Question

1460 kg weather rocket accelerates upward at 12 m/s2. It explodes 2.0 s after liftoff and breaks into two fragments, one twice as massive as the other. Photos reveal that the lighter fragment traveled straight up and reached a maximum height of 530 m. What were the speed and direction of the heavier fragment just after the explosion?

Answer 1

M = 1460
Let the two fragments be , m1 & m2
Given , m1 = 2 * m2
m1 + m2 = 1460 kg
2 m2 + m2 = 1460
m2 = 1460/3 = 486.7 kg
m1 = 973.4 kg


vf = vi + a*t
vf = 0 + 12 * 2.0
vf = 24.0 m/s

h = 1/2*12*2.0^2
h = 24 m

vf^2 = vi^2 + 2*a*h
0 = vi^2 - 2*9.8*(530 - 24)
vi = 99.58 m/s

Now,Using momentum conservation,
1460 * 24 = m1 * v1 + m2 * v2
1460 * 24 = 973.4 * v1 + 486.7 * 99.58
v1 = - 13.8 m/s

So,
Speed of the heavier fragment = 13.8 m/s
Direction = Downwards !!