JEE Advanced 2023 · Paper 1 · Physics Q4

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PhysicsSingle correct+3 / -1Question 21 of 51

A bar of mass $M = 1.00$ kg and length $L = 0.20$ m is lying on a horizontal frictionless surface. One end of the bar is pivoted at a point about which it is free to rotate. A small mass $m = 0.10$ kg is moving on the same horizontal surface with $5.00\ \mathrm{m\,s^{-1}}$ speed on a path perpendicular to the bar. It hits the bar at a distance $L/2$ from the pivoted end and returns back on the same path with speed $v$ . After this elastic collision, the bar rotates with an angular velocity $\omega$ . Which of the following statement is correct?

\omega = 6.98\ \mathrm{rad\,s^{-1}}

and

v = 4.30\ \mathrm{m\,s^{-1}}

\omega = 3.75\ \mathrm{rad\,s^{-1}}

and

v = 4.30\ \mathrm{m\,s^{-1}}

\omega = 3.75\ \mathrm{rad\,s^{-1}}

and

v = 10.0\ \mathrm{m\,s^{-1}}

\omega = 6.80\ \mathrm{rad\,s^{-1}}

and

v = 4.10\ \mathrm{m\,s^{-1}}

Answer and worked solution

Official answer: A

Setup.

Moment of inertia about the pivot: $I=\frac{ML^2}{3}=\frac{0.04}{3}\ \text{kg m}^2$ .
The impact is at $r=L/2=0.1$ m.

Angular momentum about the pivot (the ball rebounds with speed $v$ ):
$mur=-mvr+I\omega\implies0.05=-0.01v+0.01333\,\omega.$

Elastic impact. The relative speed of the contact points is reversed:
$u=v+\omega r\implies v=5-0.1\,\omega.$

Solve. Substituting:
$0.05=-0.05+0.001\,\omega+0.01333\,\omega\implies\omega=6.98\ \text{rad/s},\quad v=4.30\ \text{m/s}.$

Topics:Angular momentum and its conservation Elastic and inelastic collisions Rotational kinetic energy

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