You might have noted that the angle between the force acting on a body and the momentum of the body can be anything between zero and 2π. (But the *change* of momentum is always in the direction of force). Here is a simple question which you should be able to answer in a minute. If you find it difficult you should work harder to understand basic points in mechanics thoroughly.

**A particle moves in a plane such that the rectangular components of its momentum vary simple harmonically with the same period and amplitude, but with a constant phase difference of π/2. The angle (in radian) between the momentum of the particle and the force acting on it is **

**(a) π (b) ****–**** π ** **(c) zero (d) π/2 (e) varying between zero and 2π **

The particle is forced to move simple harmonically along two mutually perpendicular directions. In other words, this is a case of the superposition of two simple harmonic motions of the same perod and amplitude at right angles to each other. Since the phase difference is π/2, the resultant motion is *uniform circular motion*. The angle between the resultant momentum of the particle and the force acting on the particle is therefore π/2.

Now consider the following MCQ which appeared in IIT-JEE 2007 question paper:

A particle moves in the X-Y plane under the influence of a force such that its linear momentum is **p**(t) = A[**î** cos(kt) – **ĵ** sin(kt)], where A and k are constants. The angle between the force and the momentum is

(a) 0º (b) 30º (c) 45º (d) 90º

Simply by noting that the momentum vector **p **has simple harmonically varying components A cos(kt) and A sin(kt) in the X and Y directions respectively, you can conclude that this is a case of the superposition two simple harmonic motions of the same frequency and amplitude at right angles, with a constant phase difference of π/2. [The phase difference is π/2 since one is a sine function while the other is a cosine function].

So, this is uniform circular motion and the angle between momentum and force, as usual, is π/2.

Now, if you want to give a quantitative proof, you may proceed as follows:

The time rate of change of momentum is force. Therefore, force

**F** = d**p/**dt = –Ak[**î **sin(kt) + **ĵ** cos(kt)]

The angle θ between the force vector **F **and the momentum vector** p **is given by

Cosθ = **F.p/ **Fp, where the bold face letters represent vectors.

But** F.p** = –Ak[**î **sin(kt) + **ĵ** cos(kt)] **.** A[**î** cos(kt) – **ĵ** sin(kt)]** **

** = **–A^{2}k sin(kt)cos(kt) + A^{2}k cos(kt)sin(kt) = 0

Since cosθ = 0, θ = π/2.

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