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. $m \frac{v^{2}_{0}}{x^{2}_{0}}$
• 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

The Correct Option is D

Initial momentum of the system block $(C) = mv_0$. 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 $mv_{0} = \left(m+2m\right)v \Rightarrow v = \frac{v_{0}}{3}$ By the law of conservation of energy K.E. of block C = K.E. of system + P.E. of system $\frac{1}{2}mv^{2}_{0} = \frac{1}{2}\left(3m\right)\left(\frac{v_{0}}{3}\right)^{2}+\frac{1}{2}kx^{2}_{0}$ $\Rightarrow\quad \frac{1}{2}mv^{2}_{0} = \frac{1}{6} mv^{2}_{0} +\frac{1}{2}kx^{2}_{0}$ $\Rightarrow\quad \frac{1}{2}kx^{2}_{0} = \frac{1}{2}mv^{2}_{0} - \frac{1}{6}mv^{2}_{0} = \frac{mv^{2}_{0}}{3}$ $\therefore\quad k = \frac{2}{3}m\left(\frac{v_{0}}{x_{0}}\right)^{2}$