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Momentum

Two masses m1...

Question

Two masses $m_{1}$ and $m_{2}$ are connected by a massless spring of constant $k$ and unstretched length $L$ . The masses are placed on a frictionless straight channel, which is considered as x -axis. They are initially at rest at $x = 0$ and $x = L$ respectively. At $t = 0$ , a velocity of $v_{0}$ is suddenly imparted to the first particle. At a later time $t_{0}$ , the centre of mass of the two masses is at $x =$

A

\frac{m_{2} L}{m_{1} + m_{2}}

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B

\frac{m_{1} L}{m_{1} + m_{2}} + \frac{m_{2} v_{0} t}{m_{1} + m_{2}}

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C

\frac{m_{2} L}{m_{1} + m_{2}} + \frac{m_{2} v_{0} t}{m_{1} + m_{2}}

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D

\frac{m_{2} L}{m_{1} + m_{2}} + \frac{m_{1} v_{0} t}{m_{1} + m_{2}}

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Solution

The correct option is D $\frac{m_{2} L}{m_{1} + m_{2}} + \frac{m_{1} v_{0} t}{m_{1} + m_{2}}$

Velocity of COM
$V_{c o m} = \frac{m_{1} v_{0} + 0}{m_{1} + m_{2}}$

Initial position of COM
$X_{i - c o m} = \frac{m_{1} (0) + m_{2} L}{m_{1} + m_{2}};$

Final position of COM
$X_{f - c o m} = X_{i - c o m} + V_{c o m} t = \frac{m_{2} L}{m_{1} + m_{2}} + \frac{m_{1} v_{0} t}{m_{1} + m_{2}}$

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