Question 254: Work, Energy and Power & Circular Motion

Question

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

Accepted Answer

At both points, the momentum will be $\mathrm{mv}$, but will be in opposite directions.
$\begin{array}{rlrl}
	herefore & \Delta \mathrm{P}  =\mathrm{mv}-(-\mathrm{mv}) \
	herefore & \Delta \mathrm{P}=2 \mathrm{~mv}
\end{array}$
The momentum changes by $2 \mathrm{~mv}$.

Answer

the momentum changes by $\mathrm{mv}$

Answer

the momentum changes by $2 \mathrm{~mv}$

Answer

the kinetic energy changes by $\frac{1}{2} \mathrm{mv}^2$

Answer

the kinetic energy changes by $\mathrm{m v^2}$

Work, Energy and Power & Circular Motion

A particle is moving in a circle with uniform speed 'v'. In moving from a point to another diametrically opposite point

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