Q. A particle of mass m is projected with a speed $u$ from the ground at an angle $\theta=\frac{\pi}{3}$ w.r.t. horizontal (x-axis). When it has reached its maximum height, it collides completely inelastically with another particle of the same mass and velocity $u\,\hat{i}$. The horizontal distance covered by the combined mass before reaching the ground is :
Solution:
By momentum conservation,
$\frac{mu}{2} + mu = 2mv'$
$v' = \frac{3v}{4}$
Range after collision $= \frac{3v}{4}\sqrt{\frac{2H}{g}}$
$= \frac{3v}{4} \sqrt{\frac{2\cdot u^{2} \,sin^{2} \,60^{\circ}}{g2g}}$
$=\frac{3}{4} \frac{\sqrt{3}}{2}. \frac{u^{2}}{g} = \frac{3\sqrt{3}u^{2}}{8g}$
