Question

A particle of mass $m$ is projected from the ground with initial linear momentum $p$ (magnitude) so as to have maximum possible range. Its minimum kinetic energy will be

• A. $\frac{p^2}{2m}$
• B. $\frac{p^2}{4m}$
• C. $\frac{p^2}{m}$
• D. $\frac{p^2}{6m}$

Answer 1

The correct option is B $\frac{p^2}{4m}$


Range will be maximum when angle of projection is ${45}^{\circ }$. Let $u$ be the initial velocity with which the particle is projected.
Initial linear momentum is given by $p=mu$
At the highest point, velocity is minimum so linear momentum will also be minimum.
$v_{min}=u\cos {45}^{\circ }\Rightarrow p_{min}=\frac{mu}{\sqrt{2}}=\frac{p}{\sqrt{2}}{K}_{min}=\frac{p_{min}^2}{2m}=\frac{(\frac{p}{\sqrt{2}}{)}^2}{2m}=\frac{p^2}{4m}$
Hence, the correct answer is option (b)