Question

A particle of mass $M =02 \,kg$ is initially at rest in the $xy$-plane at a point $( x =-\ell, y =- h )$, where $\ell=10 \,m$ and $h =1\, m$ The particle is accelerated at time $t =0$ with a constant acceleration $a =10\, m / s ^{2}$ along the positive $x$-direction Its angular momentum and torque with respect to the origin, in SI units, are represented by $\vec{ L }$ and $\vec{\tau}$ respectively $\hat{ i }, \hat{ j }$ and $\hat{ k }$ are unit vectors along the positive $x , y$ and $z$-directions, respectively If $\hat{k}=\hat{i} \times \hat{j}$ then which of the following statement(s) is(are) correct?

• A. The particle arrives at the point $(x=\ell, y=-h)$ at time $t=2 s$.
• B. $\vec{\tau}=2 \hat{k}$ when the particle passes through the point $(x=\ell, y=-h)$
• C. $\vec{ L }=4 \hat{ k }$ when the particle passes through the point $( x =\ell, y =- h )$
• D. $\vec{\tau}=\hat{k}$ when the particle passes through the point $(x=0, y=-h)$

Answer 1

The Correct Option is C

Given:

• Mass of particle: \( M = 0.2\, \text{kg} \)
• Initial position: \( (-\ell, -h) = (-10, -1) \)
• Constant acceleration along +x: \( a = 10\, \text{m/s}^2 \)
• Initial velocity = 0

We are to calculate: Angular momentum \( \vec{L} \) and torque \( \vec{\tau} \) w.r.t. origin, especially at the point \( (\ell, -h) = (10, -1) \)

At t = 0:

• Initial position: \( \vec{r}_0 = -10\hat{i} - 1\hat{j} \)
• Initial velocity: \( \vec{v}_0 = 0 \)
• Initial angular momentum: \( \vec{L}_0 = \vec{r}_0 \times M\vec{v}_0 = 0 \)
• Torque: \( \vec{\tau} = \vec{r}_0 \times M\vec{a} = (-10\hat{i} - \hat{j}) \times (0.2 \cdot 10\hat{i}) = 0 \)
• ✓ Angular momentum and torque at t = 0 are both zero

At the point (x = +10, y = -1):

This is a displacement of 20 m along x-axis from the starting point, so using:

\[ x = ut + \frac{1}{2}at^2 \Rightarrow 20 = 0 + \frac{1}{2} \cdot 10 \cdot t^2 \Rightarrow t^2 = 4 \Rightarrow t = 2\,\text{s} \]

Velocity at this point: \( v = at = 10 \cdot 2 = 20\, \text{m/s} \)

Position vector: \( \vec{r} = 10\hat{i} - 1\hat{j} \)

Linear momentum: \( \vec{p} = M\vec{v} = 0.2 \cdot 20\hat{i} = 4\hat{i} \)

Angular momentum:

\[ \vec{L} = \vec{r} \times \vec{p} = (10\hat{i} - \hat{j}) \times (4\hat{i}) = -\hat{j} \times 4\hat{i} = 4\hat{k} \]

✓ Correct: Option C — Angular momentum is \( \vec{L} = 4\hat{k} \) at \( (10, -1) \)

Correct Answer: Option C