Question

A body of mass m moving with velocity v makes a head on elastic collision with another body of mass 2m which is initially at rest. The loss of kinetic energy of the colliding body (mass m ) is(a) $\frac{1}{2}$ of its initial kinetic energy(b) $\frac{1}{9}$ of its initial kinetic energy(c) $\frac{8}{9}$ of its initial kinetic energy(d) $\frac{1}{4}$ of its initial kinetic energy

Answer 1

let suppose v is the velocity of mass m & v1 is the final velocity of mass 2m &m

mv = 3mv1

v1=v/3 loss of energy = (1/2mv^2 - 1/2mv/2^2)/1/2mv^2

= 8/9

Answer 2

Answer:

(c).The loss of kinetic energy of the colliding  first body is $\dfrac{8}{9}$ of its initial kinetic energy.

Explanation:

Given that,

A body of mass m moving with velocity v makes a head on elastic collision with another body of mass 2 m which is initially at rest.

Mass of first body = m

Velocity of first body = v

Mass of second body = 2 m

Velocity of second body = 0

Using law of conservation of energy

$\dfrac{1}{2}m_{1}u_{1}^2+\dfrac{1}{2}m_{2}u_{2}^2=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2$

For perfectly elastic collision

The final velocity of first body is

$v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}u_{2}}{m_{1}+m_{2}})$

Here, $u_{2}=0$

Because second body is on rest.

Therefore,

$v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}$

$v_{1}=(\dfrac{m-2m}{m+2m})u_{1}$

$v_{1}=\dfrac{-u_{1}}{3}\ m/s$

The loss of kinetic energy = initial kinetic energy -final kinetic energy

$\Delta K.E=\dfrac{1}{2}mu_{1}^2-\dfrac{1}{2}mv_{1}^2$

The ratio of loss of kinetic energy and initial kinetic energy is

$\dfrac{\Delta K.E}{K.E_{i}}=\dfrac{\dfrac{1}{2}mu_{1}^2-\dfrac{1}{2}mv_{1}^2}{\dfrac{1}{2}mu_{1}^{2}}$

$\dfrac{\Delta K.E}{K.E_{i}}=1-\dfrac{v_{1}^2}{u_{1}^2}$

$\dfrac{\Delta K.E}{K.E_{i}}=1-\dfrac{(\dfrac{-u_{1}}{3})^2}{u_{1}^2}$

$\dfrac{\Delta K.E}{K.E_{i}}=1-\dfrac{1}{9}$

$\dfrac{\Delta K.E}{K.E_{i}}=\dfrac{8}{9}$

$\Delta K.E=\dfrac{8}{9}\times K.E_{i}$

Hence, The loss of kinetic energy of the colliding  first body is $\dfrac{8}{9}$ of its initial kinetic energy.