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

Options

A. `mva hate_(x)`
B. `2 mva hate_(x)`
C. `ymv hate_(x)`
D. ` 2 ymv hate_(x)`

Answer

B

Explanation

the intial velocity is `v_(1) = V hate _(y)` and after refectio from the wall the final velocity is `V_(1)=- v hate_(y)` . The trajectory is described as position vector `r=yhate_(y)+ ahate_(z)`
Hence , the change in angular momentum is `rxxm(v_(f)-v_(j))=2 mva hate _(x)`.