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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Infinity Learn · Accepted Answer

Let final velocity of the block of mass $$4m=v$$'Initial velocity of block of mass $$4$$ $$m=0$$ Final velocity of block of mass $$m=0According$$ to law of conservation of linear momentum,$$mv+4m$$×$$0=4mv$$'$$+0v$$'$$=v4Coefficient$$ of restitution,$$e=$$ Relative velocity of separation  Relative velocity of approach $$=v/4v$$

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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