A block of mass m=1 kg slides with velocity v=6 m/s


A block of mass $m=1 \mathrm{~kg}$ slides with velocity $v=6 \mathrm{~m} / \mathrm{s}$ 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 $M=2 \mathrm{~kg}$, and length $l=1 \mathrm{~m}$, the value of $\theta$ is approximately: (take $g=10 \mathrm{~m} / \mathrm{s}^{2}$ )

  1. $63^{\circ}$

  2. $55^{\circ}$

  3. $69^{\circ}$

  4. $49^{\circ}$

Correct Option: 1


(1) Using conservation of angular momentum

$m v l=\left(m l^{2}+\frac{2 m l^{2}}{3}\right) \omega \Rightarrow m v l=\frac{5}{3} m l^{2} \omega \Rightarrow \omega=\frac{3 v}{5 l}$

or, $\omega=\frac{3 \times 6}{5 \times 1}=\frac{18}{5} \mathrm{rad} / \mathrm{s}$

Now using energy conservation, after collision

$\frac{1}{2} I \omega^{2}=2 m g \frac{l}{2}(1-\cos \theta)+m g l(1-\cos \theta)$

$\Rightarrow \frac{1}{2}\left(\frac{5}{3} m l^{2}\right) \frac{9 v^{2}}{25 l^{2}}=2 m g l(1-\cos \theta)$

$\Rightarrow \frac{3}{5 \times 2} m v^{2}=2 m g l(1-\cos \theta)$

$\frac{3}{10} \times \frac{36}{2 \times 10}=1-\cos \theta \Rightarrow 1-\frac{27}{50}=\cos \theta$

or, $\cos \theta=\frac{23}{50}$                $\therefore \theta \simeq 63^{\circ}$

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