question_answer 34)

                  A particle of mass m is moving in yz-plane with a uniform velocity \[\upsilon \] with its trajectory running parallel to +ve y-axis and intersecting z-axis at \[z=a\] (Fig.). The change in its angular momentum about the origin as it bounces elastically from a wall at \[y=\] constant is: (a) \[m\upsilon a\,\,\hat{e}\,x\]                 (b) \[2m\upsilon a\,\,\hat{e}x\] (c) \[ym\upsilon \,\,\hat{e}x\]              (d) \[2ym\upsilon \,\,\hat{e}x\]                

Answer:

                  (b) Change in momentum, \[\Delta \vec{p}\,=\,2mv\,\hat{e}y\]                 \[\vec{r}\,=\,y\hat{e}y\,+\,a\hat{e}z\]                 \[\therefore \]\[\vec{L}\,=\vec{r}\,\times \,\Delta \vec{p}\,\,=\,(y\hat{e}{{ & }_{y}}+\,a{{\hat{e}}_{z}})\,\times \,2\,mv{{\hat{e}}_{y}}\]                 \[=2mv\,\,a\,{{\hat{e}}_{x}}\]               \[(\because \,\hat{e} & {{ & }_{z}}\times \,{{\hat{e}}_{y}}={{\hat{e}}_{x}})\]