A bar of mass $M=1.00 \mathrm{~kg}$ and length $L=0.20 \mathrm{~m}$ 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 $m=0.10 \mathrm{~kg}$ is moving on the same horizontal surface with $5.00 \mathrm{~m} \mathrm{~s}^{-1}$ 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?

(A) $\omega=6.98 \mathrm{rad} \mathrm{s}^{-1}$ and $\mathrm{v}=4.30 \mathrm{~m} \mathrm{~s}^{-1}$

(B) $\omega=3.75 \mathrm{rad} \mathrm{s}^{-1}$ and $\mathrm{v}=4.30 \mathrm{~m} \mathrm{~s}^{-1}$

(C) $\omega=3.75 \mathrm{rad} \mathrm{s}^{-1}$ and $\mathrm{v}=10.0 \mathrm{~m} \mathrm{~s}^{-1}$

(D) $\omega=6.80 \mathrm{rad} \mathrm{s}^{-1}$ and $\mathrm{v}=4.10 \mathrm{~m} \mathrm{~s}^{-1}$

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Rotational Mechanics

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 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

ω

. Which of the following statement is correct?

Answer: (A)

Applying conservation of angular momentum about the hinge:

m u \cdot \frac{L}{2} = - m v \cdot \frac{L}{2} + I ω

where

I = \frac{M L ^{2}}{3}

for a rod pivoted at one end.

m u \cdot \frac{L}{2} + m v \cdot \frac{L}{2} = \frac{M L ^{2}}{3} ω

... (1)

For elastic collision, coefficient of restitution

e = 1

1 = \frac{ω \frac{L}{2} + v}{u}

u = ω \frac{L}{2} + v

... (2)

Substituting values:

m = 0.1

kg,

M = 1.0

kg,

L = 0.2

u = 5

m/s.

From (1):

0.1 \times 5 \times 0.1 + 0.1 \times v \times 0.1 = \frac{1.0 \times 0.04}{3} ω

0.05 + 0.01 v = 0.0133 ω

... (3)

From (2):

5 = 0.1 ω + v

... (4)

Solving (3) and (4):

v \approx 4.30

m/s and

ω \approx 6.98

rad/s.

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Single correct option