A particle moves in the X-Y plane under the influence of a force such that its linear momentum p(t)=A[**i**cos(kt)-**j**sin(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[**i**cos(kt)-**j**sin(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.