the rocket's mass.
Exercise 8.12 A bat strikes a 0.145{\rm kg} baseball. Just before impact, the ball is traveling horizontally to the right at 60.0 {\rm m/s} , and it leaves the bat traveling to the left at an angle of 35 {\rm ^\circ} above horizontal with a speed of 55.0 {\rm m/s} . https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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The ball and bat are in for 1.85 {\rm ms} .
Part A Find the horizontal component of the average force on the ball. Take the xdirection to be positive to the right Express your answer using two significant figures. ANSWER: F_{\rm x} = 8200 {\rm N}
Answer Requested
Part B Find the vertical component of the average force on the ball. Express your answer using two significant figures. ANSWER: F_{\rm y} = 2500 {\rm N}
Correct
Exercise 8.17 The expanding gases that leave the muzzle of a rifle also contribute to the recoil. A .30 caliber bullet has mass 7.20×10−3 {\rm kg} and a speed of 601 {\rm m/s} relative to the muzzle when fired from a rifle that has mass 2.90 {\rm kg} . The loosely held rifle recoils at a speed of 2.05 {\rm m/s} relative to the earth.
Part A Find the momentum of the propellant gases in a coordinate system attached to the earth as they leave the muzzle of the rifle. ANSWER: p = 1.63 {\rm kg \cdot m/s}
Correct
A Girl on a Trampoline A girl of mass m_1 = 60.0 kilograms springs from a trampoline with an initial upward velocity of v_i = 8.00 meters per https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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second. At height h = 2.00 meters above the trampoline, the girl grabs a box of mass m_2 = 15.0 kilograms. For this problem, use g = 9.80 meters per second per second for the magnitude of the acceleration due to gravity.
Part A What is the speed \texttip{v_{\rm before}}{v_before} of the girl immediately before she grabs the box? Express your answer numerically in meters per second.
Hint 1. How to approach the problem Use conservation of energy. Find the initial kinetic energy \texttip{K_{\rm i}}{K_i} of the girl as she leaves the trampoline. Then find her gravitational potential energy \texttip{U_{\rm before}}{U_before} just before she grabs the box (define her initial potential energy to be zero). According to the principle of conservation of energy, K_i = U_{\rm before} + K_{\rm before}. Once you have \texttip{K_{\rm before}}{K_before}, use the definition of translational kinetic energy to find the girl's speed \texttip{v_{\rm before}}{v_before}.
Hint 2. Initial kinetic energy What is the girl's initial kinetic energy \texttip{K_{\rm i}}{K_i} as she leaves the trampoline? Express your answer numerically in joules. ANSWER: \texttip{K_{\rm i}}{K_i} = 1920 \rm{J}
Hint 3. Potential energy at height \texttip{h}{h} What is the girl's gravitational potential energy \texttip{U_{\rm before}}{U_before} immediately before she grabs the box? Express your answer numerically in joules. ANSWER: \texttip{U_{\rm before}}{U_before} = 1180 \rm{J}
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ANSWER: \texttip{v_{\rm before}}{v_before} = 4.98 \rm{m/s}
Correct
Part B What is the speed \texttip{v_{\rm after}}{v_after} of the girl immediately after she grabs the box? Express your answer numerically in meters per second.
Hint 1. How to approach the problem Think of the process of grabbing the box as a collision. Though the girl and the box don't collide as such, any interaction between two objects that takes place extremely fast can be thought of as a collision. To find the velocity at a later time, which of the following principles could you use? ANSWER: conservation of momentum alone conservation of energy alone both conservation of momentum and conservation of energy Newton's second law
Hint 2. Total initial momentum What is the total momentum before the collision? Answer in kilogram meters per second. ANSWER: \texttip{p_{\rm before}}{p_before} = 299 \rm kg \, m/s
ANSWER: \texttip{v_{\rm after}}{v_after} = 3.98 \rm m/s
Correct
Part C Is this "collision" elastic or inelastic? https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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Hint 1. Definition of an inelastic collision If two objects move together with the same velocity after a collision, the collision is said to be inelastic. ANSWER: elastic inelastic
Correct In inelastic collisions, some of the system's kinetic energy is lost. In this case the kinetic energy lost is converted to heat energy in the girl's muscles as she grabs the box, and sound energy.
