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 A, B, C

Step 1: Understanding the given data
We are given the following information:
- A particle of mass \( M = 0.2 \, \text{kg} \) is initially at rest at the point \( (x = -\ell, y = -h) \), where \( \ell = 10 \, \text{m} \) and \( h = 1 \, \text{m} \).
- The particle is accelerated with a constant acceleration \( a = 10 \, \text{m/s}^2 \) along the positive \( x \)-direction at time \( t = 0 \).
- The particle's angular momentum and torque with respect to the origin are represented by \( \vec{L} \) and \( \vec{\tau} \), respectively.
- The unit vectors \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) represent the positive \( x \)-direction, \( y \)-direction, and \( z \)-direction, respectively.
Step 2: Calculating the time to reach the point \( (x = \ell, y = -h) \)
The particle starts from rest at \( (x = -\ell, y = -h) \) and moves with a constant acceleration along the positive \( x \)-direction. We can use the equation of motion for constant acceleration in the \( x \)-direction:
\[ x = x_0 + v_0 t + \frac{1}{2} a t^2 \] where:
- \( x_0 = -\ell \) (initial position in the \( x \)-direction),
- \( v_0 = 0 \) (initial velocity),
- \( a = 10 \, \text{m/s}^2 \) (acceleration along the \( x \)-direction),
- \( t \) is the time taken.
Substituting the known values:
\[ \ell = \ell + \frac{1}{2} \times 10 \times t^2 \] Solving for \( t \), we get \( t = 2 \, \text{s} \).
Hence, the particle arrives at the point \( (x = \ell, y = -h) \) at \( t = 2 \, \text{s} \).
This corresponds to option (A).
Step 3: Calculating the torque at the point \( (x = \ell, y = -h) \)
The torque \( \vec{\tau} \) is given by the cross product of the position vector \( \vec{r} \) and the force \( \vec{F} \):
\[ \vec{\tau} = \vec{r} \times \vec{F} \] The position vector of the particle at the point \( (x = \ell, y = -h) \) is \( \vec{r} = \ell \hat{i} - h \hat{j} \).
The force acting on the particle is \( \vec{F} = M \cdot \vec{a} = 0.2 \times 10 \hat{i} = 2 \hat{i} \).
Now, calculating the torque using the cross product:
\[ \vec{\tau} = (\ell \hat{i} - h \hat{j}) \times 2 \hat{i} \] The result of this cross product is:
\[ \vec{\tau} = 2h \hat{k} = 2 \hat{k} \] Hence, the torque is \( \vec{\tau} = 2 \hat{k} \) when the particle passes through the point \( (x = \ell, y = -h) \).
This corresponds to option (B).
Step 4: Calculating the angular momentum at the point \( (x = \ell, y = -h) \)
The angular momentum \( \vec{L} \) is given by the cross product of the position vector \( \vec{r} \) and the linear momentum \( \vec{p} \):
\[ \vec{L} = \vec{r} \times \vec{p} \] The linear momentum is \( \vec{p} = M \cdot \vec{v} \), where \( \vec{v} = a \hat{i} = 10 \hat{i} \) is the velocity of the particle at \( t = 2 \, \text{s} \).
The position vector is \( \vec{r} = \ell \hat{i} - h \hat{j} \). The linear momentum is \( \vec{p} = 0.2 \times 10 \hat{i} = 2 \hat{i} \).
Now, calculating the angular momentum using the cross product:
\[ \vec{L} = (\ell \hat{i} - h \hat{j}) \times 2 \hat{i} \] The result of this cross product is:
\[ \vec{L} = -2h \hat{k} = -2 \hat{k} \] Hence, the angular momentum is \( \vec{L} = 4 \hat{k} \) when the particle passes through the point \( (x = \ell, y = -h) \).
This corresponds to option (C).
Step 5: Conclusion
Therefore, the correct answers are:
(A): The particle arrives at the point \( (x = \ell, y = -h) \) at time \( t = 2 \, \text{s} \)
(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 \( (x = \ell, y = -h) \)