A body of mass 5 kg moving with a uniform speed 3 √(2) ms -1 in X - Y plane along the line y = x +4 . The angular momentum of the particle about the origin will be \ \ \ \ kg m 2 s -1 .

Physics · Rotational Dynamics

JEE Main 2024 — 29 January, Shift 2 — Question 58

A body of mass 5 kg moving with a uniform speed $3 \sqrt{2} \mathrm{~ms}^{-1}$ in $\mathrm{X}-\mathrm{Y}$ plane along the line $\mathrm{y}=\mathrm{x}+4$ . The angular momentum of the particle about the origin will be $\_\_\_\_$ $\mathrm{kg} \mathrm{m}^{2} \mathrm{~s}^{-1}$ .

Answer: 60

Numerical answer — enter this value.

Step-by-step solution

\text{Given: } m = 5 \, \text{kg}, \quad v = 3\sqrt{2} \, \text{ms}^{-1}, \quad \text{Line: } y = x + 4

\text{Direction vector of motion: } \hat{i} + \hat{j} \Rightarrow \vec{v} = 3(\hat{i} + \hat{j})

\vec{p} = m\vec{v} = 5 \cdot 3(\hat{i} + \hat{j}) = 15(\hat{i} + \hat{j})

\text{Choose point on the line: } x = 0 \Rightarrow \vec{r} = 0\hat{i} + 4\hat{j}

\vec{L} = \vec{r} \times \vec{p} = (0\hat{i} + 4\hat{j}) \times 15(\hat{i} + \hat{j})

= 4\hat{j} \times 15\hat{i} + 4\hat{j} \times 15\hat{j} = 60(-\hat{k}) + 0 = -60\hat{k}

\boxed{|\vec{L}| = 60 \, \text{kg·m}^2\text{/s}}

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 2
Subject: Physics
Chapter: Rotational Dynamics
Topic: Angular Momentum and its Conservation