Part D What is the maximum height \texttip{h_{\rm max}}{h_max} that the girl (with box) reaches? Measure \texttip{h_{\rm max}}{h_max} with respect to the top of the trampoline. Express your answer numerically in meters.
Hint 1. How to approach the problem Use conservation of energy. From Part B you know the velocity of the girl/box system just after the girl grabs the box. Therefore, you can compute the kinetic energy \texttip{K_{\rm after}}{K_after} of the girl/box system just after the collision. You can also compute the gravitational potential energy \texttip{U_{\rm after}} {U_after} of the girl/box system at this point. The sum of these two quantities must equal the gravitational potential energy of the girl/box system at the height \texttip{h_{\rm max}}{h_max} (where their velocity, and therefore kinetic energy, will be zero).
Hint 2. Finding \texttip{U_{\rm after}}{U_after} What is the girl/box system's gravitational potential energy \texttip{U_{\rm after}}{U_after} immediately after she grabs the box? Express your answer numerically in joules. ANSWER: \texttip{U_{\rm after}}{U_after} = 1470 \rm J
Hint 3. Finding \texttip{K_{\rm after}}{K_after} What is the girl/box system's kinetic energy \texttip{K_{\rm after}}{K_after} immediately after she grabs the box? Express your answer numerically in joules. ANSWER:
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\texttip{K_{\rm after}}{K_after} = 594 \rm J
ANSWER: \texttip{h_{\rm max}}{h_max} = 2.81 \rm m
Correct
Conservation of Momentum in Inelastic Collisions Learning Goal: To understand the vector nature of momentum in the case in which two objects collide and stick together. In this problem we will consider a collision of two moving objects such that after the collision, the objects stick together and travel off as a single unit. The collision is therefore completely inelastic. You have probably learned that "momentum is conserved" in an inelastic collision. But how does this fact help you to solve collision problems"
Part A What physical quantities are conserved in this collision? ANSWER: the magnitude of the momentum only the net momentum (considered as a vector) only the momentum of each object considered individually
Correct
Part B Two cars of equal mass collide inelastically and stick together after the collision. Before the collision, their speeds are \texttip{v_{\rm 1}}{v_1} and \texttip{v_{\rm 2}}{v_2}. What is the speed of the twocar system after the collision?
Hint 1. How to approach the problem Think about how you would calculate the final speed of the two cars with the information provided and using the idea of conservation of momentum. Better yet, try the calculation out. What do you get? ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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v_1+v_2 v_1v_2 v_2v_1 \sqrt{v_1v_2} \large{\frac{v_1+v_2}{2}} \sqrt{{v_1}^2+{v_2}^2} The answer depends on the directions in which the cars were moving before the collision.
Correct
Part C Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are \texttip{p_{\rm 1}}{p_1} and \texttip{p_{\rm 2}}{p_2}. After the collision, what is the magnitude of their combined momentum?
Hint 1. How to approach the problem Think about how you would calculate the final momentum of the two cars using the information provided and the idea of conservation of momentum. Better yet, try the calculation out. What do you get? Keep in mind that momentum is a vector, but you are asked about the magnitude of the momentum, which is a scalar. ANSWER: p_1+p_2 p_1p_2 p_2p_1 \sqrt{p_1p_2} \large{\frac{p_1+p_2}{2}} \sqrt{{p_1}^2+{p_2}^2} The answer depends on the directions in which the cars were moving before the collision.
Correct
Part D Two cars collide inelastically and stick together after the collision. Before the collision, their momenta are \texttip{\vec{p}_{\rm 1}}{p_1_vec} and \texttip{\vec{p}_{\rm 2}}{p_2_vec}. After the collision, their combined momentum is \texttip{\vec{p}}{p_vec}. Of what can one be certain?
