Question

Two bodies $A$ and $B$ of mass $m$ and $2m$ 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. $m\frac{v_0}{2x_0}$
• C. $2m\frac{v_0}{x_0}$
• D. $\frac{2}{3}m{\left(\frac{v_0}{x_0}\right)}^2$

Answer 1

Answer: (D)

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 v when compression in spring is $x_0$.
By the law of conservation of linear momentum
$m{v}_0=\left(m+2m\right)v\Rightarrow v=\frac{v_0}{3}$
By the law of conservation of energy K.E. of block C = K.E. of system + P.E. of system
$\frac{1}{2}m{v}_0^2=\frac{1}{2}\left(3m\right){\left(\frac{v_0}{3}\right)}^2+\frac{1}{2}k{x}_0^2$
$\Rightarrow \frac{1}{2}m{v}_0^2=\frac{1}{6}m{v}_0^2+\frac{1}{2}k{x}_0^2$
$\Rightarrow \frac{1}{2}k{x}_0^2=\frac{1}{2}m{v}_0^2-\frac{1}{6}m{v}_0^2=\frac{m{v}_0^2}{3}$
$\therefore k=\frac{2}{3}m{\left(\frac{v_0}{x_0}\right)}^2$