Work, energy and power for Cambridge IGCSE Physics

Reading cue. Both 1.8 m and 1.2 m run from the ground up to the ball, not from one to the other, and 1.2 m is the top of the rebound, where the ball is momentarily at rest.

1. aIdentify Identify the point on the path drawn in Fig. 5.1 at which the ball has its greatest kinetic energy, and state the height of the ball above the ground there.

   recall2 marks

   Check answer · 2 marks

   1. just before it reaches the ground, at the bottom of the fall (1)
   2. height = 0 m, at ground level (1)

2. bDetermine Determine the speed of the ball just before it strikes the ground. Take the gravitational field strength as 9.8 N/kg and ignore air resistance.

   routine3 marks

   Check answer · 3 marks

   1. kinetic energy gained equals gravitational potential energy lost: (1/2)mv² = mgh (1)
   2. v² = 2 × 9.8 × 1.8 = 35.28 (m/s)² (1)
   3. v = 5.9 m/s (1)

3. cCalculate Calculate the percentage of its kinetic energy at the ground that the ball still has immediately after the bounce, and show that this percentage does not depend on the mass of the ball.

   demanding3 marks

   Check answer · 3 marks

   1. energy immediately after the bounce is mgh with h = 1.2 m; energy at the ground before the bounce is mgh with h = 1.8 m (1)
   2. ratio = 1.2/1.8 = 0.67, so 67% (1)
   3. m and g appear in both expressions and cancel, so only the two marked heights decide the percentage (1)

4. dEstimate The ball is allowed to bounce a second time. Assuming the same percentage of energy is lost at every bounce, estimate the height of the ball at the top of its second rebound, and suggest one reason why the true height would be lower than your estimate.

   top of the paper3 marks

   Check answer · 3 marks

   1. height = 1.2 x (1.2/1.8) (1)
   2. = 0.80 m (1)
   3. reason, any one: air resistance also removes energy during the flight; the fraction lost at each bounce is not really constant, since the ball deforms differently at different speeds (1)

Reading cue. In the upper panel the 2.5 m/s belongs to A alone, B at rest carrying none; in the lower panel v is the speed of the joined 2.0 kg, not of either trolley on its own.

1. aState State the momentum of trolley B before the collision, and give the reason from the upper panel of Fig. 8.1.

   recall2 marks

   Check answer · 2 marks

   1. zero (0 kg m/s) (1)
   2. B is at rest, so its velocity is zero, and momentum is mass x velocity (1)

2. bDetermine After the collision the joined trolleys move to the right at 1.0 m/s. Determine the change in momentum of trolley A, and give its direction.

   routine3 marks

   Check answer · 3 marks

   1. momentum of A before = 0.80 × 2.5 = 2.0 kg m/s to the right (1)
   2. momentum of A after = 0.80 × 1.0 = 0.80 kg m/s to the right (1)
   3. change = 1.2 kg m/s, directed to the left, that is opposite to A's motion (1)

3. cCalculate The two trolleys are in contact for 0.15 s. Calculate the average force that trolley B exerts on trolley A during the collision.

   demanding3 marks

   Check answer · 3 marks

   1. force = change in momentum / time taken (1)
   2. = 1.2 / 0.15 (1)
   3. = 8.0 N, acting to the left on A (1)

4. dExplain Using both panels of Fig. 8.1, explain why the momentum gained by trolley B is exactly equal to the momentum lost by trolley A, even though B is the heavier trolley and its velocity changes by less.

   top of the paper4 marks

   Check answer · 4 marks

   1. the force B exerts on A is equal in size and opposite in direction to the force A exerts on B (1)
   2. the two trolleys are in contact for the same length of time, so force x time is the same for both and the changes in momentum are equal and opposite (1)
   3. A's velocity changes by 1.5 m/s, giving 0.80 × 1.5 = 1.2 kg m/s; B's changes by 1.0 m/s, giving 1.2 × 1.0 = 1.2 kg m/s (1)
   4. B's larger mass is offset exactly by its smaller change in velocity, so the total momentum of the two trolleys is unchanged (1)