Question

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

$[$Take $g=10{\text{ms}}^{-2}]$

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

Answer 1

The correct option is A ${63}^{\circ }$


Using conservation of angular momentum, about point $\text{O}$ is,

$mvl=(m{l}^2+\frac{M{l}^2}{3})\omega$

$\Rightarrow \omega =\frac{3mv}{3ml+Ml}$

Now using energy conservation, after collision

$\frac{1}{2}{I}_{rod+mass}{\omega }^2=Mg\frac{l}{2}(1-\cos \theta )+mgl(1-\cos \theta )$

$\Rightarrow \frac{1}{2}(m{l}^2+\frac{M{l}^2}{3}){\left(\frac{3mv}{3ml+Ml}\right)}^2=l(1-\cos \theta )(\frac{Mg}{2}+m)$

$\Rightarrow \frac{l}{2}\times \frac{3m{v}^2}{3ml+Ml}=\frac{l}{2}(1-\cos \theta )\times (2mg+Mg)$

On putting the values we get,

$\frac{108}{200}=1-\cos \theta$

$\cos \theta =0.46$

$\Rightarrow \theta ={63}^{\circ }$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.