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A particle strikes a horizontal friction...

A particle strikes a horizontal frictionless floor with a speed u, at an angle `theta` to the vertical, and rebounds with a speed `v`, at an angle `phi` to the vertical. The coefficient of restitution between the particle and the floor is e. The magnitude of v is

A

`eu`

B

`(1-e)u`

C

`usqrt(e^(2)sin^(2) theta+cos^(2)theta)`

D

`usqrt(sin^(2() theta+e^(2)cos^(2) theta)`

Text Solution

Verified by Experts

The correct Answer is:
D
Initial momentum of the body = 0
The new momentum of the two parts `= m_(1)v_(1) +m_(2) ( -v_(2))`
`:.` From the principle of conservation of moment` 0 = m_(1)v_(1) - m_(2)v_(2)` `:. m_(1)v_(1) = m_(2)v_(2)`
`:. ( v_(1))/(v_(2)) = ( m_(2))/(m_(1))`
`:. ` The ratio of their `K.E. = ((1)/(2) m_(1)v_(1)^(2))/( (1)/(2) m_(2)v_(2)^(2))`
`:. (K_(1))/( K_(2)) =( m_(1))/( m_(2)) xx((m_(2))/( m_(1)))^(2) = ( m_(2))/( m_(1))`
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