Question

A projectile of mass 100 kg is lunched from a tower 100 m above the ground in horizontal direction with initial velocity 100m/s. At some time, explosion splits the projectile in two pieces. One piece, of mass 60 kg is observed to hit the ground 400 m away from the tower after 4 s. Find the location of the second fragment at the same time.

Answer 1

The projectile will hits the ground travelling a vertical distance 100 m (if there is no explsion) after a time, t.

y = u_y + 1/2 * a*t²

-100 = 0 - 1/2 * g*t² (because velocity in the vertical direction is zero, u_y = 0)

t = Sqrt [200/9.81] = 4.515 s

The horizontal distance travelled by the projectile (if there is no explosion) is given by

x = u_x t

x_cm = 100 * 4.515 = 451.5 m (Since u_x = 100 m/s)

This is the location of center of mass of the two fragments is at x_cm = 451.5 m

If the position of fragment with mass m₁ = 60 kg is x₁ = 400 m and position of fragment with mass m₂ = 40 kg is x₂

x_cm = [m₁x₁ + m₂x₂] / [m₁ + m₂]

451.5 = [60*400 + 40*x₂] / [ 60 + 40 ]

x₂ = 528.75 m

Location of second fragment is 528.75 m