Ncert Exercises & Solutions

A particle of mass m is moving in yz-plane with a uniform velocity v with its trajectory running parallel to the +ve y-axis and intersecting z-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. mvae^x
2. 2mvae^x
3. ymvae^x
4. 2ymvae^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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