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

Answer: (A)

Initial momentum of the system is zero i.e. $P _{ i }=0$ Let 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 $P _{ f }= m _{1} v _{1}- m _{2} v _{2}$ Using conservation of momentum :
$P _{ i }= P _{ f }$ $ \therefore 0= m _{1} v _{1}- m _{2} v _{2} $ $ \begin{array}{l} \Longrightarrow m _{1} v _{1}= m _{2} v _{2} \quad\left( v =\frac{ dx }{ dt }\right) \\ \Longrightarrow m _{1} x _{1}= m _{2} x _{2} \\ \Longrightarrow \frac{ x _{1}}{ x _{2}}=\frac{ m _{2}}{ m _{1}} \end{array} $