# Chapter 9 - Worked Solutions

### to questions from the OUP text Senior Physics - Concepts in Context by Walding, Rapkins and Rossiter

My thanks to Peter Finch from St Joseph's College, Gregory Terrace, for these (pdf) solutions.

Solutions-Ch9-Finch.pdf

The rest are by me, Richard Walding:

Question 34

(a) pi = pf
m1u1 + m2u2 = m1v1 + m2v2
1.6 x 5.5 + 2.4 x 2.5 = 1.6 v1 + 2.4 x 4.9
14.8 = 1.6 v1 + 11.76
v1 = 1.9 ms-1 (assuming the positive direction is to the right).

(b) Initial kinetic energy EK(i) = ½ m1u12 + ½ m2u22 = ½ x1.6 x 5.52 + ½ 2.4 x 2.52 = 31.7 J
Final kinetic energy EK(f) = ½ m1v12 + ½ m2v22 = ½ x 1.6 x 1.92 + ½ x 2.4 x 4.92 = 31.7 J
As EK(i) = EK(f) the collision is elastic.

(c) No. Mathematically v1 would equal - 3.9 ms-1 and v2 would equal + 3.8 ms-1.

PROOF: u1 = +5.5 ms-1; u2 = -6 ms-1

Conservation of momentum:
pi = pf
m1u1 + m2u2 = m1v1 + m2v2
2.8 = 1.6 v1 + 2.4 v2
v1 = 1.75 - 1.5 v2

Conservation of kinetic energy:
EK(i) = EK(f)
½ m1u12 + ½ m2u22 = ½ m1v12 + ½ m2v22
39.6 = v12 + 1.5v22
39.6 = (1.75 - 1.5 v2)2 + 1.5 v22
v22 - 1.39 v2 = 9.74 = 0

Using the quadratic formula there are two solutions: v1 = +2.5 ms-1 and hence v2 = -0.5 ms-1.
This is impossible as the balls would have to had jumped over each other.

The second solution: v1 = -3.9 ms-1 and hence v2 = +3.8 ms-1. In this case Ball 1 is travelling in the negative direction and hence can't be travelling in the positive direction.

Question 35
Part (a) Prove vertical displacement is 7.3 cm: Solving for side of triangle (200 - h) = 192.7 cm, hence h = 7.3 cm

Part (b) Calculate velocity: Which is quite fast. (A big thank you to Chris Smith at Sunshine Coast Grammar School for pointing out my poor proofreading).

Question 36
GPE = mgh = 5.5 x 106 x 10 x 50 = 2.75 x 109 J (= 2750 MJ)
At 2 cents per MJ this comes to \$55 per second or \$4.75 million per day.

Question 37
(a) 1 MJ = 1/31 litres = 0.0322 L
Distance = 100/11.5 x 0.0322 = 0.28 km

(b) 60 km/h = 60/100 x 11.5 litres/hour = 6.9 L/hr
Power = 6.9 x 31 MJ per hour = 214 MJ/hr = 2.14 x 108 J/hr
Joules per second = J/hr divided by 3600
= 2.14 x 108/3600 =5.94 x 104 J/s = 5.94 x 104 Watts).

Question 38
EK(i) = ½ mv2 = ½ x 0.63 x 142 = 61.74 J
GPE = mgh = 0.63 x 10 x 8.1 = 51.03 J
Work done = 61.74 - 51.03 = 10.7 J

Question 39
assume all GPE converted to EPE
mgh = ½ kx2
1587 x 10 x 180 = ½ k (180 - 30)2
k = 254 Nm

Question 40 Conservation of Momentum: pi = pf
m1u1 + m2u2 = m1v1 + m2v2
2 u = 2u/4 + mv2
3/2 u = mv2
u = 2 mv2/3

Conservation of Kinetic Energy: EK(i) = EK(f)
½ m1u12 + ½ m2u22 = ½ m1v12 + ½ m2v22
½ x u2 + 0 = ½ x 2 x (u/4)2 + ½ mv22
2u2 = 2u2/16 + mv22
2u2 - 2/16 u2 = mv22
30/16 u2 = mv22
u2 = 16mv22/30

From momentum: u = 2 mv2/3, therefore u2 = 4 m2v22/9
Hence 4 m2v22/9 = 16mv22/30
Cancel v22 from both sides:
4 m2/9 = 16 m/30
120 m = 144
m = 1.2 kg

Question 41
D KE = D GPE
D GPE = mgh = 1586 x 103 x 10 x 1.2 = 1.9 x 107 J
D KE = ½ m (v2 - u2) = - 1.9 x 107 J
½ x 1586 x 103 (v2 - (28000/3600)2) = - 1.9 x 107 J
793000 (v2 - 60.5) = - 1.9 x 107 J
793000 v2 - 4.8 x 107 = - 1.9 x 107 J
v = 6.0 ms-1 (21.7 km h-1
Question 42
EPE = ½ kx2 = ½ mv2
½ x 292 x (3.2 x 10-2)2 = ½ x 12 x 10-3 x v2
v = 5 ms-1 Return to Physics Text Home Page