Hint 1. Momentum is a vector https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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Momentum is a vector quantity, and conservation of momentum holds for twodimensional and three dimensional collisions as well as for onedimensional collisions. ANSWER: \vec{p}=\vec{p_1}+\vec{p_2} \vec{p}=\vec{p_1}\vec{p_2} \vec{p}=\vec{p_2}\vec{p_1}
Correct You can decompose the vector equation that states the conservation of momentum into individual equations for each of the orthogonal components of the vectors.
Part E Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are \texttip{p_{\rm 1}}{p_1} and \texttip{p_{\rm 2}}{p_2}. After the collision, the magnitude of their combined momentum is \texttip{p}{p}. Of what can one be certain?
Hint 1. How to approach the problem mathematically Momentum is a vector quantity. It is impossible to make exact predictions about the direction of motion after a collision if nothing is known about the direction of motion before the collision. However, one can put some bounds on the values of the final momentum. Start with the expression for \texttip{\vec{p}}{p_vec} from Part D: \vec{p} = \vec{p_1} + \vec{p_2}. Therefore, |p| = \left|\vec{p_1} + \vec{p_2}\right| = \sqrt{|p_1|^2 + |p_2|^2 + 2 \vec{p_1} \cdot \vec{p_2}} = \sqrt{|p_1|^2 + |p_2|^2 + 2|p_1||p_2|\cos{\theta}}, where \texttip{\theta }{theta} is the angle between \texttip{\vec{p}_{\rm 1}}{p_1_vec} and \texttip{\vec{p}_{\rm 2}}{p_2_vec}. (To derive the above, you would have to break each vector into components.) So the value of |p| is controlled by \texttip{\theta }{theta}.
Hint 2. How to approach the problem empirically Consider the directions for the initial momenta that will give the largest and smallest final momentum. ANSWER: p_1+p_2\ge p\ge \sqrt{p_1 p_2} \large{p_1+p_2\ge p\ge \frac{p_1+p_2}{2}} p_1+p_2\ge p\ge |p_1p_2| p_1+p_2\ge p\ge \sqrt{{p_1}^2+{p_2}^2}
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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Correct When the two cars collide, the magnitude of the final momentum will always be at most p_1+p_2 (a value attained if the cars were moving in the same direction before the collision) and at least |p_1p_2| (a value attained if the cars were moving in opposite directions before the collision).
Colliding Cars In this problem we will consider the collision of two cars initially moving at right angles. We assume that after the collision the cars stick together and travel off as a single unit. The collision is therefore completely inelastic. Two cars of masses \texttip{m_{\rm 1}}{m_1} and \texttip{m_{\rm 2}}{m_2} collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of \texttip{v_{\rm 1}}{v_1}, and car 2 was traveling northward at a speed of \texttip{v_{\rm 2}}{v_2}. After the collision, the two cars stick together and travel off in the direction shown.
Part A First, find the magnitude of \texttip{\vec{v}}{v_vec}, that is, the speed \texttip{v}{v} of the twocar unit after the collision. Express \texttip{v}{v} in of \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, and the cars' initial speeds \texttip{v_{\rm 1}}{v_1} and \texttip{v_{\rm 2}}{v_2}.
Hint 1. Conservation of momentum Recall that conservation of linear momentum may be expressed as a vector equation, \vec{p}_{\rm initial}=\vec{p}_{\rm final}. Each vector component of linear momentum is conserved separately.
Hint 2. x and y components of momentum The momentum of the twocar system immediately after the collision may be written as \vec{p}=p_x\hat{i}+p_y\hat{j}, where the x and y directions are the eastward and northward directions, respectively. Find \texttip{p_{\mit x}}{p_x} and \texttip{p_{\mit y}}{p_y} https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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Express the two components, separated by a comma, in of \texttip{m_{\rm 1}}{m_1}, \texttip{m_{\rm 2}}{m_2}, \texttip{v_{\rm 1}}{v_1} and \texttip{v_{\rm 2}}{v_2}. ANSWER: \texttip{p_{\mit x}}{p_x}, \texttip{p_{\mit y}}{p_y} = m_{1} v_{1},m_{2} v_{2}
Hint 3. A vector and its components Recall that the square of the magnitude of a vector is given by the Pythagorean formula: p^2=p_x^2+p_y^2.
