Question

A particle of mass $M =0.2 \,kg$ 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

Answer: (A), (B), (C)

Time taken to reach from $(-\ell,- h )$ to $(\ell,- h )$ is given by
$2 \ell=\frac{1}{2} at ^{2} $
$20=\frac{1}{2}(10) t ^{2} $
$ t =2\, sec $
$\vec{ L }= mvh \hat{ k }= math \hat{ k }$
$\vec{ L }=(0.2)(10) 2(1) \hat{ k }=4 \hat{ k }$ [ when particle passes through $ (\ell,- h )] $
$\vec{\tau}=\frac{ d \vec{ L }}{ dt }= mah \hat{ k }=(0.2)(10)(1) \hat{ k }=2 \hat{ k } $ (always)