Question

A particle of mass m is projected from the ground with an initial speed $u_0$ at an angle $\alpha$ with the horixontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed $u_0$. The angle that the composite system makes with the horizontal immediately after the collision is

• A. $\frac{\pi }{4}+\alpha$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{2}-\alpha$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is D $\frac{\pi }{4}$


Speed of first particle at highest point = $u_0\cos \alpha$
Speed of second particle at highest point =$\sqrt{u_0^2-2gH}$
Now, $H=\frac{u_0^2{\sin }^2\alpha }{2g}$
$\Rightarrow$ Speed of $2^{nd}$ particle =$u_0\cos \alpha$
$\therefore$ Final momentum = $m{u}_0\cos \alpha \hat{j}+m{u}_{0}cos\alpha \hat{j}$
$\therefore \text{Angle}=\frac{\pi }{4}$