First slide

Coefficient of restitution

Question

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' [NEET 2018]

Moderate

A

0.8
B

0.25
C

0.5
D

0.4

Solution

Since, the collision mentioned is an elastic head on collision. Thus, according to the law of conservation of linear
momentum, we have
   $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
where, m₁ and m₂are the masses of the two blocks respectively, u₁ and u₂ are their initial velocities and v₁ and v₂
are their final velocities, respectively.
Given,    $m_{1} = m, m_{2} = 4 m$
$u_{1} = v, u_{2} = 0 and v_{1} = 0$
$∴ mv + 4 m \times 0 = 0 + 4 {mv}_{2}$
$\Rightarrow$ $mv = 4 {mv}_{2}$ or $v_{2} = \frac{v}{4}$ …(i)
Now, the coefficient of restitution,
$e = \frac{relative velocity of separation}{relative velocity of approach}$
   $= - \frac{v_{2} - v_{1}}{u_{2} - u_{1}} = - \frac{\frac{v}{4} - 0}{0 - v}$ [from Eq.(i)]
= 1/4
$∴$ e = 0.25

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