Question

A block of mass $m=1kg$ slides with velocity $v=6m/s$ on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about $O$ and swings as a result of the collision making angle $\theta$ before momentarily coming to rest. If the rod has mass $M=2kg,$ and length $l=1m,$ the value of $\theta$ is approximately :
(Take $g=10m/s^2)$

• A. $69^{\circ }$
• B. $63^{\circ }$
• C. $55^{\circ }$
• D. $49^{\circ }$

Answer 1

Answer: (B)

Angular momentum conservation
$mvl=\frac{M{l}^2}{3}\omega +m{l}^2\omega$
$\Rightarrow \omega =\frac{1\times 6\times 1}{\frac{2}{3}+1}$
$=\frac{18}{5}$
Now using energy consevation
$\frac{1}{2}\left(M\frac{l^2}{3}\right){\omega }^2+\frac{1}{2}\left(m{l}^2\right){\omega }^2$
$=(m+M)r_{cm}(1-\cos \theta )$
$=(m+M)\left(\frac{ml+\frac{Ml}{2}}{m+M}\right)g(1-\cos \theta )$
$\frac{5}{6}\times {\left(\frac{18}{5}\right)}^2=20(1-\cos \theta )$
$\Rightarrow 1-\cos \theta =\frac{18}{5}\times \frac{3}{20}$
$\cos \theta =1-\frac{27}{50}$
$\cos \theta =\frac{23}{50}$
$\Rightarrow \theta \simeq 63^{\circ }$