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Rat's New Year Dynamics Question
Posted on 2012-12-31
Here is a new Maths question for 2013
A spaceship is moving with constant velocity in deep space when an onboard explosion splits the ship into two.
If the two parts of the ship, masses M and m, still continue in the same direction and the kinetic energy released by the explosion was E, what is the difference in velocity of the two parts?
And a happy new year to all!
A spaceship is moving with constant velocity in deep space when an onboard explosion splits the ship into two.
If the two parts of the ship, masses M and m, still continue in the same direction and the kinetic energy released by the explosion was E, what is the difference in velocity of the two parts?
And a happy new year to all!
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Active today
Assuming classical dynamics..
Change the observable frame of reference to the CM of the spaceship. Then, its velocity is 0, and its KE and P are both 0. After the explosion,
E = KE1 + KE2 = 1/2(M*v1^2 + m*v2^2)
P = 0 = M*v1 + m*v2
---
v1 = -m/M*v2
---
Solve for v1, v2 in terms of E, and answer is |v2 - v1|
Change the observable frame of reference to the CM of the spaceship. Then, its velocity is 0, and its KE and P are both 0. After the explosion,
E = KE1 + KE2 = 1/2(M*v1^2 + m*v2^2)
P = 0 = M*v1 + m*v2
---
v1 = -m/M*v2
---
Solve for v1, v2 in terms of E, and answer is |v2 - v1|
Active 2 days ago
phoffric: I personally would not bother to change any frame of reference, since I think it comes out easier if you don't. But in "Solve for v1, v2 in terms of E, and answer is |v2 - v1|" I expect YOU to solve your own equations (and in fact in terms of E,M and m) and the answer required is the velocity not the speed.
ozo: as always a terse answer. It might be of some interest in how you came about this instead of me having to prove you right or wrong. Alsjeblieft?
ozo: as always a terse answer. It might be of some interest in how you came about this instead of me having to prove you right or wrong. Alsjeblieft?
Active today
After explosion, m moving with velocity A and M with velocity B
m ---> A M ---> B
From the diagram, the speeds are shown as B > A
The difference in velocity is B - A whose direction is pointing right since B > A.
Using CM approach as noted in previous post..
For the magnitude of the difference, B - A:
(1) 0 = mA + MB ==> B = -m/M A
(2) B = -m/M A (Note: we're in 1D, so vector direction is known by +/- sign)
(2A) B^2 = (m/M)^2 A^2
(2B) B-A = -m/M A -A = -(m+M)/M * A
(2C) (B-A)^2 = (m+M)^2/M^2 * A^2
(3) 2E = m A^2 + M B^2
(4) 2E = m A^2 + M (m/M)^2 A^2
(4a) 2E = (M + m) m/M A^2
(4b) A^2 = (M/m)2E/(M+m)
(5) (B-A)^2 = (m+M)^2/M^2 * [ (M/m)2E/(M+m) ]
= 2E (m+M)/(Mm)
(5a) (B-A) = sqrt[ 2E (m+M)/(Mm) ]
m ---> A M ---> B
From the diagram, the speeds are shown as B > A
The difference in velocity is B - A whose direction is pointing right since B > A.
Using CM approach as noted in previous post..
For the magnitude of the difference, B - A:
(1) 0 = mA + MB ==> B = -m/M A
(2) B = -m/M A (Note: we're in 1D, so vector direction is known by +/- sign)
(2A) B^2 = (m/M)^2 A^2
(2B) B-A = -m/M A -A = -(m+M)/M * A
(2C) (B-A)^2 = (m+M)^2/M^2 * A^2
(3) 2E = m A^2 + M B^2
(4) 2E = m A^2 + M (m/M)^2 A^2
(4a) 2E = (M + m) m/M A^2
(4b) A^2 = (M/m)2E/(M+m)
(5) (B-A)^2 = (m+M)^2/M^2 * [ (M/m)2E/(M+m) ]
= 2E (m+M)/(Mm)
(5a) (B-A) = sqrt[ 2E (m+M)/(Mm) ]
Active 2 days ago
This question comes from the Cambridge University's Mathematical Tripos Examination set in 1902. Ebenezer Cunningham, who was senior wrangler then, http://en.wikipedia.org/wiki/Ebenezer_Cunningham, must have solved it. The space ship was simply termed "a body" in the original question.
Let u be the velocity of the ship before the explosion, and v and V the velocities of m and M afterwards. The total kinetic energy after the explosion is mv²/2+mV²/2 and this must be equal to that before the explosion (m+M)u²/2 plus the released energy E.
