Question

Two bodies A and B of mases m and 2m respectively are placed on a smooth floor. They 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 as shown in figure. At a certain instant of time $t_0$ after collision, it is found that the instantaneous velocities of A and B are the same. Further at this instant the compression of the spring is found to be $x_0$. Determine (a) the common velocity of A and B at time $t_0$ and (b) the spring constant.

• A. $v=\frac{v_0}{3},k=\frac{2}{3}\frac{m{v}_0^2}{x_0^2}$
• B. $v=\frac{v_0}{3},k=\frac{3}{4}\frac{m{v}_0^2}{x_0^2}$
• C. $v=\frac{v_0}{2},k=\frac{2}{3}\frac{m{v}_0^2}{x_0^2}$
• D. $v=\frac{v_0}{2},k=\frac{3}{4}\frac{m{v}_0^2}{x_0^2}$

Answer 1

The correct option is A

$v=\frac{v_0}{3},k=\frac{2}{3}\frac{m{v}_0^2}{x_0^2}$

Upon collision, C transfers its entire momentum ($m{v}_0$) and kinetic energy ($\frac{1}{2}m{v}_0^2$) to block C and comes to rest. Let the common velocity of A and B at $x_0$ compression be $v$.
Applying conservation of momentum, $m{v}_0=mv+2mv\Rightarrow v=\frac{v_0}{3}$
From energy conservation, $\frac{1}{2}m{v}_0^2=\frac{1}{2}m{v}^2+\frac{1}{2}\left(2m\right)v^2+\frac{1}{2}k{x}_0^2$
Solving we get, $k=\frac{2}{3}\frac{m{v}_0^2}{x_0^2}$