Question

A uniform thin rod of length $L$, mass $m$ is lying on a smooth horizontal table. A horizontal impulse $P$ is suddenly applied perpendicular to the rod at one end. The total energy of the rod after the impulse is

• A. $\frac{P^{2}}{M}$
• B. $\frac{7 P^{2}}{8 M}$
• C. $\frac{13 P^{2}}{2 M}$
• D. $\frac{2 P^{2}}{M}$

Answer 1

Answer: (D)

$P=(m v-0)$
angular impulse $J=P \times \frac{L}{2}=L_{f}-0$
$L_{f}=\frac{P L}{2} $
$K E=K_{r}+K_{t}=\frac{P^{2}}{2 m}+\frac{L_{f}^{2}}{2 I} $
$=\frac{P^{2}}{2 m}+\frac{\left(P \frac{L}{2}\right)^{2}}{2 \times \frac{m L^{2}}{12}}$
$=\frac{P^{2}}{2 m}+\frac{P^{2} L^{2}}{4 \times \frac{m L^{2}}{6}}$
$=\frac{P}{2 m}+\frac{3 P^{2}}{2 m}=\frac{2 P^{2}}{m}$