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Q.

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 as shown in figure. The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is  [NCERT Exemplar)

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a

mvae^x

b

2mvae^x

c

ymve^x

d

2ymve^x

answer is B.

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Detailed Solution

The initial velocity is  vi=ve^y  and after reflection from the wall, the final velocity is  vf=-ve^y The trajectory is described as position vector  r=ye^y+ae^zHence, the change in angular momentum=r×mvf-vi=ye^y+ae^z×-mve^y-mve^y=y^y+ae^z×-2mve^y  ∵e^y×e^y=0 and e^z×e^y=-e^x=2mvae^x
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