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
√2mv
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C
2mv
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D
0
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Solution
The correct option is C2mv Assuming leftward direction (in the figure) as −vex axis. Initially, the ball is moving along −vex− 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=mΔv=m[vf−vi] ⇒J=m[(+v)−(−v)]=+2mv
Hence direction of impulse is along the +vex− axis direction.