(a) p_{i}= p_{f }m_{1}u_{1}+ m_{2}u_{2}= m_{1}v_{1}+ m_{2}v_{2 }1.6 x 5.5 + 2.4 x 2.5 = 1.6 v_{1}+ 2.4 x 4.9

14.8 = 1.6 v_{1}+ 11.76

v_{1}= 1.9 ms^{-1}(assuming the positive direction is to the right).

(b) Initial kinetic energy E_{K(i)}= ½ m_{1}u_{1}^{2}+ ½ m_{2}u_{2}^{2}= ½ x1.6 x 5.5^{2}+ ½ 2.4 x 2.5^{2}= 31.7 J

Final kinetic energy E_{K(f)}= ½ m_{1}v_{1}^{2}+ ½ m_{2}v_{2}^{2}= ½ x 1.6 x 1.9^{2}+ ½ x 2.4 x 4.9^{2}= 31.7 J

As E_{K(i)}= E_{K(f)}the collision is elastic.

(c) No. Mathematically v_{1}would equal - 3.9 ms^{-1}and v_{2}would equal + 3.8 ms^{-1}.

PROOF: u_{1}= +5.5 ms^{-1}; u_{2}= -6 ms^{-1 }

Conservation of momentum:p_{i}= p_{f }m_{1}u_{1}+ m_{2}u_{2}= m_{1}v_{1}+ m_{2}v_{2 }2.8 = 1.6 v_{1}+ 2.4 v_{2 }v_{1}= 1.75 - 1.5 v_{2 }Conservation of kinetic energy:EUsing the quadratic formula there are two solutions: v_{K(i)}= E_{K(f) }½ m_{1}u_{1}^{2}+ ½ m_{2}u_{2}^{2}= ½ m_{1}v_{1}^{2}+ ½ m_{2}v_{2}^{2 }39.6 = v_{1}^{2}+ 1.5v_{2}^{2 }39.6 = (1.75 - 1.5 v_{2})^{2}+ 1.5 v_{2}^{2 }v_{2}^{2}- 1.39 v_{2}= 9.74 = 0

_{1}= +2.5 ms^{-1}and hence v_{2}= -0.5 ms^{-1}.

This is impossible as the balls would have to had jumped over each other.

The second solution: v_{1}= -3.9 ms^{-1}and hence v_{2}= +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.

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 10^{6}x 10 x 50 = 2.75 x 10^{9}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 10^{8}J/hr

Joules per second = J/hr divided by 3600

= 2.14 x 10^{8}/3600 =5.94 x 10^{4}J/s = 5.94 x 10^{4}Watts).

Question 38

E_{K(i)}= ½ mv^{2}= ½ x 0.63 x 14^{2}= 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 kx^{2}

1587 x 10 x 180 = ½ k (180 - 30)^{2}

k = 254 Nm

Question 40

Conservation of Momentum:p_{i}= p_{f 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 10^{3}x 10 x 1.2 = 1.9 x 10^{7}J

D KE = ½ m (v^{2}- u^{2}) = - 1.9 x 10^{7}J

½ x 1586 x 10^{3}(v^{2}- (28000/3600)^{2}) = - 1.9 x 10^{7}J

793000 (v^{2}- 60.5) = - 1.9 x 10^{7}J

793000 v^{2}- 4.8 x 10^{7}= - 1.9 x 10^{7}J

v = 6.0 ms^{-1}(21.7 km h^{-1 }Question 42

EPE = ½ kx^{2}= ½ mv^{2 }½ x 292 x (3.2 x 10^{-2})^{2}= ½ x 12 x 10^{-3}x v^{2 }v = 5 ms^{-1}

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