Hint 4. Velocity and momentum Find \texttip{v}{v}, the magnitude of the final velocity. Express \texttip{v}{v} in of the magnitude of the final momentum \texttip{p}{p} and the masses \texttip{m_{\rm 1}}{m_1} and \texttip{m_{\rm 2}}{m_2}. ANSWER: \texttip{v}{v} = \large{\frac{p}{m_{1} + m_{2}}}
ANSWER: \texttip{v}{v} = \large{{\frac{1}{m_{1}+m_{2}}} \sqrt{\left(m_{1}{^2}v_{1}{^2}\right)+\left(m_{2}{^2}v_{2}{^2}\right)}}
Correct
Part B Find the tangent of the angle \texttip{\theta }{theta}. Express your answer in of the momenta of the two cars, \texttip{p_{\rm 1}}{p_1} and \texttip{p_{\rm 2}}{p_2}. ANSWER: \tan(\theta) = \large{\frac{p_{2}}{p_{1}}}
Correct
Part C Suppose that after the collision, \tan \theta=1; in other words, \texttip{\theta }{theta} is 45^\circ. This means that before the collision: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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ANSWER: The magnitudes of the momenta of the cars were equal. The masses of the cars were equal. The velocities of the cars were equal.
Correct
Exercise 8.42 A bullet of mass 4.00 {\rm g} is fired horizontally into a wooden block of mass 1.13 {\rm kg} resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.150. The bullet remains embedded in the block, which is observed to slide a distance 0.210 {\rm m} along the surface before stopping.
Part A What was the initial speed of the bullet? ANSWER: v = 223 {\rm m/s}
Correct
An Exciting Encounter An atom of mass \texttip{M}{M} is initially at rest, in its ground state. A moving (nonrelativistic) electron of mass \texttip{m_{\rm e}}{m_e} collides with the atom. The atom+electron system can exist in an excited state in which the electron is absorbed into the atom. The excited state has an extra, "internal," energy \texttip{E}{E} relative to the atom's ground state.
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Part A Find the kinetic energy \texttip{K_{\rm e}}{K_e} that the electron must have in order to excite the atom. Express your answer in of \texttip{E}{E}, \texttip{m_{\rm e}}{m_e}, and \texttip{M}{M}.
Hint 1. How to approach the problem Depending on the initial kinetic energy of the electron, there are two possible outcomes to the collision between the electron and the atom. If the electron's kinetic energy is too low or too high, the collision will be elastic. If the electron's kinetic energy is just right, the collision will be perfectly inelastic: The electron will be absorbed into the atom, and the two will stick together. In both cases, total momentum must be conserved. In the case of an elastic collision, total kinetic energy is conserved. However, in the case of a perfectly inelastic collision, the final kinetic energy of the system is less than the initial kinetic energy. The loss of kinetic energy goes into exciting the atom, so total energy is conserved. From the perspective of Newtonian mechanics, you can think of the excited atom as having an internal potential energy. For example, it is as if the electron has compressed a spring in the atom and is held there by a catch mechanism. Thus, you can think of the system as having acquired a final potential energy \texttip{E}{E}. The usual expression for energy conservation then becomes K_{\rm final} + E = K_{\rm initial}.
Hint 2. Find the final kinetic energy in of the initial kinetic energy of the electron Find the kinetic energy \texttip{K_{\rm final}}{K_final} that the atom+electron system would have after a perfectly inelastic collision. Express your answer in of the initial kinetic energy of the electron \texttip{K_{\rm initial}} {K_initial}, \texttip{m_{\rm e}}{m_e}, and \texttip{M}{M}. The initial velocity of the electron, \texttip{v_{\rm 0}}{v_0}, should not appear in your answer.
