Question

A block of mass $m$ moving with a velocity ${\nu }_0$ collides with a stationary block of mass $M$ to which a spring of stiffness $k$ is attached, as shown in Fig. Choose the correct alternative.

• A. The velocity of the centre of mass is ${\nu }_0$
• B. The initial kinetic energy of the system in the centre of mass frame is $\frac{1}{4}\left(\frac{mM}{M+m}\right){\nu }_0^2$.
• C. The maximum compression in the spring is ${\nu }_0\sqrt{\left(\frac{mM}{m+M}\frac{1}{k}\right)}$
• D. When the spring is in the state of maximum compression, the kinetic energy in the centre of mass frame is zero.

Answer 1

The correct option is D When the spring is in the state of maximum compression, the kinetic energy in the centre of mass frame is zero.

A block of mass m moving with velocity $v_0$ collide with a stationary

block of mass $M$ to which a spring of stiffness $k$ is attached.

From the given diagram in the question,

The velocity of the center of mass is,

$v_{cm}=\frac{m{v}_0}{m+M}$. . . . .(1)

In the reference of center of mass,

Kinetic energy is $K=\frac{1}{2}m(v_0-v_{cm}{)}^2$

Center of mass is moving so the kinetic energy is now

$K^{\prime }=\frac{1}{2}M{v}_{cm}^2$

The kinetic energy of the system is,

$K.E=K+K^{\prime }$

$K.E=\frac{1}{2}m(v_0-v_{cm}{)}^2+\frac{1}{2}M{v}_{cm}^2$

substitue the value of $v_{cm}$ in the above equation and we get,

$K.E=\frac{1}{2}\frac{mM}{m+M}{v}_0^2$

when the spring is in the state of maximum compression, kinetic energy is equal to spring potential energy,

$K.E=\frac{1}{2}k{x}^2$ where, $k=$ spring constant

$\frac{1}{2}\frac{mM}{m+M}{v}_0^2=\frac{1}{2}k{x}^2$