A particle of mass m is moving in YZ-plane with a uniform velocity `v` with its trajectory running parallel to `+ve` Y-axis and intersecting Z-axis at `z=a` in figure. The change in its angular momentum about the origin as it bounces elastically form a wall at y=constant is

Answer

b

Explanation

The initial velocity is
`vecv_(i) = vhatey`
After reflection from the wall, the final velocity is
`vecv_(f) = -vhatey`
The trajectory is given as
`vecr= yhate_(y) + ahate_(z)`
Hence, the change in angular momentum is
`DeltavecL=vecr xx m(vecv_(f) - vecv_(i))`
`=(yhate_(y) + avece_(z)) xx (-2mhatey)`
`=2"mva"vece_(x) [therefore hate_(y) xx hate_(y) = 0` and `hate_(z) xx hate_(y) = -hate_(x)]`