Question 127 of 130

Two bodies $A$ and $B$ of mass $m$ and $2 m$ respectively are placed on a smooth floor. They are connected by a spring of negligible mass. $A$ third body $C$ of mass $m$ is placed on the floor. The body $C$ moves with a velocity $v_0$ along the line joining $A$ and $B$ and collides elastically with $A$. At a certain time after the collision it is found that the instantaneous velocities of $A$ and $B$ are same and the compression of the spring is $x_0$. The spring constant $k$ will be

A$m \frac{v_0^2}{x_0^2}$B$m \frac{v_0}{2 x_0}$C$2 m \frac{v_0}{x_0}$D$\frac{2}{3} m\left(\frac{v_0}{x_0}\right)^2$

Show answer and solution▾

Answer: (D) $\frac{2}{3} m\left(\frac{v_0}{x_0}\right)^2$

Initial momentum of the system block $(C)$ $=m v_0$. After striking with $A$, the block $C$ comes to rest and now both block A and B moves with velocity $\mathrm{v}$ when compression in spring is $x_0$.
By the law of conservation of linear momentum
$$m v_0=(m+2 m) v \Rightarrow v=\frac{v_0}{3}$$
By the law of conservation of energy K.E. of block $C=$ K.E. of system + P.E. of system
$$\begin{aligned} & \frac{1}{2} m v_0^2=\frac{1}{2}(3 m)\left(\frac{v_0}{3}\right)^2+\frac{1}{2} k x_0^2 \\ & \Rightarrow \quad \frac{1}{2} m v_0^2=\frac{1}{6} m v_0^2+\frac{1}{2} k x_0^2 \\ & \Rightarrow \quad \frac{1}{2} k x_0^2=\frac{1}{2} m v_0^2-\frac{1}{6} m v_0^2=\frac{m v_0^2}{3} \\ & \end{aligned}$$

$$\begin{aligned} & \Rightarrow \quad \frac{1}{2} k x_0^2=\frac{1}{2} m v_0^2-\frac{1}{6} m v_0^2=\frac{m v_0^2}{3} \\ & \therefore \quad k=\frac{2}{3} m\left(\frac{v_0}{x_0}\right)^2 \end{aligned}$$