A particle moves in the x - y plane under the action of a force F such that the coordinates of its linear…

A particle moves in the x-y plane under the action of a force F such that the coordinates of its linear momentum P→ at any time t is px=2 cos t,py= 2sin⁡t
The angel θ between F→ and P→ at a given time t will be

  1. 90°
  2. 0°
  3. 180°
  4. 30°

Solution

P=px2+py2

=2cos⁡t2+2sin⁡t2=2

If m be the mass of the body, then kinetic energy

=p22m=222m=2m

Since kinetic energy does not change with time, both work done and power are zero

Now Power = Fvcos⁡θ=0

As F≠0,v≠0

∴ cos⁡θ=0

Or θ=90°

As direction of p→ is same that v→(∵p→=mv→) hence angle between F→ and p→ is equal to 90°

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