Question

Two point masses connected by a light inextensible string are lying on a frictionless surface as shown in figure. An impulse of magnitude $10\text{kg m/s}$ is given to the $5\text{kg}$ block. Then:

• A. Velocity of $10\text{kg}$ mass immediately after impulse is given is $\frac{1}{3}\text{m/s}$.
• B. Velocity of $10\text{kg}$ mass immediately after impulse is given is $2\text{m/s}$.
• C. Velocity of $5\text{kg}$ mass immediately after impulse is given is $\sqrt{\frac{28}{9}}\text{m/s}$.
• D. Velocity of $5\text{kg}$ mass immediately after impulse is given is $\frac{\sqrt{28}}{9}\text{m/s}$.

Answer 1

Answer: (A), (C)

The correct options are
A Velocity of $10\text{kg}$ mass immediately after impulse is given is $\frac{1}{3}\text{m/s}$.
C Velocity of $5\text{kg}$ mass immediately after impulse is given is $\sqrt{\frac{28}{9}}\text{m/s}$.

Since the string is inextensible, velocities of both bodies along the string after impact should be the same.


Let's assume the x-axis along the string. After impact, the velocities of the two masses will be as shown in the diagram. An impulse of $J$ will be applied by the string on both masses.


Along horizontal for $10\text{kg}$ block,
$J=\Delta P=10V_x-0\dots \left(i\right)$
For $5\text{kg}$ block along horizontal
$5-J=5V_x\dots \left(ii\right)$
Using (i) and (ii),
$V_x=\frac{1}{3}{\text{ms}}^{-1}$
For $5\text{kg}$ block along vertical
$5\sqrt{3}=5V_y-0$
$V_y=\sqrt{3}{\text{ms}}^{-1}$
Final velocity of $10\text{kg},\overrightarrow{V_{10}}=\frac{1}{3}\hat{i}{\text{ms}}^{-1}\dots \left(i\right)$
Final velocity of $5\text{kg},\overrightarrow{V_5}=\frac{1}{3}\hat{i}+\sqrt{3}\hat{j}{\text{ms}}^{-1}$
or $|\overrightarrow{V_5}|=\sqrt{3+\frac{1}{9}}=\sqrt{\frac{28}{9}}{\text{ms}}^{-1}$
Hence, options (a) and (c) are correct.