Question

A uniform rod of length $l$ lies on a frictionless horizontal surface on which it is pivoted about its centre. A ball moving with speed $v$ as shown in figure collides with the rod at one of the ends. After hitting, the ball sticks to the rod. There is a similar uniform rod of length $l$, pivoted at one of its ends and a similar ball with same speed strikes the other end of the rod and sticks to it. The ratio of final angular momentum in both cases is:

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{8}{1}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $\frac{1}{2}$

Rod pivoted about its centre:


As we know,
$L_i=L_f$
Initial angular momentum $L_i=mv{R}_{\perp }=\frac{mvl}{2}$
$\therefore L_i=L_f=\frac{mvl}{2}$ ... (1) (pivoted at $A$)

Rod pivoted about its end:


As we know,
$L_i=L_f$
Initial angular momentum about end $B$ $=mv{r}_{\perp }=mvl$
$\therefore L_i=L_f=mvl$ (pivoted at $B$)

Hence, ratio in final angular momentum in the two cases
$\frac{(L_f{)}_A}{(L_f{)}_B}=\frac{(L_i{)}_A}{(L_i{)}_B}=\frac{mvl/2}{mvl}$
i.e $\frac{(L_f{)}_A}{(L_f{)}_B}=\frac{1}{2}$