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

Question:

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 $\mathrm{O}$ and swings as a result of the collision making angle $\theta$ before momentarily coming to rest. If the rod has mass $\mathrm{M}=2 \mathrm{~kg}$, and length $l=1 \mathrm{~m}$, the value of $\theta$ is approximately :

(Take $g=10 \mathrm{~m} / \mathrm{s}^{2}$ )

1. $69^{\circ}$

2. $63^{\circ}$

3. $55^{\circ}$

4. $49^{\circ}$

Correct Option: , 2

Solution:

Angular momentum conservation

$\mathrm{mv} l=\frac{\mathrm{M} l^{2}}{3} \omega+\mathrm{m} l^{2} \omega$

$\Rightarrow \omega=\frac{1 \times 6 \times 1}{\frac{2}{3}+1}=\frac{18}{5}$

Now using energy consevation

$\frac{1}{2}\left(\mathrm{M} \frac{l^{2}}{3}\right) \omega^{2}+\frac{1}{2}\left(\mathrm{~m} l^{2}\right) \omega^{2}$

$=(\mathrm{m}+\mathrm{M}) \mathrm{r}_{\mathrm{cm}}(1-\cos \theta)$

$=(\mathrm{m}+\mathrm{M})\left(\frac{\mathrm{m} l+\frac{\mathrm{M} l}{2}}{\mathrm{~m}+\mathrm{M}}\right) \mathrm{g}(1-\cos \theta)$

$\frac{5}{6} \times\left(\frac{18}{5}\right)^{2}=20(1-\cos \theta)$

$\Rightarrow 1-\cos \theta=\frac{18}{5} \times \frac{3}{20}$

$\cos \theta=1-\frac{27}{50}$

$\cos \theta=\frac{23}{50} \Rightarrow \theta=63^{\circ}$

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