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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:

$(a)\; mva\; \hat{e}_{x}$

$(b)\; 2mva\; \hat{e}_{x}$

$(c)\; ymv\; \hat{e}_{x}$

$(d)\;2 ymv\; \hat{e}_{x}$

Answers (1)

The answer is the option (b). Angular momentum of an object is given by $\vec{L}=\vec{r}\times m\vec{v}$ and direction by Right-hand thumb rule

Now, $\vec{r}=y\; \widehat{e}_{y}+a\; \widehat{e}_{z}$

$V_{i}=v\; \widehat{e}_{y}\; \; \; \; \; \; \; \; \; \; \; V_{f}=-v\; \widehat{e}_{y}$

$\vec{L}=\vec{r}\times \vec{p}$

As the $\vec{r}$ , $\vec{v}$ and $\vec{p}$ are in plane of $y-z,$ L will be in the plane of $+x$

$L=(y\; \widehat{e}_{y}+a\; \widehat{e}_{z})\times m(v_{f}-v_{i})$

$L=(y\; \widehat{e}_{y}+a\; \widehat{e}_{z})\times m (-v\; \widehat{e}_{y}-v\; \widehat{e}_{y})=(y\; \widehat{e}_{y}+a\; \widehat{e}_{z})\times m (-2\; v\; \widehat{e}_{y})$

$y\widehat{e}\times \widehat{e_{y}}=y\; \sin\; 0=0$

$a\; \widehat{e_{z}}\times \widehat{e_{y}}=a\; \sin\; (90)(-\widehat{e_{x}})$

$L=-a\; \widehat{e_{x}}mv(-2)=2amv\widehat{e_{x}}$

Hence option b is the correct answer.

Posted by

infoexpert23

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