Question

A grenade that is falling vertically explodes into two equal fragments when it is at a height of 2000 m and has a downward velocity of 60 m/s. Immediately after the explosion one of the fragments is moving downward at 80 m/s. Find the position of the second fragment and the position of the center of mass of the system 10 s after the explosion.

Answer 1

There is no new force on the center of mass of the system so it keeps accelerating downward at 9.8 m/s^2 so do the second part first.
h = 2000 - 60(10) - 4.9(100)

First part, let m1 = 1 kg and m2 = 1 kg
before
p = (m1+m2)60 = 120 kg m/s
after
p = m1 V1 + m2 V2 = V1 + 80
p before = p after
120 = V1 + 80
V1 = 40 m/s
so
we have m1 at 2000 meters moving at 40 m/s down and accelerating down at 9.8 m/s^2

h = 2000 - 40(10) - (4.9)(100)

h = 2000 - 400 - 490

Answer 2

thank you so much !