Question

Two bodies $A$ and $B$ of masses $m$ 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.

Answer 1

(a) Collision between $A$ and $C$ is elastic and mass of both the blocks is same. Therefore, they will exchange their velocities i.e., $C$ will come to rest and $A$ will be moving will velocity $v_0$. Let $v$ be the common velocity of $A$ and $B$, then from conservation of linear momentum, we have

$m_A{v}_0=\left(m_A+m_B\right)v$
or $m{v}_0=(m+2m)v$
or $v=\frac{v_0}{3}$
(b) From conservation of energy, we have
$\frac{1}{2}{m}_A{v}_0^2=\frac{1}{2}\left(m_A+m_B\right)v^2+\frac{1}{2}k{x}_0^2$
or $\frac{1}{2}m{v}_0^2=\frac{1}{2}(3m){\left(\frac{v_0}{3}\right)}^2+\frac{1}{2}k{x}_0^2$
or $\frac{1}{2}k{x}_0^2=\frac{1}{3}m{v}_0^2$ or $k=\frac{2m{v}_0^2}{3x_0^2}$