Momentum and CollisionHardBloom L3

Question

A ball of mass 1kg strikes a heavy platform, elastically, moving upwards with a velocity of 5m/s. The speed of the ball just before the collision is 10m/s downwards. Then the impulse imparted by the platform on the ball is

Options

A.15 N – s
B.10 N – s
C.20 N – s
D.30 N – s

Solution

Sol. Let velocity of ball just after the collision is V ↑

In elastic collision

Velocity of separation = Velocity of approach

V – 5 = 15

V = 20 m/s

Ip = change in linear momentum of ball

= (1×20) – (1 × ( – 10))

= 30 NS

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