Question

In a collinear collision, a particle with an initial speed $v_0$ strikes a stationary particle of the same mass. If the final total kinetic energy is $50\%$ grater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is -

• A. $\frac{v_0}{4}$
• B. $\sqrt{2}{v}_0$
• C. $\frac{v_0}{2}$
• D. $\frac{v_0}{\sqrt{2}}$

Answer 1

Answer: (B)

It is a case of superelastic collision
$m{v}_0=m{v}_1+m{v}_2\dots .(i)$
$\Rightarrow v_1+v_2=v_0$
$\frac{1}{2}m\left(v_1^2+v_2^2\right)=\frac{3}{2}\left(\frac{1}{2}m{v}_0^2\right)$
$\Rightarrow \left(v_1^2+v_2^2\right)=\frac{3}{2}{v}_0^2...(ii)$
$\Rightarrow {\left(v_1+v_2\right)}^2=v_1^2+v_2^2+2v_1{v}_2$
$\Rightarrow v_0^2=\frac{3v_0^2}{2}+2v_1{v}_2$
$\Rightarrow 2v_1{v}_2=-\frac{v_0^2}{2}...(iii)$
$\therefore {\left(v_1-v_2\right)}^2={\left(v_1+v_2\right)}^2-4v_1{v}_2=v_0^2+v_0^2$
$\Rightarrow v_1-v_2=\sqrt{2}{v}_0$