Answer:
v_h = 9 m/s
Explanation:
given,
mass of the rocket = 1500 kg
accelerates = 10 m/s²
time = 4 s
one fragment is twice as massive as the other.
maximum height reached by the lighter fragment = 530 m
now,
using equation of motion to calculate the velocity
initial velocity of rocket,u = 0 m/s
v = u + a t
v = at.......(1)
total mass of the rocket
M = m + m'
m' is the heavier particle mass.
m' = 2 m (from the statement given in the question)
M = 3 m
m = M/3
now, again using equation of motion to calculate the initial velocity of the lighter particle.
v² = u² + 2 a s
0² = u² + 2 g h
u = √2gh.....(2)
now,
using conservation of momentum,
2 v_h = 18
v_h = 9 m/s
speed of the heavy fragment is equal to 9 m/s
