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 :

Options

A. `mvahate`
B. `2mvahat(e)_(x)`
C. `y mvhat(e)_(x)`
D. `2y mv hat(e)_(x)`

Answer

B

Explanation

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)`