Question

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

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

Answer 1

The correct option is D $\sqrt{2}{v}_0$

By conservation of linear momentum
${mv}_0+0={mv}_1+{mv}_2$
$v_0=v_1+v_2..........\left(1\right)$
$\frac{3}{2}\left[\frac{1}{2}{mv}_0^2\right]=\frac{1}{2}{mv}_1^2+\frac{1}{2}{mv}_2^2$
$\frac{3}{2}{v}_0^2=v_1^2+v_2^2\cdots \cdots \cdots \cdot \left(2\right)$

Solving equation (1) and (2)

$v_1=\frac{v_0}{2}(1+\sqrt{2})$
$v_2=\frac{v_0}{2}(1-\sqrt{2})$
${\overline{v}}_{rel}={\overline{v}}_1-{\overline{v}}_2$
$\frac{v_0}{2}[1+\sqrt{2}-1+\sqrt{2}]$
$=\frac{v_0}{2}\times 2\sqrt{2}$
$=\sqrt{2}{v}_0$