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A particle of mass m is projected from the ground with an initial speed u at an angle a with the horizontal. At the highest point of its trajectory it makes a complete inelastic collision with another identical particle which was thrown upward from ground with same intial speed u. The angle thatthe compoite system make with the horizontal immediately after collision is?

A particle of mass m is projected from the ground with an initial speed u at an angle a with the horizontal. At the highest point of its trajectory it makes a complete inelastic collision with another identical particle which was thrown upward from ground with same intial speed u. The angle thatthe compoite system make with the horizontal immediately after collision is?
 

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1 Answers

Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
9 years ago
Dear student,
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 )
Regards

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