Question

A particle of mass $m$ is moving in a circle with uniform speed $v$. In moving from a point to another diametrically opposite point,

• A. the momentum changes by $mv$
• B. the momentum changes by $2mv$
• C. the kinetic energy changes by $\left(\frac{1}{2}\right)m{v}^2$
• D. the kinetic energy changes by $m{v}^2$

Answer 1

The correct option is B the momentum changes by $2mv$

Let the particle of mass $m$ be moving in anticlockwise sense and $A$ and $B$ are the diametrically opposite points as shown in the figure.


Now we have
initial momentum of the particle at $A$ is ${\overrightarrow{p}}_i=mv\hat{j}$
and final momentum of the particle at $B$ is ${\overrightarrow{p}}_f=-mv\hat{j}$
So, change in momentum is ${\overrightarrow{p}}_f-{\overrightarrow{p}}_i=(-mv\hat{j})-\left(mv\hat{j}\right)=-2mv\hat{j}$

Also, we have
Initial kinetic energy $K.E_i=\frac{1}{2}m{v}^2$
Final kinetic energy $K.E_f=\frac{1}{2}m{v}^2$
So, change in kinetic energy is $K.E_f-K.E_i=0$

Hence, momentum changes by $2mv$ in magnitude but kinetic energy remains same.