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  4. 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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Q. 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

J & K CETJ & K CET 2019Work, Energy and Power

Solution:

According to law of conservation of linear momentum,
$es 0=4 m v'+0$
$v'=\frac{v}{4}$
$e=\frac{\text { Relative velocity of separation }}{\text { Relative velocity of approach }}=\frac{\frac{v}{4}}{v}$
$e=\frac{1}{4}=0.25$