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

The Correct Option is B

It is a case of superelastic collision
$m v_{0}=m v_{1}+m v_{2} \,\,\,\, \ldots .(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}+2 v_{1} v_{2} $
$\Rightarrow v_{0}^{2}=\frac{3 v_{0}^{2}}{2}+2 v_{1} v_{2} $
$\Rightarrow 2 v_{1} v_{2}=-\frac{v_{0}^{2}}{2}\,\,\,\,\,\,\,\,...(iii)$
$\therefore \left(v_{1}-v_{2}\right)^{2}=\left(v_{1}+v_{2}\right)^{2}-4 v_{1} v_{2}=v_{0}^{2}+v_{0}^{2}$
$\Rightarrow v_{1}-v_{2}=\sqrt{2} v_{0}$