Question

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 origin as it bounces elastically from a wall at y = constant is.

• A. $mva{\widehat{e}}_{X}$
• B. $2mva{\widehat{e}}_{X}$
• C. $ymv{\widehat{e}}_{X}$
• D. $2ymv{\widehat{e}}_{X}$

Answer 1

Answer: (B)

$2mva{\widehat{e}}_{X}$

Answer 2

The initial velocity is

${\overset{\rightarrow}{v}}_{i} = v{\widehat{e}}_{y}$

After reflections from the wall, the final velocity is ${\overset{\rightarrow}{v}}_{f} = - v{\widehat{e}}_{y}$

The trajectory is given as

$\overset{\rightarrow}{r} = y{\widehat{e}}_{y} + a{\widehat{e}}_{z}$

Hence, the change in angular momentum is

$\Delta\overset{\rightarrow}{L} = \overset{\rightarrow}{r} \times m({\overrightarrow{v}}_{f} - {\overrightarrow{v}}_{i})$

$= (y{\widehat{e}}_{y} + a{\widehat{e}}_{z}) \times ( - mv{\widehat{e}}_{y} - mv{\widehat{e}}_{y}) = (y{\widehat{e}}_{y} + a{\widehat{e}}_{z}) \times ( - 2mv{\widehat{e}}_{y})$

$= 2mva{\widehat{e}}_{x}\lbrack\therefore{\widehat{e}}_{y} \times {\widehat{e}}_{y} = 0and{\widehat{e}}_{z} \times {\widehat{e}}_{y} = - {\widehat{e}}_{x}\rbrack$