- Home
- Standard 11
- Physics
5.Work, Energy, Power and Collision
medium
An object of mass $m$ is projected with a momentum $p$ at such an angle that its maximum height is $\frac{1}{4}$ th of its horizontal range. Its minimum kinetic energy in its path will be
A$\frac{p^2}{8\,m}$
B$\frac{p^2}{4\,m}$
C$\frac{3 p^2}{4\,m}$
D$\frac{p^2}{m}$
Solution
(b)
$H=\frac{R}{4}$
$\therefore \quad \frac{u^2 \sin ^2 \theta}{2\,g}=\frac{2 u^2 \sin \theta \cos \theta}{4\,g}$
$\text { or } \tan \theta =1$
$\therefore \theta =45^{\circ}$
At highest point momentum will remain $\frac{p}{\sqrt{2}}$.
$\therefore \quad K=\frac{(p / \sqrt{2})^2}{2\,m}=\frac{p^2}{4\,m}$
$H=\frac{R}{4}$
$\therefore \quad \frac{u^2 \sin ^2 \theta}{2\,g}=\frac{2 u^2 \sin \theta \cos \theta}{4\,g}$
$\text { or } \tan \theta =1$
$\therefore \theta =45^{\circ}$
At highest point momentum will remain $\frac{p}{\sqrt{2}}$.
$\therefore \quad K=\frac{(p / \sqrt{2})^2}{2\,m}=\frac{p^2}{4\,m}$
Standard 11
Physics
