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

Question

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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