How to find rocket thrust, force impulse of energy method?

[virgil123]

Homework Statement

The 3rd and 4th stages of a rocket, mass 40kg and 60kg respectively, coast in space with a velocity of 15,000kmh when the 4th stage ignites, with thrust T and causes separation. If the velocity of the 4th stage is 10m/s greater than the 3rd stage at the end of the 0.5 second separation interval calculate the average thrust during this period. Assume the entire blast impinges against the third stage with a force equal and opposite to T.

Homework Equations

Ft=m(v₂-v₁)

(m₁u₁+m₂u₂) = (m₁v₁+m₂v₁)

The Attempt at a Solution

I have used the ft=m(v₂-v₁) formula, answer 2000N, I don't believe this is correct.

 

[kuruman]

I have used the ft=m(v₂-v₁) formula, answer 2000N, I don't believe this is correct.

You believe correctly. Can you show how you got 2000 N?

 

[virgil123]

You believe correctly. Can you show how you got 2000 N?

Yes, I was afraid of that!

F=m(v2-v1)/t
F=100(10)/0.5

Seems far to simplistic.

 

[kuruman]

In equation F = m Δv/Δt symbol "m" stands for the mass of one of the objects and "Δv" stands for the change in its velocity. You cannot use m for the sum of the masses and Δv for the difference in their velocities. Newton's Third law in this context says

m₁(Δv₁/Δt) + m₂(Δv₂/Δt) = 0

Start from there.

 

[virgil123]

In equation F = m Δv/Δt symbol "m" stands for the mass of one of the objects and "Δv" stands for the change in its velocity. You cannot use m for the sum of the masses and Δv for the difference in their velocities. Newton's Third law in this context says

m₁(Δv₁/Δt) + m₂(Δv₂/Δt) = 0

Start from there.

Wouldn't Δv₁ = 0, hence making the LHS = 0?

 

[kuruman]

Wouldn't Δv₁ = 0, hence making the LHS = 0?

Why would it? The two stages push against each other. Both their velocities must change; you cannot change the velocity of one but not the other.

 

[virgil123]

Why would it? The two stages push against each other. Both their velocities must change; you cannot change the velocity of one but not the other.

I am a bit lost here, before the release they are traveling together at the same velocity, which is 4166.67 m/s. After the release stage 4 is 10m/s faster than stage 3 after 0.5 seconds, so does Δv1 refer to after the explosion?

I think I am getting confused with, (m1u1+m2u2) = (m1v1+m2v1)

 

[kuruman]

OK. Say m₁ = 60 kg (stage 4) and m₂ = 40 kg (stage 3). If both stages have the same initial velocity v₀ (=15000 km/hr), you can write their final velocities as
v₁ = v₀ + Δv₁
v₂ = v₀ + Δv₂

You are told that the difference v₁ - v₂ is 10 m/s. Can you find a relation between Δv₁ and Δv₂?

 

[virgil123]

OK. Say m₁ = 60 kg (stage 4) and m₂ = 40 kg (stage 3). If both stages have the same initial velocity v₀ (=15000 km/hr), you can write their final velocities as
v₁ = v₀ + Δv₁
v₂ = v₀ + Δv₂

You are told that the difference v₁ - v₂ is 10 m/s. Can you find a relation between Δv₁ and Δv₂?

I will attempt to tackle this tomorrow. Thank you very much for your help, I have to get the kids to bed now.

 

[virgil123]

I will attempt to tackle this tomorrow. Thank you very much for your help, I have to get the kids to bed now.

All I can think of is Newton's third law, equal and opposite reactions.
Therefore for a 10m/s difference stage 3 would be repulsed by 5m/s whilst stage 4 would gain 5m/s hence the 10m/s difference.
Putting this into the equation would give a thrust of 83833.34n.

 

[kuruman]

Look at the two equations I gave you in posting #8. What do you get when you subtract the bottom equation (for v₂) from the top equation (for v₁)?

 

[virgil123]

Look at the two equations I gave you in posting #8. What do you get when you subtract the bottom equation (for v₂) from the top equation (for v₁)?

Not sure I follow, please excuse me, but v1-v2 = 10

 

[kuruman]

Correct. Now look at the other side of the equation. If you put it together, you get
Δv₁-Δv₂ = 10.
You also have
m₁(Δv₁/Δt) + m₂(Δv₂/Δt) = 0
which can be simplified to
m₁Δv₁ + m₂Δv₂ = 0
which is really the momentum conservation equation. You have a system of two equations and two unknowns. Solve it to find Δv₁ and then use
T₁ = m₁(Δv₁/Δt)
to find the thrust T₁ on m₁. You could just as well find Δv₂ and follow the same method to find the thrust T₂ on m₂. It should be no surprise to see that T₁+T₂=0 in agreement with Newton's Third Law.

 

[virgil123]

Correct. Now look at the other side of the equation. If you put it together, you get
Δv₁-Δv₂ = 10.
You also have
m₁(Δv₁/Δt) + m₂(Δv₂/Δt) = 0
which can be simplified to
m₁Δv₁ + m₂Δv₂ = 0
which is really the momentum conservation equation. You have a system of two equations and two unknowns. Solve it to find Δv₁ and then use
T₁ = m₁(Δv₁/Δt)
to find the thrust T₁ on m₁. You could just as well find Δv₂ and follow the same method to find the thrust T₂ on m₂. It should be no surprise to see that T₁+T₂=0 in agreement with Newton's Third Law.

Many thanks.