Question

Two bodies $A$ and $B$ of masses $m$ and $2m$ respectively placed on a smooth floor are connected by a spring. A third body $C$ of mass $m$ moves with velocity $v_0$ along the line joining $A$ and $B$ and collides elastically with $A$. At a certain instant of time $t_0$ after collision it is found that the instantaneous velocities of $A$ and $B$ are same. The compression in the spring at $t_0$ is $x_0$, then:

• A. The common velocity of $A$ and $B$ at time $t_0$ is $\frac{v_0}{3}$.
• B. The spring constant is $k=\frac{3m{v}_0^2}{2x_0^2}$.
• C. The spring constant is $k=\frac{2m{v}_0^2}{3x_0^2}$ .
• D. None

Answer 1

Answer: (A), (C)

The correct options are
A The common velocity of $A$ and $B$ at time $t_0$ is $\frac{v_0}{3}$.
C The spring constant is $k=\frac{2m{v}_0^2}{3x_0^2}$ .

Applying momentum conservation law,
$p_i=p_f$
$3m{v}^{{}^{\prime }}=m{v}_0$
$\therefore v^{{}^{\prime }}=\frac{v_0}{3}$
Now,
$\frac{1}{2}m{v}_0^2=\frac{1}{2}m{\left(\frac{v_0}{3}\right)}^2+\frac{1}{2}\left(2m\right){\left(\frac{v_0}{3}\right)}^2+\frac{1}{2}k{x}_0^2$
$=\frac{1}{2}\frac{m{v}_0^2}{9}+\frac{1}{2}2m\frac{v_0^2}{9}+\frac{1}{2}k{x}_0^2$
$\Rightarrow m{v}_0^2=\frac{m{v}_0^2}{3}+k{x}_0^2$
$\therefore k{x}_0^2=\frac{2}{3}m{v}_0^2$ $\therefore k=\frac{2m{v}_0^2}{3x_0^2}$