A body of mass 40kg having velocity 4 m/s collides with another body of mass 60kg having velocity 2 the collision is inelastic then loss in kinetic energy will be
A
440J
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B
392J
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C
48J
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D
144J
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Solution
The correct option is C48J
Given,
m1=40kg,v1=4m/s,m2=60kg,,v2=2m/s
Since the body is inelastic, both masses would move together at the velocity v,
According to the conservation of momentum,
m1v1+m2v2=(m1+m2)v
⇒40×4+60×2=(40+60)v
⇒280=100v
⇒v=2.8m/s
So initial KE=0.5×m1(v21)+0.5m2(v2)2=0.5×40×16+0.5×60×4=440J
The final KE=0.5(40+60)2.82=392J
Thus the difference in the kinetic energy =440−392=48J