Question
A particle of mass $$m$$ is projected from the ground with an initial speed $${u_0}$$ at an angle $$\alpha $$ with the horizontal. 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.
$$\frac{\pi }{4}$$
B.
$$\frac{\pi }{4} + \alpha $$
C.
$$\frac{\pi }{2} - \alpha $$
D.
$$\frac{\pi }{2}$$
Answer :
$$\frac{\pi }{4}$$
Solution :
Activity B to M for particle thrown upwards
$$\eqalign{ & v_1^2 - u_0^2 = 2( - g)\quad \left[ {\frac{{u_0^2{{\sin }^2}\alpha }}{{2g}}} \right] \cr & \therefore v_1^2 = u_0^2\left( {1 - {{\sin }^2}\alpha } \right) = u_0^2{\cos ^2}\alpha \cr & \therefore {v_1} = {u_0}\cos \alpha \,......\left( {\text{i}} \right) \cr} $$
Applying conservation of linear momentum in $$Y$$-direction
$$2mv\sin \theta = m{v_1} = m{u_0}\cos \alpha \,......\left( {{\text{ii}}} \right)\,[{\text{from}}\,\left( {\text{i}} \right)]$$
Applying conservation of linear momentum in $$X$$-direction
$$2mv\cos \theta = m{u_0}\cos \alpha \,.....\left( {{\text{iii}}} \right)$$
on dividing (ii) and (iii) we get
$$\tan \theta = 1\,\therefore \theta = \frac{\pi }{4}$$
Activity B to M for particle thrown upwards
$$\eqalign{ & v_1^2 - u_0^2 = 2( - g)\quad \left[ {\frac{{u_0^2{{\sin }^2}\alpha }}{{2g}}} \right] \cr & \therefore v_1^2 = u_0^2\left( {1 - {{\sin }^2}\alpha } \right) = u_0^2{\cos ^2}\alpha \cr & \therefore {v_1} = {u_0}\cos \alpha \,......\left( {\text{i}} \right) \cr} $$
Applying conservation of linear momentum in $$Y$$-direction
$$2mv\sin \theta = m{v_1} = m{u_0}\cos \alpha \,......\left( {{\text{ii}}} \right)\,[{\text{from}}\,\left( {\text{i}} \right)]$$
Applying conservation of linear momentum in $$X$$-direction
$$2mv\cos \theta = m{u_0}\cos \alpha \,.....\left( {{\text{iii}}} \right)$$
on dividing (ii) and (iii) we get
$$\tan \theta = 1\,\therefore \theta = \frac{\pi }{4}$$
