[physics] 2010 Itute Unit 3 Solutions 5p222g

2010 Physics Trial Exam 1 Solutions © Copyright 2010 itute.com

Area of study 1 − Motion in one and two dimensions Q1 Speed = circumference / time taken =

Q2 Fnet = ma = m

π × 120 60

≈ 6.28 ms-1.

v2 6.282 = 75 × ≈ 49 N r 60

Q11 4.20 × v = 14.7 , v = 3.5 ms-1. Q12 Work done by Fav = change in kinetic energy of the rifle: 1 Fav × 0.025 = × 4.20 × 3.52 , Fav ≈ 1.03 × 103 N. 2 Q13 Action: Rifle on cushion. Reaction: Cushion on rifle. Q14 The net force on the ball must be zero for it to move in a straight line at constant speed. Corner C provides a reaction force on the ball opposite to the force of gravity.

Q3 Centripetal force, W.

Q15 Weight W = force of gravity ∝

20 × 1000 Q4 Force constant k = gradient = = 4.0 × 10 4 Nm-1. 0.5

At r = 1R , W = 9.6N. 1 At r = 2 R , W = × 9.6 = 2.4 N. 4

Q5 Gravitational potential energy changes to elastic potential 1 1 energy: mgh = kx 2 , 40 × 10 × (2 + x ) = × 4.0 × 10 4 × x 2 , 2 2 50 x 2 − x − 2 = 0 , x = 0.21 m = 21 cm. Q6 It is possible for the skier to move under gravity only, i.e. zero reaction force from the ski slope along the entire parabolic path QR. This occurs when the path of the skier’s projectile motion is the same as the parabolic path QR. In this situation the skier experiences weightlessness.

Q16

2M X aB g M = Y = Y = =2. aA g X MX MX

Q17

WB m B g Y 2 m A × 2 = = =4. WA m A g X mA 4π 2 r 4π 2 r , T2 = . 2 a T 1 1 T g 1 g = 1Y = X = , ∴ B = ≈ 0.71 . T g 2 2 Y A gX

Q18 a = 2

∴ Q7 At R, increase in kinetic energy = decrease in gravitational 1 1 potential energy: mv 2 − mV 2 = mgh , v 2 − V 2 = 2 gh , 2 2

1 . r2

TB 2 TA

v = V 2 + 40 . V 2 + 40 ms-1. 2

Vertical component: v y = V 2 + 40 sin 45° = 2

2

Q8 Vertical component: v y = u y + 2as y , 2

Area of study 2 − Electronics and photonics Q1 V2 =

v y = u y + 2as y = 0 + 2 × 10 × 2 = 40 , ∴

V 2 + 40 = 40 , 2

R2 R4 × 9.0 = 6.0 V; V4 = × 9.0 = 3.6 V. R1 + R2 R3 + R4

VPQ = 6.0 − 3.6 = 2.4 V.

V = 40 = 6.3 ms-1.

1 15 = kΩ. 1 + 3+ 2 13 9.0 = 15 = 7.8 × 10 −3 A = 7.8 mA. 3 × 10 13

Q2 Rtotal = Q9 Horizontal component: Vt = 4 , t =

4 40

Alternatively, vertical component: s y = u y +

= 0.63 s. 1 2 at , 2

1 2 = 0 + × 10 × t 2 , t = 0.63 s. 2

r r Q10 ∆p = m∆v = 0.0525 + 280 − 0 = + 14.7 kg ms-1.

(

)

I=

V Rtotal

1 0.5+1

Q3 Ptotal = VI = 9.0 × 7.8 × 10 −3 ≈ 0.070 W. Q4 When V = 0.50 V, I = 20 mA. R =

2010 Physics Trial Exam 1 Solutions © Copyright 2010 itute.com

0.50 = 25 Ω 20 × 10 −3

Q5

Detailed study 2− − Investigating materials and their use in structures 1 A

2 A

7 C

3 B

8 D

9 B

4 D 10 D

5 C 11 C

6 A 12 A

13 A

Q1 The combined weight of the cable and the mass is greatest at point P. A Q2 Since the tension is greatest at point P, .: the stress is greatest at point P. Q3 E = gradient = When I = 29 mA, Vresistor + Vdiode = 0.725 + 0.775 = 1.5 V. Q4 σ = Eε , ε =

∆v −5 Q6 Voltage gain = o = = −50 . ∆vi + 0.1

0.275 ≈ 69 GPa. 0.004

σ E

. For constant σ ,

A B

ε al E = steel . ε steel Eal

D

Q5

C

Q6

A

Q7

C

Q8

D

Q9

B

Q7

Q10 σ =

8900 ≈ 18 kPa. π × 0.40 2

Q11 The door is kept in equilibrium, .: net torque is zero.

C

Q12 τ = r⊥ F = 0.75 sin 60° × 3 = 1.95 Nm clockwise.

A

Q13 Net torque is zero. .: the torque of the spring-closer is 1.95 Nm anticlockwise.

A

Q8 25°C. Q9

R 8.0 = , R = 30 × 103 Ω = 30 kΩ. 3 4 .0 15 × 10

Q10 The resistance of the thermistor will be higher than 15kΩ when the temperature is below 15°C, the voltage across it (output voltage) will then be above 4.0V and the alarm will remain on.

D

Please inform [email protected] re conceptual, mathematical and/or typing errors

Q11 C Q12 B and D.

2010 Physics Trial Exam 1 Solutions © Copyright 2010 itute.com