Question

A projectile is fired at angle is fired with velocity v₀ at an angle 60⁰ with the horizontal. At top of its trajectory, it explodes into two fragments of equal mass. If one fragment retraces the path then the speed of the other fragment is?

Answer 1

Initial horizontal velocity = v₀cosθ = v₀/2

According to principle of conservation of momentum:

Initial horizontal momentum= Final horizontal momentum

(we take horizontal velocity because it stays constant throughout projectile motion)

$P_i = P_f$

Since one mass retraces its path, it travels back along the same path, with the same velocity.  Its momentum is: mv₀cosθ. Since it travels in opposite direction, the sign becomes -ve

Thus,

$2m\frac{v_0}{2} = -m\frac{v_0}{2} + mv'\\\\v' = \frac{3v_0}{2}$