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. `mvahat(e)_(x)`
B. `2mvahat(e)_(x)`
C. `ymvhat(e)_(x)`
D. `2ymvhat(e)_(x)`

Answer

B

Explanation

The initial velocity is `v_(i)=vhat(e)_(y)` and after reflection from the wall, the final velocity is `v_(f)=-vhat(e)_(y)` . The trajectory is described as position vector `vec(r)=yhat(e)_(y)+ahat(e)_(z)`
Hence, the change in angular momentum is `vec(r)xxm(v_(f)-v_(j))=2mvahat(e)_(x)` .