JEE Advanced 2023 — Paper 1 — PHY-II solution

JEE Advanced2023Paper 1PHY-II

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?

A
$ω = 6.98 rad s^{- 1}$ and $v = 4.30 m s^{- 1}$
B
$ω = 3.75 rad s^{- 1}$ and $v = 4.30 m s^{- 1}$
C
$ω = 3.75 rad s^{- 1}$ and $v = 10.0 m s^{- 1}$
D
$ω = 6.80 rad s^{- 1}$ and $v = 4.10 m s^{- 1}$

Solution

Angular momentum about the pivot. The small mass approaches with speed $u = 5$ m/s and returns with speed $v$ along the same line; its lever arm about the pivot is $L /2$ . The bar's moment of inertia about the end is $I = \frac{M L ^{2}}{3}$ . Taking the incoming direction as positive: $m u \frac{L}{2} = - m v \frac{L}{2} + \frac{M L ^{2}}{3} ω . (1)$

Elastic collision condition. The point on the bar at distance $L /2$ has velocity $ω L /2$ after collision. With coefficient of restitution unity: $\frac{ω L /2 - ( - v )}{u} = 1 \Rightarrow u = \frac{ω L}{2} + v . (2)$

Solve. Substitute the numbers $M = 1$ , $m = 0.1$ , $L = 0.2$ , $u = 5$ . From (1): $0.1 \cdot 5 \cdot 0.1 = - 0.1 v \cdot 0.1 + \frac{0.04}{3} ω$ , i.e. $0.05 = - 0.01 v + 0.01333 ω$ . From (2): $5 = 0.1 ω + v$ , so $v = 5 - 0.1 ω$ .

Substitute: $0.05 = - 0.01 (5 - 0.1 ω) + 0.01333 ω = - 0.05 + 0.001 ω + 0.01333 ω$ . Thus $0.10 = 0.01433 ω$ , giving $ω \approx 6.98 rad s^{- 1}$ and $v = 5 - 0.1 \cdot 6.98 \approx 4.30 m s^{- 1}$ .

Answer: (A).

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