First slide

Second law

Question

A particle moves in the x-y plane under the influence of a force such that its linear momentum is $\vec{p} (t) = A [\hat{i} \cos (kt) - \hat{j} \sin (kt)]$ , where A and k are constants. Angle between the force and the momentum is:

Easy

A

0°
B

30°
C

45°
D

90°

Solution

$\vec{F} = \frac{d \vec{p}}{dt} = Ak [- \hat{i} \sin kt - \hat{j} \cos kt]$
Let angle between $\vec{F} and \vec{p} be θ$ , then
$\cos θ = \frac{\vec{F} \cdot \vec{p}}{Fp}$
But $\vec{F} \cdot \vec{p} = 0 so \cos θ = 0$
$\Rightarrow θ = 90^{\circ}$

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