Question

A particle of mass m is projected from the ground with an initial speed u not at an angle alpha with horizontal at the highest point of 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 not the angle that the composite system makes with the horizontal immediately after the collision is

Answer 1

At the highest point, the projectile would be moving horizontal with a speeducos\alpha. Hence after inelastic collision, we have:

mucos\alpha=2mvcos\theta;

m\left ( u-g\left ( \frac{usin\alpha}{g} \right ) \right )=mu(1-sin\alpha)=2mvsin\theta

Here, we assume that after collision, the combination starts moving such that it makes an angle theta with the horizontal and the final velocity being v.

Thus, we get:

\tan\theta=\frac{\left ( 1-\sin\alpha \right )}{\cos\alpha}\Rightarrow \theta=tan^{-1}\left (\frac{\left ( 1-\sin\alpha \right )}{\cos\alpha}\right )

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