Question

A uniform rod of mass $m$ and length $l$ is resting on a smooth horizontal surface. A particle of mass $m/2$ travelling with a speed $v_0$ hits the rod normally and elastically. Then,

• A. final velocity of the particle is $-\frac{2}{15}{v}_0$
• B. final velocity of the particle is $-\frac{1}{15}{v}_0$
• C. angular velocity of the rod $\frac{8v_0}{5\ell }$
• D. angular velocity of the rod $\frac{6v_0}{5\ell }$

Answer 1

Answer: (B), (C)

The correct options are
B angular velocity of the rod $\frac{8v_0}{5\ell }$
C final velocity of the particle is $-\frac{1}{15}{v}_0$

After impact, let particle velocity be $v^{{}^{\prime }}$, rod linear velocity be $v$ and it's angular velocity be $w$

Energy conservation:

$\frac{1}{2}\frac{m}{2}{v}_o^2=\frac{1}{2}\frac{m}{2}{v}^{{}^{\prime }2}+\frac{1}{2}m{v}^2+\frac{1}{2}I{w}^2$

Here $I$ is about C and is given by, $I=\frac{1}{12}m{l}^2$

Conservation of angular momentum about C:

$\frac{m}{2}{v}_o\frac{l}{4}=\frac{m}{2}{v}^{{}^{\prime }}\frac{l}{4}+Iw$

Conservation of linear momentum:

$\frac{m}{2}{v}_o=\frac{m}{2}{v}^{{}^{\prime }}+mv$

Solving, we get,

from linear momentum conservation,

$v=\frac{v_o-v^{{}^{\prime }}}{2}$

from angular momentum conservation,

$w=\frac{3(v_o-v^{{}^{\prime }})}{2l}$

Substituting in energy conservation,

$\frac{1}{4}(v_o^2-v^{{}^{\prime }2})=\frac{7}{32}(v_o-v^{{}^{\prime }}{)}^2$

Since, $v_o\ne v^{{}^{\prime }}$

$(v_o+v^{{}^{\prime }})=\frac{7}{8}(v_o-v^{{}^{\prime }})$

$v^{{}^{\prime }}=-\frac{1}{15}{v}_0$ and $w=\frac{8v_o}{5l}$

Answer: B,C