A block of mass 1kg slides with velocity 6 m/s on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown in fig. The rod is pivoted about O and swings as a result of the collision making angle $θ$ before momentarily coming to rest. If the rod has mass $M = 2 k g$ and length $l = 1 m$ , the value of $θ$ is approximately $[g = 10 m s^{- 2}]$

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$63^{0}$

$55^{0}$

$69^{0}$

$49^{0}$

answer is A.

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Detailed Solution

Conservation of angular momentum,

$\begin{array}{l} m v l = [m L^{2} + \frac{2 m L^{2}}{3}] ω \\ \Rightarrow ω = \frac{3 v}{5 l} \\ \Rightarrow ω = \frac{3 \times 6}{5 \times 1} = \frac{18}{5} r a d / s \end{array}$

Now, using energy conservation, after collision

$\begin{array}{l} \frac{1}{2} I ω^{2} = 2 m g (\frac{l}{2}) [1 - \cos θ] + m g l (1 - \cos θ) \\ \Rightarrow \frac{1}{2} (\frac{5}{3} m l^{2}) \frac{9 v^{2}}{25 l^{2}} = 2 m g l (1 - \cos θ) \\ \Rightarrow \frac{3}{10} \times \frac{36}{2 \times 10} = 1 - \cos θ \\ \Rightarrow \cos θ = \frac{23}{50} \\ ∴ θ = 63^{0} \end{array}$