Hint 1. Find the final velocity of the atom Suppose the initial velocity of the electron is \texttip{v_{\rm 0}}{v_0}. Use conservation of momentum to find the velocity \texttip{v_{\rm final}}{v_final} of the atom after it has absorbed the electron. Express your answer in of \texttip{v_{\rm 0}}{v_0}, \texttip{m_{\rm e}}{m_e}, and \texttip{M}{M}. ANSWER: \texttip{v_{\rm final}}{v_final} = \large{\frac{m_{e} v_{0}}{M+m_{e}}}
Hint 2. Find the final kinetic energy in of initial velocity In of \texttip{v_{\rm 0}}{v_0}, what is \texttip{K_{\rm final}}{K_final}, the kinetic energy of the atom after it has absorbed the electron? Express \texttip{K_{\rm final}}{K_final} in of \texttip{m_{\rm e}}{m_e}, \texttip{M}{M}, and \texttip{v_{\rm 0}}{v_0} (but without \texttip{K_{\rm initial}}{K_initial}). ANSWER: \texttip{K_{\rm final}}{K_final} = \large{\frac{1}{2} {v_{0}}^{2} \frac{{m_{e}}^{2}}{M+m_{e}}}
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ANSWER: \texttip{K_{\rm final}}{K_final} = \large{\frac{m_{e}}{M+m_{e}} K_{\rm{initial}}}
ANSWER: \texttip{K_{\rm e}}{K_e} = \large{{\frac{M+m_{e}}{M}}E}
Correct
Part B We can use the result from Part A to study a process of interest in atomic physics: a collision of two atoms that causes one of the atoms to ionize (lose an electron). In this case, \texttip{E}{E} is the energy needed to ionize one of the atoms, called the ionization energy. The most efficient way to ionize an atom in a collision with another atom is for the collision to be completely inelastic (atoms stick together after the collision). If the collision were perfectly elastic, then translational kinetic energy would be conserved, and there would be no energy left over for exciting the atom. If the collision were partially elastic, then some of the initial kinetic energy would be converted into internal energy, but not as much as in a perfectly inelastic collision. In practice, interatomic collisions are never perfectly inelastic, but analyzing this case can give a lower bound on the amount of kinetic energy needed for ionization. Is it possible to ionize an atom of {}^{133}\rm Cs, initially at rest, by a collision with an atom of {}^{16}\rm O that has kinetic energy \texttip{K_{\rm initial}}{K_initial} of 4.0 electron volts? The ionization energy of the cesium atom is 3.9 electron volts. It doesn't matter what energy units you choose, as long as you are consistent. For this question, it is most convenient to use electron volts (\rm {eV}) throughout. Note that 1\ \rm {eV} = 1.60 \times 10^{19} \rm {J}, which you maybe more familiar with. You can take the mass of the oxygen atom to be 16 atomic mass units and that of the cesium atom to be 133 atomic mass units. It doesn't matter what mass units you choose, as long as you are consistent. For this question, it is most convenient to use atomic mass units, since these are the numbers you are provided with.
Hint 1. How to approach the problem The cesium atom can be ionized only if a sufficient amount of kinetic energy is available. That is, the initial kinetic energy of the oxygen minus the final kinetic energy of the oxygen and cesium must be greater than the ionization energy of the cesium atom. The final kinetic energy of the oxygen and cesium will be minimal in a perfectly inelastic collision. Think about how this situation corresponds to the one you analyzed in Part A.
Hint 2. What is the correspondence with Part A? The oxygen atom in this collision corresponds to which of the two particles in the collision in Part A? ANSWER:
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the electron the atom
Hint 3. The masses of cesium and oxygen For this part of the problem, all you need to know is that the cesium atom is less than ten times as massive as the oxygen atom. ANSWER: yes no
Correct
Part C What is the least possible initial kinetic energy \texttip{K_{\rm min}}{K_min} the oxygen atom could have and still excite the cesium atom? Express your answer in electron volts, to one decimal place.
