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