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Question

A smooth sphere of mass m=5 kg is moving on a horizontal plane with a velocity (5^i+^j) m/s when it collides with a vertical wall which is parallel to the vector ^j. If the coefficient of restitution between the sphere and the wall is 12, find the loss in kinetic energy caused by the impact.

A
46.875 J
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B
0 J
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C
32.5 J
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D
49.375 J
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Solution

The correct option is A 46.875 J
We must separate the velocity components parallel and perpendicular to the wall. The component of velocity parallel to the wall remains unchanged, while component perpendicular to the wall changed according to.
e=speed of separationspeed of approach
Let v=vx^i+vy^j be the velocity after collision.
Then, vx=eux=52 m/s
and vy=1 m/s

So, v=52^i+^j ...(i)



Hence the velocity of sphere after the impact is v=52^i+^j

Loss in kinetic Energy = Initial Kinetic Energy - Final Kinetic Energy
=12mv212mu2
=12×5×(52+12)12×5[(52)2+12]
K.E.=3758=46.875 J

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