Question

Explanation A 5000 kg rocket is at rest in deep space. The rocket burns fuel pushing 10kg of exhaust gases rearward at 4000 m/s. This process takes 10 s. During the time interval what average force is applied to the rocket?

Answer 1

4000 N

Step-by-step explanation:

mass of fuel, m = 10 kg

velocity of ejected fuel, v = 4000 m/s

time, t = 10 s

force, F = velocity x mass of fuel burning per second

F = 4000 x 10 / 10 = 4000 N

Thus, the force is 4000 N.

Answer 2

F = 4000 N

Step-by-step explanation:

given,

mass of rocket (M)= 5000 Kg

10 Kg gas burns at speed (m)= 4000 m/s

time = 10 s

average force = ?

at the end the rocket is at rest

by conservation of momentum

M v + m v' = 0

5000 x v - 10 x 4000 = 0

5000 v = 40000

v = 8 m/s

speed of rocket = 8 m/s

now,

we know

change in momentum = F x Δ t

`F = (m(v_i-v_f))/(\Delta t)`

`F = (5000(8-0))/(10)`

F = 4000 N

Hence, the average force applied to the rocket is equal to F = 4000 N