Question

a 1500kg weather rocket is launched upward from the ground and accelerates at 10m/s it explodes 4s after lifeoff 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

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,

$M v = m u + m' v_h$

$M at = \dfrac{M}{3}u + \dfrac{2M}{3}v_h$

$at = \dfrac{u}{3}+ \dfrac{2v_h}{3}$

$10\times 4 = \dfrac{\sqrt{2gh}}{3}+ \dfrac{2v_h}{3}$

$120= \sqrt{2\times 9.8 \times 530}+2v_h$

   2 v_h = 18

        v_h = 9 m/s

speed of the heavy fragment is equal to 9 m/s