Question

A plank of mass $5\text{kg}$ is placed on a frictionless horizontal plane. Further a block of mass $1\text{kg}$ is placed over the plank. A massless spring of natural length $2\text{m}$ is fixed to the plank by its one end. The other end of spring is compressed by the block by half of spring's natural length. The system is now released from the rest, velocity of the plank when block leaves the plank is

[Assume, there is no friction between the two blocks]

• A. $\sqrt{\frac{5}{3}}\text{m/s}$
• B. $\sqrt{\frac{7}{3}}\text{m/s}$
• C. $\sqrt{\frac{8}{3}}\text{m/s}$
• D. $\sqrt{\frac{10}{3}}\text{m/s}$

Answer 1

The correct option is D $\sqrt{\frac{10}{3}}\text{m/s}$

Given,
mass of block, $m_1=1\text{kg}$
mass of plank, $m_2=5\text{kg}$
spring constant, $k=100\text{N/m}$
compression in spring, $x=1\text{m}$.

Let us suppose block moves with speed $v_1$ and plank with speed $v_2$ when block leaves the plank.

On applying law of conservation of mechanical energy before and after block leaves the plank,

$\text{P.E of spring}=\text{K.E of block}+\text{K.E of plank}$

$\frac{1}{2}k{x}^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$

$\Rightarrow 100\times 1^2=1\times v_1^2+5\times v_2^2$

$\Rightarrow v_1^2=100-5v_2^2...\left(1\right)$

Also, velocity of centre of mass will be zero all the times as initially both block and plank were at rest and there is no external forces acting on the system.
Applying the conservation of momentum principle,

$\therefore \frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}=0$

$\Rightarrow m_1{v}_1+m_2{v}_2=0$

$\Rightarrow v_1=-\frac{m_2}{m_1}{v}_2$

On squaring both sides,

$\Rightarrow v_1^2={(-\frac{m_2}{m_1})}^2{v}_2^2...\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we have

$\Rightarrow 100-5v_2^2={(-\frac{5}{1})}^2{v}_2^2$

$\Rightarrow 100-5v_2^2=25v_2^2$

$\Rightarrow 30v_2^2=100$

$\therefore v_2=\sqrt{\frac{10}{3}}\text{m/s}$

Hence, option $\left(d\right)$ is correct answer.