A uniform rod of length L lies on a smooth horizontal table. A particle moving on the table strikes the rod perpendicularly at an end and stops. Find the distance travelled by the centre of the rod by the time it turns through a right angle. Show that if the mass of the rod is four times that of the particle, the collision is elastic.
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Let the mass of the particle = m & the mass of the rod = M.Let the particle strikes the rod with a velocity V.
If we take the two body to be a system.
Therefore the net external torque & net external force = 0
Therefore Applying laws of conservation of linear momentum MV' = mV (V' = velocity of the rod after striking)
$\Rightarrow\frac{\text{V}'}{\text{V}}=\frac{\text{m}}{\text{M}}$
Again applying laws of conservation of angular momentum
$\Rightarrow\frac{\text{mVR}}{2}=\ell\omega$
$\Rightarrow\frac{\text{mVR}}{2}=\frac{\text{MR}^2}{12}\times\frac{\pi}{2\text{t}}$
$\Rightarrow\text{t}=\frac{\text{MR}\pi}{\text{m}\times12\times\text{V}}$
Therefore distance travelled:
$\text{V}'\text{t}=\text{V}'\Big(\frac{\text{MR}\pi}{\text{m}\times12\times\pi}\Big)$
$=\frac{\text{m}}{\text{M}}\times\frac{\text{M}}{\text{m}}\times\frac{\text{R}\pi}{12}=\frac{\text{R}\pi}{12}$
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