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Momentum Questions

Q6.

(a) A cricketer throws a ball vertically upwards so that the ball leaves his hands at a speed of 25 ms-1. If air resistance can be neglected, calculate

(i) the maximum height reached by the ball,

v2 = u2 + 2as

0 = 252 – 2 × 9.81 × s

19.6 s = 625 so,

s = 32 m

(ii) the time taken to reach maximum height,

v = u + at, but v = 0 m/s so

t = 25/9.81 = 2.5 s

(iii) the speed of the ball when it is at 50% of the maximum height.

v2 = u2 + 2as = 252 – 2 × 9.81 × 16

v = 18 m s–1

(4 marks maximum)

(b) When catching the ball, the cricketer moves his hands for a short distance in the direction of travel of the ball as it makes contact with his hands.

Explain why this technique results in less force being exerted on the cricketer's hands.

The time taken to stop the ball is greater∴ rate of change of momentum is less

[OR work done on ball is the same but greater distance ∴ less force ]

(2 marks)

(6 Marks Total)