Question

Two identical blocks $A$ and $B$ each of mass $m$ resting on the smooth horizontal floor are connected by a light spring of natural length $L$ and spring constant $k$. A third block $C$ of mass $m$ moving with a speed $v$ along the line joining $A$ and $B$ collides elastically with $A$. The maximum compression in the spring is:

• A. $\sqrt{\frac{mv}{2k}}$
• B. $\sqrt{\frac{m}{2k}}$
• C. $\sqrt{\frac{mv}{k}}$
• D. $v\sqrt{\frac{m}{2k}}$

Answer 1

The correct option is D

$v\sqrt{\frac{m}{2k}}$

Step 1: Given data

1. Two blocks $A$ and $B$ each of mass $m$ resting on the smooth horizontal floor connected by a light spring of natural length $L$ and spring constant $k$.
2. A third block $C$ of mass $m$ moving with a speed $v$ along the line joining $A$ and $B$ collides elastically with $A$.

Step 2: To find

The maximum compression of the spring

Step 3: Formula and calculation

In this case the spring will compress until $v_A>v_B$. Therefore, maximum compression is when $v_A=v_B$

Since, here we have to find the maximum compression, let $v_A=v_B=v_1$ be the velocity of blocks $A$ and $B$ at the maximum compression of the spring

From the law of conservation of momentum,

total momentum before collision = total momentum after collision.

C moves with a velocity $v$ and hits the block A and then A starts to move with a velocity $v_A$ and since block A is connected to block B, B starts to move with a velocity $v_B$. After collision C comes to rest.

$$\begin{gathered} mv=m{v}_A+m{v}_B \\ \Rightarrow mv=m{v}_1+m{v}_1 \\ \Rightarrow v=2v_1 \end{gathered}$$

From the law of conservation of energy,

Kinetic energy of C = sum of kinetic energies of the blocks A and B + the energy of the spring connecting A and B.

$$\begin{gathered} \frac{1}{2}{m}_C{v_C}^2=\frac{1}{2}{m}_A{v}_A^2+\frac{1}{2}{m}_B{v}_B^2+\frac{1}{2}k{x}^2(v_A=v_B=v_1,m_A=m_B=m_C=m) \\ \Rightarrow \frac{1}{2}m{v}^2=\left(\frac{1}{2}\left(m+m\right)v_1^2\times 2\right)+\frac{1}{2}k{x}^2 \\ \Rightarrow \frac{1}{2}m{v}^2=\left(\frac{1}{2}\left(2m\right){\left(\frac{v}{2}\right)}^2\right)+\frac{1}{2}k{x}^2 \\ \Rightarrow \frac{1}{2}m{v}^2=\frac{1}{4}m{v}^2+\frac{1}{2}k{x}^2 \\ \Rightarrow \frac{1}{4}m{v}^2=\frac{1}{2}k{x}^2 \\ \Rightarrow m{v}^2=2k{x}^2 \\ \Rightarrow x=\sqrt{\frac{m{v}^2}{2k}} \\ \Rightarrow x=v\sqrt{\frac{m}{2k}} \end{gathered}$$

The maximum compression of the spring, $x=v\sqrt{\frac{m}{2k}}$

Hence, option D is correct.