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A particle moves in the X-Y plane under the influence of a force such that its linear momentum p(t)=A[icos(kt)-jsin(kt)] where A and k are constants. Find the angle between the force and the momentum.

Solution:

In the given problem, linear momentum p(t)=A[icos(kt)-jsin(kt)]

We know, force$$F=\frac{dp}{dt}$$
Differentiating p(t) with respect to t, we get

$$F=\frac{dp}{dt}=A[-iksin(kt)-jkcos(kt)]$$
Taking dot product of F and p,
$$F.p=-ksin(kt)cos(kt)+kcos(kt)sin(kt) \Rightarrow 0$$
Since, the dot product is zero, we can conclude that $$cos\theta=0.$$ where θ is the angle between the force and momentum.
Hence, it is implied that $$\theta=90^\circ$$
Thus, the angle between the force and momentum will be ninety degrees.