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A particle of mass M = 0.2 kg is initially at rest ....

A particle of mass M = 0.2 kg is initially at rest in the xy-plane at a point (x = -l, y = -h), where l = 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 L and $\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 = l, y = -h) at time t = 2 s.
(B) $\vec τ = 2 \hat k $ when the particle passes through the point (x = l, y = -h)
(C) $\vec L = 4 \hat k $ when the particle passes through the point (x = l, y = -h)
(D) $\vec τ = \hat k $ when the particle passes through the point (x = 0, y = -h)

Solution


$s_x =u_x t + \frac {1}{2} a_x t^2 $

$\Rightarrow 2l = 0+\frac {1}{2} 10 t^2 $ at (l, -h) or (10, -1)

$\Rightarrow 2\times 10 =5t^2$

$\Rightarrow t = 2 $ sec

$\vec L = \vec r \times \vec p = (l\hat i - h\hat j)\times Mv_x \hat i = Mhv_x \hat k = Mha_x t \hat k$

at (10, -1) $\vec L = 0.2\times 1 \times 10 \times 2 \hat k = 4 \hat k$

$\vec \tau  = \frac{{d\vec L}}{{dt}} = Mh{a_x}\hat k$

$=0.2\times 1 \times 10 \hat k = 2 \hat k$

The above answer for $\tau $ is independent of x-coordinate.

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