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Question

A man throws a ball of mass m on a vertical wall with speed v. If the ball bounces back with the same speed, find out the magnitude of impulse on the ball, imparted by normal reaction from the wall.


A
mv
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B
2 mv
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C
2 mv
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D
0
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Solution

The correct option is C 2 mv
Assuming leftward direction (in the figure) as ve x axis.
Initially, the ball is moving along ve x axis.
Initial velocity of the ball v1=v ^i
After collision with the wall, the ball reverses its direction
Final velocity of ball v2=v ^i

Impulse imparted by normal reaction on the ball:
J=m Δv=m[vfvi]
J=m[(+v)(v)]=+2mv

Hence direction of impulse is along the +ve x axis direction.

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