Solve the following :

Question:

A uniform rod of length $L$ lies on a smooth horizontal table. A particle moving on a 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.

Solution:

Momentum conservation

$P_{i}=P_{f}$

$m u+0=m(0)+M(v)$

$\mathrm{V}=\frac{\operatorname{mu}}{M}$

Angular Momentum conservation $L_{i}=L_{f}$

$0+m u\left(\frac{\Sigma}{2}\right)=\left(\frac{M L^{2}}{12}\right) \omega+0$

$\omega=\frac{6 m u}{M L}$

Time to rotate by $90^{\circ}$

$\omega=\frac{\theta}{t}$

$t=\frac{(\pi / 2)}{\left(\frac{6 m u}{M L}\right)}$

$\mathrm{t}=\frac{\pi M L}{12 \mathrm{mu}}$

distance travelled by rod

distance $=$ speed $x$ time

$=\left(\frac{m U}{M}\right) \times\left(\frac{\pi M L}{12 m u}\right)$

$\mathrm{S}=\frac{\pi L}{12}$

Now, if $\mathrm{M}=4 \mathrm{~m}$ then

$K E_{i}=\frac{1}{2} m u^{2}$

$K E_{f}=\frac{1}{2} M v^{2}+\frac{1}{2} I \omega^{2}$

$\frac{1}{2}(4 m)\left(\frac{m u}{4 m}\right)^{2}+\frac{1}{2}\left(\frac{M L^{2}}{12}\right)\left(\frac{6 m u}{M L}\right)^{2}$

$K E_{f}=\frac{1}{2} m u^{2}$

Hence proved.

 

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