A ball falls from a height of 5m and strikes the roof of a lift. If at time of collision, lift is moving in the upward direction with a velocity of 1ms−1. Then the velocity with which the ball rebounds after collision will be :
A
13ms−1 upwards
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B
12ms−1 downwards
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C
12ms−1 upwards
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D
11ms−1 downwards
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Solution
The correct option is C12ms−1 upwards Velocity of ball before collision=√2gh=√2×10×5=√100=10m/s Velocity of approach=10−(−1)=11m/s
Velocity of separation=(v−1)m/s Since collision is elastic, e=1. 11=v−1v=12m/s