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-The-change-in-its-angular-momentum-about-the-

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Question Number 24786 by Tinkutara last updated on 26/Nov/17

$$\mathrm{A}\mathrm{particle}\mathrm{of}\mathrm{mass}m\mathrm{is}\mathrm{moving}\mathrm{in}yz-\mathrm{plane}$$$$\mathrm{with}\mathrm{a}\mathrm{uniform}\mathrm{velocity}v\mathrm{with}\mathrm{its}$$$$\mathrm{trajectory}\mathrm{running}\mathrm{parallel}\mathrm{to}+\mathrm{ve}y-$$$$\mathrm{axis}\mathrm{and}\mathrm{intersecting}z-\mathrm{axis}\mathrm{at}z=a.$$$$\mathrm{The}\mathrm{change}\mathrm{in}\mathrm{its}\mathrm{angular}\mathrm{momentum}$$$$\mathrm{about}\mathrm{the}\mathrm{origin}\mathrm{as}\mathrm{it}\mathrm{bounces}\mathrm{elastically}$$$$\mathrm{from}\mathrm{a}\mathrm{wall}\mathrm{at}y=\mathrm{constant}\mathrm{is}:$$$$\left(1\right)mva{\stackrel{\wedge }{e}}_x$$$$\left(2\right)2mva{\stackrel{\wedge }{e}}_x$$$$\left(3\right)ymv{\stackrel{\wedge }{e}}_x$$$$\left(4\right)2ymv{\stackrel{\wedge }{e}}_x$$

Commented by Tinkutara last updated on 26/Nov/17

Answered by ajfour last updated on 26/Nov/17

$$\overline{l}=\overline{r}\times \overline{p}$$$$△\overline{l}=\overline{r}\times △\overline{p}$$$$=(y\hat{j}+a\hat{k})\times (-2mv\hat{j})$$$$=2mva\hat{i}\equiv 2mva{\hat{e}}_x$$$$option\left(2\right).$$

Commented by Tinkutara last updated on 26/Nov/17

$$Butwhatis{\stackrel{\wedge }{e}}_x?$$

Commented by Tinkutara last updated on 26/Nov/17

$$Isitsameas\stackrel{\wedge }{i}\stackrel{}{?}$$

Commented by ajfour last updated on 26/Nov/17

$$yes.$$

Commented by Tinkutara last updated on 26/Nov/17

$$\mathrm{Thank}\mathrm{you}\mathrm{Sir}!$$

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