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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 (v) and vec (a) ` both are perpendicular to ` vec (r)`

B

` vec (v) and vec (a)` both are parallel to ` vec (r)`

C

` vec (v) ` is perpendicular to ` vec (r) and vec (a)` is directed away from the origin

D

` vec (v)` is perpendicular to ` vec (r) and vec (a)` is directed towards the origin

Text Solution

Verified by Experts

The correct Answer is:
D
`vec (r) = cos omega t hat(i) + sin omega t hat(j)`
`vec (v) = (d vec (r))/( dt ) = omega (- sin omega t hat(i) + cos omega t hat(j))`
` vec (a) = ( d vec (v))/( dt ) = - omega ^(2) ( cos omega t hat(i) + sin omega t hat(j))`
`vec (a) = - omega ^(2) vec (r) ` `:. vec (a) ` is anti - parallel to ` vec (r)`
And ` vec (v) . vec (r) = omega (- sin omega t cos omega t + cos omega sin omega t ) = 0 `
So ` vec (v) bot vec (r)`
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