A projectile of mass 6 kg is fired with an initial speed of

A projectile of mass 6 kg is fired with an initial speed of 67 m/s at an angle of 24 degree with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 3.5 kg and 2.5 kg. The 2.5 kg fragment lands on the ground directly below the point of explosion 1.6 s after the explosion. The acceleration due to gravity is 9.81 m/s^2. Find the magnitude of the velocity of the 3.5 kg fragment immediately after the explosion.

Solution

p_ix = p_fx

m₁ v_1x = m₂ v_2x + m₃ v_3x

where   v_1x= v_ox= v_o sin 24

v_3x= –v_1x= –v_o sin 24

6 kg (67 m/s * sin 24) = -2.5 kg (67 m/s * sin 24) + 3.5 v_2x

v_2x= 66.22 m/s

However, this speed is relative to original piece

(v_2x)_ground = v_2x+ v_1x= 66.22 m/s + (67 m/s * sin 24)  

v_2x= 93.49 m/s

(v_3x)_ground= v_3x+ v_1x= 0