Ans.

Let us consider a particle of mass \( m \) moving with velocity \( u_1 \) collides head-on with a particle of same mass \( m \) which is at rest, \( u_2=0 \).

Let \( v_1 \) and \( v_2 \) be their respective velocities after collision.

Since the collision is perfectly elastic collision,

According to the conservation of linear momentum,

\( mu_1+mu_2=mv_1+mv_2 \)

or, \( mu_1=mv_1+mv_2 \), since \( u_2=0 \)

or, \( u_1=v_1+v_2 \)

or, \( u_1-v_1=v_2\tag{1} \)

According to the conservation of energy,

\( \displaystyle{\frac{1}{2}mu_1^2+\frac{1}{2}mu_2^2=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2} \)

or,\( \displaystyle{\frac{1}{2}mu_1^2=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2} \), since \( u_1=0 \)

or, \( u_1^2-v_1^2=v_2^2\tag{2} \)

Now, dividing the equation (2) by equation (1), we get

\( u_1+v_1=v_2 \)

or, \( v_1=v_2-u_1\tag{3} \)

From equations (1) & (3) we get,

\( u_1=v_2-u_1+v_2 \)

or, \( 2u_1=2v_2 \)

or, \( u_1=v_2 \tag{4}\)

From equation (1) & (4) we get

\( v_1=u_1-v_2 \)

or, \( v_1=u_1-u_1 \)

or, \( v_1=0 \)

So after collision, the colliding particles exchange their velocities.