Question

Model rocket engines are rated by the impulse that they deliver when they fire. A particular engine is rated to deliver an impulse of 3.5 kg⋅m/s. The engine powers a 120g rocket, including the mass of the engine. Part A What is the final speed of the rocket once the engine has fired? (Ignore the change in mass as the engine fires and ignore the weight force during the short duration of the engine firing.)

Answer 1

Step-by-step explanation:

The given data is as follows.

J = 3.5 kg-m/s, m = 120 g = 0.120 kg (as 1 g = 0.001 kg)

It is known that formula to calculate impulse is as follows.

J =
`m(v_(2) - v_(1))`

`0.120(v_(2) - v_(1))` = 3.5

`v_2 - v_1` = 29 m/s

So,
`v_(2) = 29 + v_(1)`

= 29 + 0 = 29 m/s

Thus, we can conclude that final speed (
`v_(2)`) is 29 m/s.

Answer 2

The final speed of the rocket once the engine has fired is 29.16 m/s.

Step-by-step explanation:

By the principle of momentum and amount of movement, we match the momentum data to the amount of movement and find the value of speed.

I=3.5 kg*m/s

m=120g= 0.12kg

V=?

I=F*t

F*t=m*V

I/m=V

(3.5 kg*m/s) / 0.12kg = V

V=29.16 m/s