Question

A particle of mass $m$ moving with a velocity $v$ makes a head on elastic collision with another particle of the same mass initially at rest. The velocity of the first particle after collision is
A. $v$
B. $\dfrac{v}{2}$
C. $2v$
D. $0$

Answer 1

Hint: We know that in case of head on elastic, the kinetic energy always remains constant. Also, in this case, the velocities and the speeds are interchanged. Here, we will use the formula of law of conservation of linear momentum to calculate the velocity of the particle.

Formula used:
In case of head-on collision, conservation of linear momentum is given by
${m_1}{\vec v_1} = {m_1}{\vec v'_1} + {m_2}{\vec v'_2}$
Here, ${m_1}$ is the mass of the first particle and ${m_2}$ is the mass of the second particle, ${v_1}$ is the velocity of the particle before collision, ${v'_1}$ is the velocity of first particle after collision and ${v'_2}$ is the velocity of the second collision after collision.

Complete step by step answer:
We know that law of conservation of linear momentum in case of head on collision of the particles is given by
$${m_1}{\vec v_1} = {m_1}{\vec v'_1} + {m_2}{\vec v'_2}$$
Now, it is given in the question that the mass of both the particles are the same. Therefore, we will take the mass of both the particles as $m$ . Also, the velocity of the particle before collision is given as $v$ , we will take the velocities of particles after collisions as ${v_1}$ and ${v_2}$ . Hence, the above equation will become
$mv = m{v_1} + m{v_2}$
$ \Rightarrow \,mv = m\left( {{v_1} + {v_2}} \right)$
$v = {v_1} + {v_2}$
Now, as we know that in case of elastic collision, the coefficient of restitution is one as given by $e = \dfrac{{velocity\,of\,separation}}{{velocity\,before\,collision}} = 1$
$\dfrac{{{v_2} - {v_1}}}{v} = 1$
$ \Rightarrow \,{v_2} - {v_1} = v$
$ \Rightarrow \,{v_2} = v + {v_1}$
Now, putting the value of $v$ in the above equation, we get
$mv = m{v_1} + mv + m{v_1}$
$ \therefore \,{v_1} = 0$
Therefore, we get ${v_2} = v$, the velocity of the particle before collision is zero.

Hence, option D is the correct option.

Note: As it is given in the question, the particle of mass $m$ is initially at rest. Therefore, the force on the particle will be zero. Also, the momentum of the particle will also be zero. That is why, we have taken the coefficient of restitution as one.