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

         

1. \(mva~\hat e_{x}\) 2. \(2mva~\hat e_{x}\)
3. \(ymva~\hat e_{x}\) 4. \(2ymva~\hat e_{x}\)
(b) Hint: After rebounding, only the direction of angular momentum changes.
Step 1: Find the initial and final angular momentum.
The initial velocity is vi=ve^y, and after reflection from the wall, the final velocity is vt=ve^y. The trajectory is described as position vector r=ye^y+ae^z.
Step 2: Find the change in angular momentum.The magnitude of the angular momentum
vector is
Hence, in angular momentum is r×m(vfvj)=2mvaex.