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प्रश्न
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 (Figure). The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is ______.
पर्याय
`mva hate_x`
`2mva hate_x`
`ymv hate_x`
`2ymv hate_x`
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उत्तर
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 (Figure). The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is `underline(2mva hate_x)`.
Explanation:
The initial velocity is `vhati = vhate_y` and, after reflection from the wall, the final velocity is `v_f = - vhate_y`. The trajectory is described as `r = yhate_y + ahate_z`. Hence the change in angular momentum is `r xx m(v_f - v_i) = 2mvahate_x`.
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