A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial…

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A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be

MARKS Content Team · Accepted Answer

Correct answer: (B) 0.25. By conversation of linear momentum mv=4mv^(′)⇒v^(′)=v/4 Coefficient of restitution ( e ) =(Velocity of separation)/(Velocity of approach) = v/4-0v-0=1/4=0.25

A moving block having mass $m,$ collides with another stationary block having mass $4 m .$ The lighter block comes to rest after collision. When the initial velocity of the lighter block is $v,$ then the value of coefficient of restitution $(e)$ will be

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