# A particle is projected with velocity U makes an angle theta with respect to horizontal now It Breaks into two identical parts at the highest point of trajectory 1part retraces its path then velocity of the other part is?

Mar 17, 2018

We know that at the highest point of its motion a projectile has only its horizontal component of velocity i.e $U \cos \theta$

So,after breaking,one part can retrace its pathway if it will have the same velocity after the collsion in the opposite direction.

So,applying law of conservation of momentum,

Initial momentum was $m U \cos \theta$

After the collsion momentum became, $- \frac{m}{2} U \cos \theta + \frac{m}{2} v$ (where,$v$ is the velocity of the other part)

So,equating we get,

$m U \cos \theta = - \frac{m}{2} U \cos \theta + \frac{m}{2} v$

or, $v = 3 U \cos \theta$