A body of mass m moving with velocity v makes a head-on collision with another body of mass 2m which is initially at rest. The loss of kinetic energy of the colliding body (mass m) is

Correct Option: 8/9 of its initial kinetic energy

Let us analyze the collision step-by-step under the assumption of a perfectly elastic head-on collision.

Step 1: Identify the initial parameters of the two bodies
Let the first body have mass $m_1=m$ and initial velocity $u_1=v$.
Let the second body have mass $m_2=2m$ and initial velocity $u_2=0$ (since it is initially at rest).

Step 2: Find the final velocity of the colliding body (mass m) after the collision
For a perfectly elastic head-on collision, the final velocity $v_1$ of the first body of mass $m_1$ is given by the standard formula:

$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}\right)u_2$

Substituting the values $m_1=m$, $m_2=2m$, $u_1=v$, and $u_2=0$ into the formula:

$v_1=\left(\frac{m-2m}{m+2m}\right)v+0$

$v_1=-\frac{1}{3}v$

The negative sign indicates that the colliding body bounces back in the opposite direction with one-third of its original speed.

Step 3: Calculate the initial and final kinetic energy of the colliding body
The initial kinetic energy ($K_i$) of the body of mass $m$ is:

$K_i=\frac{1}{2}m{v}^2$

The final kinetic energy ($K_f$) of the body of mass $m$ after the collision is:

$K_f=\frac{1}{2}m{v_1}^2=\frac{1}{2}m{\left(-\frac{1}{3}v\right)}^2$

$K_f=\frac{1}{9}\left(\frac{1}{2}m{v}^2\right)=\frac{1}{9}{K}_i$

Step 4: Determine the loss of kinetic energy of the colliding body
The loss of kinetic energy ($\mathrm{\Delta K}$) is the difference between its initial and final kinetic energies:

$\mathrm{\Delta K}=K_i-K_f$

$\mathrm{\Delta K}=K_i-\frac{1}{9}{K}_i$

$\mathrm{\Delta K}=\frac{8}{9}{K}_i$

Thus, the loss of kinetic energy of the colliding body is indeed 8/9 of its initial kinetic energy.