Momentum and CollisionMediumBloom 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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