Sarthaks Test
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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 (Fig. 7.2). The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is:

(a) mva êx 

(b) 2mva êx 

(c) ymv êx 

(d) 2ymv êx

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The initial velocity is vi = vêy and, after reflection from the wall, the final velocity is vf = -vêy . The trajectory is described as r = yêy + aêz . Hence the change in angular momentum is r x m(vf - vi) = 2mvaêx.

Hence the answer is (b).

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