Question

A particle of mass $m$ is projected from the ground with initial linear momentum $p$ (magnitude) 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. $\frac{p^2}{3m}$

Answer 1

Answer: (B)

At the time of projection $\theta =45^{\circ }$ and momentum $p=mu$
$\Rightarrow u=\frac{p}{m}$
Minimum $KE$ will be when projectile is at the highest point.
Velocity will be is $u\cos \theta$
$\therefore K_{\min }=\frac{1}{2}m(u\cos \theta {)}^2$
$=\frac{m}{2}\left(\frac{p^2}{m^2}\right)\left(\frac{1}{2}\right)=\frac{p^2}{4m}$