Question

Initially, the spring is at its natural length and collision between blocks of mass $m\&m$ is elastic. The horizontal surface is smooth. Then, maximum compression of the spring during the motion will be:

• A. $\sqrt{\frac{m{v}^2}{k}}$
• B. $\sqrt{\frac{m{v}^2}{3k}}$
• C. $\sqrt{\frac{2m{v}^2}{k}}$
• D. $\sqrt{\frac{2m{v}^2}{3k}}$

Answer 1

The correct option is D $\sqrt{\frac{2m{v}^2}{3k}}$

Due to elastic collision between equal mass blocks $m\&m$, they will exchange velocity, hence $m$ moving initially with $v$ will come to rest immediately after collision and other block of mass $m$ will move with velocity $v$.
At maximum compression $\left(x_m\right)$ in spring, both blocks will be moving with same velocity, $v_0$.


Applying $P.C.L.M$ between initial and final states:
$P_i=P_f$
$\Rightarrow mv+0+0=m\left(0\right)+m{v}_0+2m\times v_0$
or $v_0=\frac{mv}{3m}=\frac{v}{3}...\left(i\right)$

Applying conservation of mechanical energy,
$K{E}_i+P{E}_i=K{E}_f+P{E}_f$
$\frac{1}{2}m{v}^2+0=\frac{1}{2}m{v}_0^2+\frac{1}{2}\times 2m\times v_0^2+\frac{1}{2}k{x}_m^2$
Substituting value of $v_0$ from Eq $\left(i\right)$,
$\Rightarrow m{v}^2=m\times {\left(\frac{v}{3}\right)}^2+2m\times {\left(\frac{v}{3}\right)}^2+k{x}_m^2$
$\Rightarrow k{x}_m^2=m{v}^2-\frac{3m{v}^2}{9}$
$\Rightarrow k{x}_m^2=\frac{2m{v}^2}{3}$
$\therefore x_m=\sqrt{\frac{2m{v}^2}{3k}}$
Therefore, maximum compression of spring during motion will be $\sqrt{\frac{2m{v}^2}{3k}}$.