Momentum and CollisionMediumBloom L3
Question
Two bars connected by a weightless spring of stiffness k rest on a smooth horizontal plane as shown in figure. Bar 2 is shifted a small distance x to the left and then released. The velocity of the centre of mass of the system after bar 1 breaks off the wall is -
Options
A.$\frac{x\sqrt{m_{2}k}}{m_{1} + m_{2}}$
B.$x\sqrt{\frac{k}{m_{1} + m_{2}}}$
C.zero
D.$\frac{\sqrt{m_{1}k}}{m_{1} + m_{2}}$
Solution
Sol.
$\frac{1}{2}kx^{2} = \frac{1}{2}m_{2}v^{2}$
⇒ v = $x\sqrt{\frac{k}{m_{2}}}$
Vcm = $\frac{m_{1}v_{1} + m_{2}v_{2}}{m_{1} + m_{2}}$
= $\frac{m_{1}(0) + m_{2}x\sqrt{\frac{k}{m_{2}}}}{m_{1} + m_{2}}$
= $\frac{x\sqrt{m_{2}k}}{m_{1} + m_{2}}$
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