Question

A block with mass $M$ is connected by a massless spring with stiffness constant $k$ to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude $A$ about an equilibrium position $x_0$. Consider two cases: (i) when the block is at $x_0$; and (ii) when the block is at $x=x_0+A$. In both the cases, a particle with mass $m(<M)$ is softly placed on the block after which they stick to each other. Which of the following statement(s) is(are) true about the motion after the mass $m$ is placed on the mass $M$?

• A. The amplitude of oscillation in the first case changes by a factor of $\sqrt{\frac{M}{m+M}}$, whereas in the second case it remains unchanged
• B. The final time period of oscillation in both the cases is same
• C. The total energy decreases in both the cases
• D. The instantaneous speed at $x_0$ of the combined masses decreases in both the cases

Answer 1

Answer: (A), (B), (D)

The instantaneous speed at $x_0$ of the combined masses decreases in both the cases

In case I,
When the particle of mass $m$ is placed on top of mass $M$ then there is no change in mean position, but amplitude will change because of inelastic collision and also linear momentum will conserved.
Applying momentum conservation

$M{V}_1=(M+m)V_2$

$MA\sqrt{\frac{k}{M+m}}=(M+m)A^{\prime }\sqrt{\frac{k}{M+m}}$

$A^{\prime }=A\sqrt{\frac{M}{M+m}}$.

Hence option A is correct.

In case II,
The amplitude will be same as the block is kept at extreme position and its velocity is zero
$A=A^{\prime }$.

The time period in both the cases is same as
$T=2\pi \sqrt{\frac{M+m}{k}}$ in both the cases, hence option B is correct.

Total energy $\left(\frac{1}{2}K{A}^2\right)$ decreases in first case as the amplitude decreases from $A$ to $A^{\prime }$
where as remain same in 2nd case as the amplitude remains same, so option C is incorrect

Instantaneous speed at $x_0$ decreases in both case, as $M{V}_1=(M+m)V_2$ and here we can see that $V_2$ is less than $V_1$ and for the case II, if we compare initial and final total energy $\frac{1}{2}K{A}^2=\frac{1}{2}\left(M+m\right)V_{max}^2$, here in this the energy is same but the mass is added and hence the velocity decreases, so the option D is correct.

$\therefore$ Answer is A, B and D.