Hint 1. How to approach the problem Use the equation you found in Part A. Here, you can take the mass of the (initially moving) oxygen atom to be 16 atomic mass units and that of the (initially stationary) cesium atom to be 133 atomic mass units. It doesn't matter what mass units you choose, as long as you are consistent. For this question, it is most convenient to use atomic mass units, since these are the numbers you are provided with. ANSWER: \texttip{K_{\rm min}}{K_min} = 4.4 \rm eV
Correct
Trading Momenta in a Collision Two particles move perpendicular to each other until they collide. Particle 1 has mass \texttip{m}{m} and momentum of magnitude 2p, and particle 2 has mass 2m and momentum of magnitude \texttip{p}{p}. Note: Magnitudes are not drawn to scale in any of the figures.
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Part A Suppose that after the collision, the particles "trade" their momenta, as shown in the figure. That is, particle 1 now has magnitude of momentum \texttip{p}{p}, and particle 2 has magnitude of momentum 2p; furthermore, each particle is now moving in the direction in which the other had been moving. How much kinetic energy, \texttip{K_{\rm lost}}{K_lost}, is lost in the collision? Express your answer in of \texttip{m}{m} and \texttip{p}{p}.
Hint 1. How to approach the problem To find the kinetic energy lost in the collision, compute the initial kinetic energy \texttip{K_{\rm initial}} {K_initial} (before the collision) and the final kinetic energy \texttip{K_{\rm final}}{K_final} (after the collision). Then take the difference: K_{\rm lost} = K_{\rm initial} K_{\rm final}.
Hint 2. Find the relationship between energy and momentum What is the kinetic energy \texttip{K}{K} of a particle with mass \texttip{m}{m} and magnitude of momentum \texttip{p}{p}? ANSWER: \texttip{K}{K} = \large{\frac{p^{2}}{2 m}}
Hint 3. Find the initial kinetic energy What is the initial kinetic energy \texttip{K_{\rm initial}}{K_initial} of the twoparticle system? Express your answer in of \texttip{p}{p} and \texttip{m}{m}.
Hint 1. Find the kinetic energy of particle 1 Find the kinetic energy \texttip{K_{\rm 1.initial}}{K_1.initial} of particle 1 before the collision. Express your answer in of \texttip{p}{p} and \texttip{m}{m}. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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ANSWER: \texttip{K_{\rm 1.initial}}{K_1.initial} = \large{\frac{\left(2 p\right)^{2}}{2 m}}
Hint 2. Find the kinetic energy of particle 2 Find the kinetic energy \texttip{K_{\rm 2.initial}}{K_2.initial} of particle 2 before the collision. Express your answer in of \texttip{p}{p} and \texttip{m}{m}. ANSWER: \texttip{K_{\rm 2.initial}}{K_2.initial} = \large{\frac{p^{2}}{2 \left(2m\right)}}
ANSWER: \texttip{K_{\rm initial}}{K_initial} = \large{\frac{9 p^{2}}{4 m}}
Hint 4. Find the final kinetic energy What is the final kinetic energy \texttip{K_{\rm final}}{K_final} of the twoparticle system? Express your answer in of \texttip{p}{p} and \texttip{m}{m}. ANSWER: \texttip{K_{\rm final}}{K_final} = \large{\frac{3 p^{2}}{2 m}}
ANSWER: \texttip{K_{\rm lost}}{K_lost} = \large{\frac{3p^{2}}{4m}}
Correct
Part B Consider an alternative situation: This time the particles collide completely inelastically. How much kinetic energy \texttip{K_{\rm lost}}{K_lost} is lost in this case? Express your answer in of \texttip{m}{m} and \texttip{p}{p}.
Hint 1. How to approach the problem As in Part A, to find the kinetic energy lost in the collision, compute the initial kinetic energy \texttip{K_{\rm initial}}{K_initial} (before the collision) and the final kinetic energy \texttip{K_{\rm final}}{K_final} (after the collision). Then take the difference: K_{\rm lost} = K_{\rm initial} K_{\rm final}. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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Hint 2. Definition of completely inelastic A completely inelastic collision is one in which the particles stick together after the collision.
Hint 3. Initial kinetic energy The initial kinetic energy in this case is the same as it was in the partially elastic collision.
