Question

A ball with initial velocity $1\text{m/s}$ strikes a stationary ball of the same mass. If the final total kinetic energy is $50\%$ greater than the original kinetic energy, the magnitude of the relative velocity between the two balls after collision (assuming collinear collision), is

• A. $\frac{1}{2}\text{m/s}$
• B. $\sqrt{2}\text{m/s}$
• C. $\frac{1}{\sqrt{2}}\text{m/s}$
• D. $\frac{1}{4}\text{m/s}$

Answer 1

The correct option is B $\sqrt{2}\text{m/s}$

Let $m$ be the mass of each ball.



Applying momentum conservation,

$P_{1i}+P_{2i}=P_{1f}+P_{2f}$

$m{v}_0+(m\times 0)=m{v}_1+m{v}_2$

$\Rightarrow v_0=v_1+v_2............\left(1\right)$
Let the original kinetic energy be $K_0$

From the data given in the question,
$K_{1f}+K_{2f}=1.5K_0$

Applying Kinetic energy conservation,

$K_{1i}+K_{2i}=1.5K_0$

$\Rightarrow \frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2=\frac{3}{2}\left(\frac{1}{2}m{v}_0^2\right)$

$\Rightarrow \frac{v_1^2}{2}+\frac{v_2^2}{2}=\frac{3}{4}{v}_0^2$

$\Rightarrow v_1^2+v_2^2=\frac{3}{2}{v}_0^2........\left(2\right)$

From $\left(1\right)$, by squaring on both sides, we get

$v_0^2=(v_1+v_2{)}^2$

$\Rightarrow v_1^2+v_2^2+2v_1{v}_2=v_0^2$

$\Rightarrow 2v_1{v}_2=v_0^2-(v_1^2+v_2^2)$

Substituting $\left(2\right)$ in the above equation,
$2v_1{v}_2=v_0^2-\frac{3}{2}{v}_0^2$

$\Rightarrow 2v_1{v}_2=-\frac{1}{2}{v}_0^2$

So, $(v_1-v_2{)}^2=v_1^2+v_2^2-2v_1{v}_2$

$=\frac{3}{2}{v}_0^2-(-\frac{1}{2}{v}_0^2)=2v_0^2$

$\Rightarrow v_1-v_2=\sqrt{2v_0^2}=\sqrt{2}{v}_0=\sqrt{2}\times 1=\sqrt{2}\text{m/s}$
$[\because v_0=1\text{m/s}]$

Hence, option $\left(b\right)$ is the correct answer.