A body of mass 2 m moving with velocity v makes a head-on elastic collision with another body of mass m…

1. $\frac{1}{9}$ of its initial kinetic energy
2. $\frac{1}{6}$ of its initial kinetic energy
3. $\frac{1}{2}$ of its initial kinetic energy
4. $\frac{8}{9}$ of its initial kinetic energy

Solution

Initial $K.E$ of ball of mass $2m= K_1$

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

$=m{v}^2$

Collision is elastic so both $K.E$ and momentum are conserved. Let velocities of balls are $v_1$ and $v_2$ after collision.



So, $KE$ is conserved

$\frac{1}{2}\left(2m\right)v^2=\frac{1}{2}\left(2m\right)v_1^2+\frac{1}{2}m{v}_2^2$

$\Rightarrow v^2=v_1^2+\frac{1}{2}{v}_2^2$ 

$2\left(v^2-v_1^2\right)=v_2^2\mathrm{....}\left(i\right)$

And, momentum is conserved

$\left(2m\right)v+m\left(0\right)=2m\left(v_1\right)+m{v}_2$

$\Rightarrow 2v=2v_1+v_2$ ... (ii)

Now,

$v_2=2\left(v-v_1\right)$

From (i)

$\frac{2\left(v^2-v_1^2\right)}{2\left(v-v_1\right)}=\frac{v_2^2}{v_2}$

$v+v_1=v_2$

$v-v_1=\frac{v_2}{2}$

$2v_1=\frac{v_2}{2}$

$v_1=\frac{v_2}{4}=\frac{v}{3}$

So, final $K.E$ of ball of mass $2m$ ,

$k_2=\frac{1}{2}\left(2m\right)\left(v_1^2\right)=\frac{1}{2}\times 2m\times \frac{v^2}{9}=\frac{1}{9}\left(k_1\right)$

Hence, loss of $K.E.$ of $I^{st}$ ball

$=K_1-\frac{1}{9}{K}_1=\frac{8}{9}{K}_1$

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