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

Solution

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