Question

A spring is compressed between two blocks of masses $m_1$ and $m_2$ placed on a horizontal frictionless surface as shown in the figure. When the blocks are released, they have initial velocity of $v_1$ and $v_2$ as shown. The blocks travel distances $x_1$ and $x_2$ respectively before coming to rest.
The ratio $\left(\frac{x_1}{x_2}\right)$ is

• A. $\frac{m_2}{m_1}$
• B. $\frac{m_1}{m_2}$
• C. $\sqrt{\frac{m_2}{m_1}}$
• D. $\sqrt{\frac{m_1}{m_2}}$

Answer 1

The correct option is A $\frac{m_2}{m_1}$

Initial momentum of the system is zero, i.e.

$P_i=0$

Given, the velocity acquire by masses $m_1$ and $m_2$ just after they are released be $v_1$ and $v_2$.

Final momentum of the system,

$P_f=m_2{v}_2-m_1{v}_1$

Using conservation of momentum,

$P_i=P_f$

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

$\Rightarrow m_2{x}_2-m_1{x}_1=0[\because v=\frac{dx}{dt}]$

$\Rightarrow \frac{x_1}{x_2}=\frac{m_2}{m_1}$

Hence, $\left(A\right)$ is the correct answer.