Q151 / 200 · Manipal (MET) 2011 · Physics

A particle moves in the $x y$-plane under the action of a force $F$ such that the components of its linear momentum $\mathbf{p}$ at any time $t$ are $p_x=2 \cos t, p_y=\sin t$. The angle between $\mathbf{F}$ and $\mathbf{p}$ at time $t$ is

  • Correct answer
    A
    $90^{\circ}$
  • B
    $0^{\circ}$
  • C
    $180^{\circ}$
  • D
    $30^{\circ}$

Solution

$\begin{aligned} & p_x=2 \cos t, p_y=2 \sin t \\ & \begin{aligned} & \therefore \quad \mathbf{p}=p_x \mathbf{i}+p_y \mathbf{j} \\ &=2 \cos t \mathbf{i}+2 \sin t \mathbf{j}\end{aligned} \\ & \text { Force }=\text { rate of change of momentum } \\ & \quad=\frac{d \mathbf{p}}{d t}\end{aligned}$ $\begin{aligned} & \Rightarrow \quad \mathbf{F}=-2 \sin t \mathbf{i}+2 \cos t \mathbf{j} \\ & \therefore \cos \theta=\frac{\mathbf{F} \cdot \mathbf{p}}{|\mathbf{F}| \cdot|\mathbf{p}|} \\ & =\frac{(-2 \sin t \mathbf{i}+2 \cos t \mathbf{j})(2 \cos t \mathbf{i}+2 \sin t \mathbf{j})}{\sqrt{4 \sin ^2 t+4 \cos ^2 t} \sqrt{4 \cos ^2 t+4 \sin ^2 t}} \\ & \Rightarrow \quad \cos \theta=0 \\ & \Rightarrow \quad \theta=90^{\circ}\end{aligned}$

Hence, the correct answer is option: 'A'.

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