The position vectors of two 1 kg particles,… - JEE Main 2025 Physics PYQ

JEE Main 2025 — Physics Mechanics

2025integerhard

The position vectors of two 1 kg particles, (A) and (B), are given by $\overrightarrow{\mathrm{r}}_{\mathrm{A}}=\left(\alpha_1 \mathrm{t}^2 \hat{i}+\alpha_2 \mathrm{t} \hat{j}+\alpha_3 \mathrm{t} \hat{k}\right) \mathrm{m}$ and $\overrightarrow{\mathrm{r}}_{\mathrm{B}}=\left(\beta_1 \mathrm{t} \hat{i}+\beta_2 \mathrm{t}^2 \hat{j}+\beta_3 \mathrm{t} \hat{k}\right) \mathrm{m}$ , respectively; $\left(\alpha_1=1 \mathrm{~m} / \mathrm{s}^2, \alpha_2=3 \mathrm{n} \mathrm{m} / \mathrm{s}, \alpha_3=2 \mathrm{~m} / \mathrm{s}, \beta_1=2 \mathrm{~m} / \mathrm{s}, \beta_2=-1 \mathrm{~m} / \mathrm{s}^2, \beta_3=4 \mathrm{pm} / \mathrm{s}\right)$ , where t is time, n and p are constants. At $t=1 \mathrm{~s},\left|\overrightarrow{V_A}\right|=\left|\vec{V}_B\right|$ and velocities $\vec{V}_A$ and $\vec{V}_B$ of the particles are orthogonal to each other. At $t=1 \mathrm{~s}$ , the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is $\sqrt{\mathrm{L}} \mathrm{kgm}^2 \mathrm{~s}^{-1}$ . The value of L is _______ .

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Held on 22 Jan 2025 · Verified 6 Jul 2026.

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Solution

At $t=1$ $r_{A B}=-1 \hat{i}+(3 n+1) \hat{j}+(2-4 p) \hat{k}$ At $t=1$ $\begin{aligned}
& v_A=2 \hat{i}+3 n \hat{j}+2 \hat{k} \
& v_B=2 \hat{i}-2 \hat{j}+4 p \hat{k} \
& \vec{v}A-\vec{v}B=0, \quad 4-6 n+8 p=0
\end{aligned}\begin{gathered}\left|v_A\right|=\left|v_B\right| \quad(3 n)^2+4=4+16 p^2 \ 3 n=-4 p \ 4+16 p=0 \ p=-\frac{1}{4}, n=\frac{1}{3} \ r{A B}=-\hat{i}+2 \hat{j}+3 \hat{k} \ v_A=2 \hat{i}+\hat{j}+2 \hat{k} \ \therefore L=m\left|\vec{r}{A B} \times \vec{v}_A\right|=90\end{gathered}$

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