A particle of mass $m$ moves in a circle of radius $R$ with a constant speed $v$. The change in momentum when the particle moves through an angle of $90^{\circ }$ is:

Solution

1. Initial momentum is ${\vec{p}}_i=mv\hat{i}$.
2. Final momentum after 90 degrees is ${\vec{p}}_f=mv\hat{j}$.
3. The change in momentum is $$\begin{array}{rl}\Delta \vec{p}& ={\vec{p}}_f-{\vec{p}}_i\\ & =mv(\hat{j}-\hat{i})\end{array}$$.
4. The magnitude is $$\begin{array}{rl}|\Delta \vec{p}|& =\sqrt{(mv{)}^2+(-mv{)}^2}\\ & =\sqrt{2}mv\end{array}$$.
5. Hence the answer is (B).