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,
[tex]M v = m u + m' v_h[/tex]
[tex]M at = \dfrac{M}{3}u + \dfrac{2M}{3}v_h[/tex]
[tex]at = \dfrac{u}{3}+ \dfrac{2v_h}{3}[/tex]
[tex]10\times 4 = \dfrac{\sqrt{2gh}}{3}+ \dfrac{2v_h}{3}[/tex]
[tex]120= \sqrt{2\times 9.8 \times 530}+2v_h[/tex]
2 v_h = 18
v_h = 9 m/s
speed of the heavy fragment is equal to 9 m/s
