Question

A particle of mass $1\text{kg}$ is projected from the ground with an initial speed $u$ at an angle ${60}^{\circ }$ with the horizontal. At the highest point of its trajectory, it makes a complete inelastic collision with another identical particle, which was thrown vertically upwards from the ground with the same initial speed $u$. The angle that the composite system makes with the horizontal just after collision in degrees is .

Answer 1

Velocity of the particle at the topmost point:
$V_1=u\cos {60}^{\circ }=\frac{u}{2}$ (In X-direction)


$H_{max}=\frac{u^2{\sin }^2\theta }{2g},\text{where}\theta ={60}^{\circ }$

Velocity of $2^{nd}$ particle just before collision:
$V{{}_2}^2=u^2-2\left(g\right)\left(\frac{u^2{\sin }^2\theta }{2g}\right)[\because S=H_{max}]$
$V{{}_2}^2=u^2-u^2{\sin }^2\theta$
$V{{}_2}^2=u^2(1-{\sin }^2\theta )=\frac{u^2}{4}\Rightarrow V_2=\frac{u}{2}$ (Along Y-axis)

In inelastic collision, both the particle will stick together.
From conservation of Linear momentum:
$P_i=P_f$
$\Rightarrow (P_i{)}_x=(P_f{)}_x\Rightarrow mu\cos {60}^{\circ }+0=2mu{{}_x}^{{}^{\prime }}$
$\Rightarrow u_x^{{}^{\prime }}=\frac{u}{4}$

$(P_i{)}_y=(P_f{)}_y\Rightarrow 0+m\frac{u}{2}=\left(2m\right)u_y^{{}^{\prime }}$
$\Rightarrow u_y^{{}^{\prime }}=\frac{u}{4}$
$\tan \alpha =\frac{u_y^{{}^{\prime }}}{u_x^{{}^{\prime }}}=\frac{\frac{u}{4}}{\frac{u}{4}}=1$
$\alpha ={45}^{\circ }$