Question

A body of mass M moves with velocity V and collides elastically with another body of mass m $(M > > m)$ at rest then the velocity of body of mass m is
A. V
B. \[2V \\ \]
C. $\dfrac{V}{2} \\ $
D. zero

Answer 1

Hint:In order to solve this question, we will use the concept of elastic collision in which momentum of the system remains conserved before and after the collision, and using coefficient of restitution we will solve for the velocity of the body of mass m.

Formula used:
The coefficient of restitution for perfectly elastic collision is $e = 1$ and its expression is,
$e = \dfrac{{{v_{separation}}}}{{{v_{approach}}}}$
Momentum of a body is calculated as,
$P = mv$
where m is the mass and v is the velocity of the body.

Complete step by step solution:
According to the question, we have given that initially first body of mass M was moving with velocity V and second body of mass m where $M > > m$ was at rest, so initial momentum for the system was \[{P_i} = MV\] now, let v and v’ be the velocity of two bodies of mass M and m after the collision so final momentum is given by ${P_f} = Mv + mv'$ and momentum is conserved in elastic collision so
${P_i} = {P_f} \\
\Rightarrow MV = Mv + mv' \to (i) \\ $
Now, velocity of separation is ${v_{separation}} = v' - v$ and velocity of approach before the collision was ${v_{approach}} = V - 0 = V$. Since for elastic collision
$e = \dfrac{{{v_{separation}}}}{{{v_{approach}}}} = 1 \\
\Rightarrow \dfrac{{v' - v}}{V} = 1 \\
\Rightarrow v = v' - V \to (ii) \\ $
Put the equation (ii) value in equation (i) we get,
$MV = M(v' - V) + mv' \\
\Rightarrow 2MV = v'(M + m) \\ $
since, $(M > > m)$ so $M + m \approx M$
$ \therefore v' = 2V$

Hence, the correct answer is option B.

Note: It should be remembered that, the value of coefficient of restitution is different for different kinds of collision such as for perfectly inelastic collision its value is zero whereas the value for inelastic collision is less than one and greater than zero.