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A particle of mass m is moving in yz-pla...

A particle of mass `m` is moving in yz-plane with a unifrom velocity `v` with its trajectory running parallel to +ve y-axis and intersecting z-axis at `z = a` Fig. The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is :

A

`mvahate`

B

`2mvahat(e)_(x)`

C

`y mvhat(e)_(x)`

D

`2y mv hat(e)_(x)`

Text Solution

Verified by Experts

The correct Answer is:
B
The initial velocity is `hatv_(i) =vhat(e)_(y)` and after reflection from the wall, the final velocity is `hat(v)_(y)=-vhat(e)_(g)` . The trajectory is described as position vector `bar(r)=vhat(e)_(y)+ahat(e)_(z)`. Hence, the changen in angular momentum is
`bar(r)xxm(bar(v)_(f)-bar(v)_(i))=2mvahat(e)_(x)`
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Knowledge Check

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