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A particle moves such that its momentum ...

A particle moves such that its momentum vector `vecp(t) = cos omega t hati + sin omega t hatj` where `omega` is a constant and t is time. Then which of the following statements is true for the velocity `vec v(t)` and acceleration `veca (t)` of the particle :

A

`vec` is perpendicular to `vecp and veca` is directed towards the origin

B

`vecv and veca` both are parallel to `vecp`

C

`vecv and veca` both are perpendicular to `vecp`

D

`vecv` is parallel to `vecp and veca` is perpendiuclar to `vecp`

Text Solution

Verified by Experts

The correct Answer is:
D
`vecp (t) = cos omega hati + sin omega t hatj therefore vecv = (vecP)/(m)`
On diff. we get
`vecf = - omega sin omega t hati+ omega cos omega t hatj therefore veca = (vecf)/(m), vecf * vecp = 0`
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