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

A particle of mass m is moving in y z - 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 origin as it bounces elastically

from a wall at y = constant is :

(1) m v a {\overset{\land}{e}}_{x}

(2) 2 m v a {\overset{\land}{e}}_{x}

(3) y m v {\overset{\land}{e}}_{x}

(4) 2 y m v {\overset{\land}{e}}_{x}

Commented by Tinkutara last updated on 26/Nov/17

Answered by ajfour last updated on 26/Nov/17

\bar{l} = \bar{r} \times \bar{p}

△ \bar{l} = \bar{r} \times △ \bar{p}

= (y \hat{j} + a \hat{k}) \times (- 2 m v \hat{j})

= 2 m v a \hat{i} \equiv 2 m v a {\hat{e}}_{x}

o p t i o n (2) .

Commented by Tinkutara last updated on 26/Nov/17

B u t w h a t i s {\overset{\land}{e}}_{x} ?

Commented by Tinkutara last updated on 26/Nov/17

I s i t s a m e a s \overset{\land}{i} \overset{}{?}

Commented by ajfour last updated on 26/Nov/17

y e s .

Commented by Tinkutara last updated on 26/Nov/17

Thank you Sir!