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

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virgil123
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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.
 
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kuruman said:
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.
 
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.
 
kuruman said:
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 said:
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 said:
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 said:
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 said:
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
 
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.
 
kuruman said:
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.