Centre of Mass 2 Question 5
6. A particle moves in the $x-y$ plane under the influence of a force such that its linear momentum is $\mathbf{p}(t)=A[\hat{\mathbf{i}}$ os $(k t)-\hat{\mathbf{j}} \sin (k t)]$, where, $A$ and $k$ are constants. The angle between the force and the momentum is
(2007, 3M)
(a) $0^{\circ}$
(b) $30^{\circ}$
(c) $45^{\circ}$
(d) $90^{\circ}$
Objective Question II (One or more correct option)
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Answer:
Correct Answer: 6. (d)
Solution:
$$ \begin{aligned} \mathbf{F} & =\frac{d \mathbf{p}}{d t}=-k A \sin (k t) \hat{\mathbf{i}}-k A \cos (k t) \hat{\mathbf{j}} \\ \mathbf{p} & =A \cos (k t) \hat{\mathbf{i}}-A \sin (k t) \hat{\mathbf{j}} \end{aligned} $$
Since, $\mathbf{F} \cdot \mathbf{p}=0$
$\therefore$ Angle between $\mathbf{F}$ and $\mathbf{p}$ should be $90^{\circ}$.