Hint 4. Find the final kinetic energy What is the combined kinetic energy \texttip{K_{\rm final}}{K_final} of the two particles after the perfectly inelestic collision? If you use the formula \large{K=\frac{p^2}{2m}}, that \texttip{p}{p} is the magnitude of the momentum vector. Express your answer in of \texttip{m}{m} and \texttip{p}{p}.
Hint 1. Find the magnitude of the final momentum What is the magnitude of \texttip{\vec{p}_{\rm final}}{p_final_vec}, the total momentum of the two particles after the collision? Express \texttip{p_{\rm final}}{p_final} in of \texttip{p}{p}.
Hint 1. Find the final momentum vector What is the total momentum \texttip{\vec{p}_{\rm final}}{p_final_vec} of the two particles after the collision? Take the positive x direction to be to the right in the figures and the positive y direction to be upward. Express your answer as a vector in of \texttip{p}{p} and the unit vectors \texttip{\hat{i}}{i_unit} and \texttip{\hat{j}}{j_unit}. ANSWER: \texttip{\vec{p}_{\rm final}}{p_final_vec} = 2 p \hat{i}+p \hat{j}
ANSWER: \texttip{p_{\rm final}}{p_final} = \sqrt{5} p
ANSWER: \texttip{K_{\rm final}}{K_final} = \large{\frac{5 p^{2}}{6 m}}
ANSWER: \texttip{K_{\rm lost}}{K_lost} = \large{\frac{17p^{2}}{12m}}
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Correct
Problem 8.101 A neutron at rest decays (breaks up) to a proton and an electron. Energy is released in the decay and appears as kinetic energy of the proton and electron. The mass of a proton is 1836 times the mass of an electron.
Part A What fraction of the total energy released goes into the kinetic energy of the proton? Express your answer as a percentage. ANSWER: \large{\frac{K_{\rm p}}{K_{\rm tot}}\,} = 5.44×10−2 \%
Correct
Problem 8.73 Spheres A (mass 0.020 {\rm kg}), B (mass 0.030 {\rm kg}), and C (mass 0.050 {\rm kg}), are each approaching the origin as they slide on a frictionless air table . The initial velocities of A and B are 1.50 {\rm m/s} and 0.50 .{\rm m/s} All three spheres arrive at the origin at the same time and stick together.
Part A What must the xcomponent of the initial velocity of C be if all three objects are to end up moving at 0.50 {\rm m/s} in the +xdirection after the collision? ANSWER:
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v_{Cx} = 1.75 {\rm m/s}
Correct
Part B What must the ycomponent of the initial velocity of C be if all three objects are to end up moving at 0.50 {\rm m/s} in the +xdirection after the collision? ANSWER: v_{Cy} = 0.260 {\rm m/s}
Correct
Part C If C has the velocity found in parts (A) and (B), what is the change in the kinetic energy of the system of three spheres as a result of the collision? Express your answer using two significant figures. ANSWER: \Delta K = −9.2×10−2 {\rm J}
Correct
Problem 8.109 A fireworks rocket is fired vertically upward. At its maximum height of 80.0 {\rm m} , it explodes and breaks into two pieces, one with mass \texttip{m_{\rm A}}{m_A} = 1.45 {\rm kg} and the other with mass \texttip{m_{\rm B}}{m_B} = 0.300 {\rm kg} . In the explosion, 900 {\rm J} of chemical energy is converted to kinetic energy of the two fragments.
Part A What is the speed of each fragment just after the explosion? Enter your answers numerically separated by a comma. ANSWER: v_{\rm A}, v_{\rm B} = 14.6,70.5 {\rm m/s}
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Correct
Part B It is observed that the two fragments hit the ground at the same time. What is the distance between the points on the ground where they land? Assume that the ground is level and air resistance can be ignored. ANSWER: \left|x_{\rm A}\;\;x_{\rm B}\right| = 344 {\rm m}
Correct
Exercise 8.49 Canadian nuclear reactors use heavy water s in which elastic collisions occur between the neutrons and deuterons of mass 2.0 u.
