Ncert Exercises & Solutions

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

v_{i} = v {\hat{e}}_{y}

, and after reflection from the wall, the final velocity is

v_{t} = - v {\hat{e}}_{y}

. The trajectory is described as position vector

r = y {\hat{e}}_{y} + a {\hat{e}}_{z}

Step 2: Find the change in angular momentum.The magnitude of the angular momentum
vector is
Hence, in angular momentum is

r \times m (v_{f} - v_{j}) = 2 {mvae}_{x}

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