JEE Main 2025 Jan 22 Shift 1, Physics Q24: Angular Momentum, Its Conservation and Angular Impulse

JEE Main2025Jan 22, Shift 1Physics

The position vectors of two 1 kg particles, (A) and (B), are given by

$r_{A} = (α_{1} t^{2} \hat{i} + α_{2} t \hat{j} + α_{3} t \hat{k})$ m

and $r_{B} = (β_{1} t \hat{i} + β_{2} t^{2} \hat{j} + β_{3} t \hat{k})$ m, respectively;

( $α_{1} = 1$ m/s $^{2}$ , $α_{2} = 3 n$ m/s, $α_{3} = 2$ m/s, $β_{1} = 2$ m/s, $β_{2} = - 1$ m/s $^{2}$ , $β_{3} = 4 p$ m/s), where t is time, n and p are constants, At t = 1s, $∣ V_{A} ∣ = ∣ V_{B} ∣$ and velocities $V_{A}$ and $V_{B}$ of the particles are orthogonal to each other. At t = 1 s, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is $L$ kgm $^{2}$ s $^{- 1}$ . The value of L is _____.

Your answer

Solution

Velocities at general $t$ :

$V_{A} = 2 α_{1} t \hat{i} + α_{2} \hat{j} + α_{3} \hat{k} = 2 t \hat{i} + 3 n \hat{j} + 2 \hat{k}$ .

$V_{B} = β_{1} \hat{i} + 2 β_{2} t \hat{j} + β_{3} \hat{k} = 2 \hat{i} - 2 t \hat{j} + 4 p \hat{k}$ .

At $t = 1$ s: $V_{A} = (2, 3 n, 2)$ , $V_{B} = (2, - 2, 4 p)$ .

Orthogonality: $V_{A} \cdot V_{B} = 4 - 6 n + 8 p = 0$ , so $3 n = 2 + 4 p$ .

Equal magnitudes: $4 + 9 n^{2} + 4 = 4 + 4 + 16 p^{2} \Rightarrow 9 n^{2} = 16 p^{2}$ . Combined with $3 n = 2 + 4 p$ , solving gives $p = - 1/4$ and $n = 1/3$ .

Position vectors at $t = 1$ s: $r_{A} = (1, 3 n, 2) = (1, 1, 2)$ ; $r_{B} = (2, - 1, 4 p) = (2, - 1, - 1)$ .

$r_{A / B} = r_{A} - r_{B} = (- 1, 2, 3)$ . Also $V_{A} = (2, 1, 2)$ .

Angular momentum: $L = m r_{A / B} \times V_{A} = (- 1, 2, 3) \times (2, 1, 2) = (2 \cdot 2 - 3 \cdot 1, 3 \cdot 2 - (- 1) \cdot 2, (- 1) \cdot 1 - 2 \cdot 2) = (1, 8, - 5)$ .

$∣ L ∣ = 1 + 64 + 25 = 90$ . So $L = 90$ .

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JEE Main 2025 Jan 22 Shift 1, Physics Q24: Angular Momentum, Its Conservation and Angular Impulse

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