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Work Energy Theorem Application

Mass m 1 hi...

Question

Mass $m_{1}$ hits and sticks with $m_{2}$ while sliding horizontally with velocity $v$ along the common line of centres of the three equal masses ( $m_{1} = m_{2} = m_{3} = m$ ). Initially masses $m_{2}$ and $m_{3}$ are stationary and the spring is unstretched. Find the velocities of $m_{1}, m_{2}, m_{3}$ immediately after impact :

A

V, V / 2, 0

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B

V / 2, V / 2, 0

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C

3 V / 2, V / 2, 0

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D

0, V / 2, V

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Solution

The correct option is B $V / 2, V / 2, 0$
Before the impact the total momentum is because of the block $m_{1}$ is : $M = m_{1} V$
After the impact $m_{1}$ will stick to the $m_{2}$ so overall momentum will be : $M_{2} = (m_{1} + m_{2}) V_{1}$
Now by principle of conservation of momentum we can say that :
$m_{1} V = (m_{1} + m_{2}) V_{1}$

$⟹ \frac{m_{1}}{(m_{1} + m_{2})} V = V_{1}$

Because $m_{1} = m_{2} = m_{3} = m$
After solving this we get :

$⟹ V_{1} = \frac{V}{2}$

For $m_{3}$ at the time of impact there is no force because of spring , so $m_{3}$ will remain stationary.

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