Question

Two blocks of masses $m$ and $M$ are moving with speeds $v_1$ and $v_2(v_1>v_2)$ in the same direction on the frictionless surface respectively, $M$being ahead of $m$. An ideal spring of force constant $k$ is attached to the backside of $M$ (as shown). The maximum compression of the spring when the blocks collide is

• A. $v_2\sqrt{\frac{M}{k}}$
• B. None of the above
• C. $(v_1-v_2)\sqrt{\frac{mM}{k(M+m)}}$
• D. $v_1\sqrt{\frac{m}{k}}$

Answer 1

The correct option is C $(v_1-v_2)\sqrt{\frac{mM}{k(M+m)}}$

As per the given diagram,
At maximum compression reduced mass of system
$=\left(\frac{mM}{M+m}\right)$
Initial relative velocity of approach
$=(v_1-v_2)$
Hence, by conservation of energy,
Potential Energy of the spring $=$ Kinetic Energy of the blocks
$\frac{1}{2}k{x}_{max}^2=\frac{1}{2}\left(\frac{mM}{m+M}\right)(v_1-v_2{)}^2$
Therefore,
$x_{max}=\sqrt{\frac{mM}{k(M+m)}}(v_1-v_2{)}^2$

Final Answer: (c)