Part A What is the speed of a neutron, expressed as a fraction of its original speed, after a headon, elastic collision with a deuteron which is initially at rest? Express your answer using three decimal places. ANSWER: 0.333
Correct
Part B What is its kinetic energy, expressed as a fraction of its original kinetic energy? Express your answer using three decimal places. ANSWER: 0.111
Correct
Part C https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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How many such successive collisions will reduce the speed of a neutron to 1/2190 of its original value? Express your answer as a whole number. ANSWER: 7 collisions
Correct
Problem 8.76 At a classic auto show, a 810 {\rm kg} 1955 Nash Metropolitan motors by at 8.4 {\rm m/s} , followed by a 1630 {\rm kg} 1957 Packard Clipper purring past at 4.4 {\rm m/s} .
Part A What is the ratio of the kinetic energy of the Nash to that of the Packard? ANSWER: \large{\frac{K_{\rm N}}{K_{\rm P}}} = 1.81
Correct
Part B What is the ratio of the magnitude of momentum of the Nash to that of the Packard? ANSWER: \large{\frac{p_{\rm N}}{p_{\rm P}}} = 0.949
Correct
Part C Let F_{\rm{N}} be the net force required to stop the Nash in time \texttip{t}{t}, and let F_{\rm{P}} be the net force required to stop the Packard in the same time. What is the ratio F_{\rm{N}}/F_{\rm{P}} of these two forces? ANSWER: \large{\frac{F_{\rm{N}}}{F_{\rm{P}}}} = 0.949
Correct https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266341
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Part D Now let F_{\rm{N}} be the net force required to stop the Nash in a distance d, and let F_{\rm{P}} be the net force required to stop the Packard in the same distance. What is the ratio F_{\rm{N}}/F_{\rm{P}} {\kern 1pt}? ANSWER: \large{\frac{F_{\rm{N}}}{F_{\rm{P}}} {\kern 1pt}} = 1.81
Correct
Problem 8.112 In Section 8.6, we considered a rocket fired in outer space where there is no air resistance and where gravity is negligible. Suppose instead that the rocket is accelerating vertically upward from rest on the earth's surface. Continue to ignore air resistance and consider only that part of the motion where the altitude of the rocket is small so that g may be assumed to be constant.
Part A How is equation \large{m\frac{{dv}}{{dt}} = v_{ex} \frac{{dm}}{{dt}}} modified by the presence of the gravity force? ANSWER: Essay answers are limited to about 500 words (3800 characters maximum, including spaces). 3785 Character(s) remaining (none provided)
Part B Derive an expression for the acceleration a of the rocket, analogous to equation \large{a = \frac{{dv}}{{dt}} = \frac{{v_{ex} }}{m}\frac{{dm}}{{dt}}}. ANSWER: Essay answers are limited to about 500 words (3800 characters maximum, including spaces). 3785 Character(s) remaining (none provided)
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Part C What is the acceleration of the rocket if it is near the earth's surface? In the first second of firing, the rocket ejects \large{\frac{1}{120}} of its mass with a relative speed of 2300 {\rm {\rm m/s}} . You can ignore air resistance. ANSWER: a_{\rm earth} = 10 {\rm m/s^2}
Incorrect; Try Again; 5 attempts remaining
Part D Now suppose that \large{\frac{3}{4}} of the initial mass m_0 of the rocket is fuel, so the final mass m=m_0/4 is and that the fuel is completely consumed at a constant rate in a total time 90 {\rm {\rm s}} . Find the speed of the rocket after 90 {\rm {\rm s}} if the rocket is fired from the earth's surface rather than in outer space. You can ignore air resistance. ANSWER: v_{\rm earth} =
{\rm m/s}
Part E How does your answer in previous part compare with the rocket speed calculated in outer space at the same conditions? ANSWER: \large{\frac{v_{\rm space}}{v_{\rm earth}}} =
Score Summary: Your score on this assignment is 83.0%. You received 34.03 out of a possible total of 41 points.
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