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A particle of mass m=0.2kg is initially at rest in the xy-plane at a point $(x = - l, y = - h)$ , where $l = 10$ m and h=1m. 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 $\vec{τ}$ respectively. $\hat{i}, \hat{j}, \hat{k}$ 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?

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The particle arrives at the point $x = l, y = - h$ at time t=2s.

If $\vec{τ} = \vec{r} \times \vec{m a}$ then $\vec{τ} = 2 \hat{k}$ when the particle passes through the point ( $x = l, y = - h$ )

If angular momentum $\vec{L} = \vec{r} \times \vec{m v}$ then $\vec{L} = 4 \hat{k}$ when the particle passes through the point( $x = l, y = - h$ )

$\vec{τ} = \hat{k}$ when the particle passes through the point ( $x = 0, y = - h$ )

answer is A, B, C.

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Detailed Solution

$A) S = u t + \frac{1}{2} a t^{2}$

$20 = 0 + \frac{1}{2} \times 10 \times t^{2} (u = 0)$

t= 2sec

$b) \vec{τ} = \vec{r} \times \vec{f} = r_{⊥} f = (1 \times 2) \hat{k} (f = 0.2 \times 10)$

$\vec{τ} = 2 \hat{k}$

$c) \vec{L} = \vec{r} \times \vec{p} = r_{⊥} p (v = a t)$

v=20m/s p=20x0.2=4kgm/s

$\overset{⇀}{L} = 4 \hat{k}$

$d) \vec{τ} = \vec{r} \times \vec{f} = r_{⊥} f = 1 \times 2 = 2$

$\vec{τ} = 2 \hat{k}$

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