# A particle of mass m strikes a wall at an angle of incidence 60° with velocity V elastically. Then the change in momentum is?

Apr 18, 2018

$| \Delta m a t h b f p | = 2 m V \sin 60$

#### Explanation:

Check what is meant by "incident" angle. See drawing for one interpretation. Swap 30 for 60 if it's the optics definition.

For an elastic collision we can say that:

• $m a t h b f {p}_{i} = m V \left(\begin{matrix}\sin 60 \\ \cos 60\end{matrix}\right)$

and

• $m a t h b f {p}_{f} = m V \left(\begin{matrix}- \sin 60 \\ \cos 60\end{matrix}\right)$

So, kinetic energy is preserved because $| m a t h b f {V}_{i} | = | m a t h b f {V}_{f} |$

$\implies | \Delta m a t h b f p | = | m V \left(\begin{matrix}- 2 \sin 60 \\ 0\end{matrix}\right) | = 2 m V \sin 60$