Question

A rod $AB$ of mass $M$ and length $8l$ lies on a smooth horizontal surface. A particle, of mass ${}^{\prime }{m}^{\prime }$ and velocity $v_0$, strikes the rod perpendicular to its length, as shown in Figure. As a result of the collision, the centre of mass of the rod attains a speed of $v_0/8$ and the particle rebounds back with a speed of $v_0/4$. Find the following:
a. The ratio $M/m$.
b. The angular velocity of the rod about $O$.
c. The coefficient of restitution ${}^{\prime }{e}^{\prime }$ for the collision.
d. The velocities of the ends ${}^{\prime }{A}^{\prime }$ and ${}^{\prime }{B}^{\prime }$ of the rod, namely, $v_A$ and $v_B$, respectively.

Answer 1

REF.Image.

Conserving momentum

$m\sqrt{0}=\frac{M{V}_0}{8}-\frac{m{V}_0}{4}$

$M=10m$

$\left(a\right)\frac{M}{m}=10$

$y_{com}=\frac{ml}{m+M}$

conserving angular momentum

$m{V}_0(l-y_{com})=\frac{16m{l}^2}{12}w-\frac{m{V}_0}{4}(l-y_{com})-\frac{m{V}_0}{8}{y}_{com}$

also $M=10m$,

$V_{0}l(1-\frac{1}{11}+\frac{1}{4}-\frac{1}{44}-\frac{5}{44})=\frac{160}{12}{l}^{2}w$

$\left(b\right)w=\frac{27}{352}\frac{V_0}{l}$

(C) e = rel vel of sep . rel vel of approach

$=[V_0/4+V_0/8+lw]/V_0$

$=\frac{1}{4}+\frac{1}{8}+\frac{27}{352}$

$e=\frac{159}{352}$

$V_A=\frac{v_0}{8}+4lw$

$=\frac{V_0}{8}+\frac{27}{88}\sqrt{0}$

$V_A=\frac{39}{88}\sqrt{0}\left(d\right)$

$V_B=\frac{V_0}{8}-4lw$

$=\frac{V_0}{8}-\frac{27}{88}{V}_0$

$V_B=\frac{-2}{11}\sqrt{0}$ (e)