The total momentum of the system remains unchanged, therefore the momentum after the explosion mv+MV must equal that of before the explosion (m+M)u.
This gives u as (mv+MV)/(m+M) and in substituting this in the kinetic equation and expanding we get :-
m²v² + M²V² + mMv² + mMV² = 2E(m+M) + m²v² + M²V² + 2mMvV
Cancelling and transfering to the LHS the term 2mMvV, the LHS becomes mM(v-V)² which, on dividing the RHS by mM and taking the square root, gives the result of Sqrt(2E(m+M)/mM).
Let u be the velocity of the ship before the explosion, and v and V the velocities of m and M afterwards. The total kinetic energy after the explosion is mv²/2+mV²/2 and this must be equal to that before the explosion (m+M)u²/2 plus the released energy E.
The total momentum of the system remains unchanged, therefore the momentum after the explosion mv+MV must equal that of before the explosion (m+M)u.
This gives u as (mv+MV)/(m+M) and in substituting this in the kinetic equation and expanding we get :-
m²v² + M²V² + mMv² + mMV² = 2E(m+M) + m²v² + M²V² + 2mMvV
Cancelling and transfering to the LHS the term 2mMvV, the LHS becomes mM(v-V)² which, on dividing the RHS by mM and taking the square root, gives the result of Sqrt(2E(m+M)/mM).
Active 2 days ago
Now onto the points.
Rat's questions always have a theme - in this case the Conservation of Linear Momentum. They are not just an exercise in algebra. Neither solution mentioned this, which I think is a shame. For all I know, ozo's solution could be a guess.
Nevertheless, I only asked for the solution and the first posted correct solution gets the cheese, so it all goes to ozo. Since however poffric has expanded on the maths, he gets a worthy mention.
Rat's questions always have a theme - in this case the Conservation of Linear Momentum. They are not just an exercise in algebra. Neither solution mentioned this, which I think is a shame. For all I know, ozo's solution could be a guess.
Nevertheless, I only asked for the solution and the first posted correct solution gets the cheese, so it all goes to ozo. Since however poffric has expanded on the maths, he gets a worthy mention.
Active 2 days ago
No it is not. It ought to have been mentioned BEFORE writing the equation. Furthermore it should show the momentum explicitly before and after velocity change. It should not be the job of the reader to "decode" equations, sort of "How on earth did they get that????"
Active today
>>Conservation of momentum was implicit in
Yes, ozo is right.
I said that I was changing reference to CM (which you didn't like so much). It is obvious that in CM reference the momentum is 0.
Yes, ozo is right.
I said that I was changing reference to CM (which you didn't like so much). It is obvious that in CM reference the momentum is 0.
Active 2 days ago
>>I said that I was changing reference to CM (which you didn't like so much).
I said I saw no NEED to do so, which is a different matter.
Furthermore one has to guess that CM means center of mass (?).
There is no need, indeed it is confusing, introducing extra concepts without explaination.
The reader will ask why does he change the frame of reference? It does not greatly simplify the algebra, nor does it enable you to apply some other principle of mechanics which would be necessary to solve the problem.
I said I saw no NEED to do so, which is a different matter.
Furthermore one has to guess that CM means center of mass (?).
There is no need, indeed it is confusing, introducing extra concepts without explaination.
The reader will ask why does he change the frame of reference? It does not greatly simplify the algebra, nor does it enable you to apply some other principle of mechanics which would be necessary to solve the problem.
I guess phoffric and I were imagining a reader already conversant with conservation of momentum for whom center of mass transforms were second nature, whereas BigRat was imagining an exposition of conservation of momentum.
An understandable difference in perspective of a type which is usually resolved by a dialogue between the questioner and answerer, but in a quiz situation, where the questioner does not want to prematurely give away their thought process and has a specific goal in mind, that dialogue may work a little different.
To me, since u is irrelevant to the answer, it made sense to immediately remove it from the equation, even if there turned out to be a pedagogical benefit to leaving it in.
An understandable difference in perspective of a type which is usually resolved by a dialogue between the questioner and answerer, but in a quiz situation, where the questioner does not want to prematurely give away their thought process and has a specific goal in mind, that dialogue may work a little different.
To me, since u is irrelevant to the answer, it made sense to immediately remove it from the equation, even if there turned out to be a pedagogical benefit to leaving it in.
Active 2 days ago
Netminder:
I only asked for the equation of the difference in velocity. I thought ozo ought to have all the points, since it was the first posted correct answer. The posting from phoffric ID: 38733578 doesn't provide a solution although it does suggest the correct procedure. That is why I awarded him a few points as an honorable mention.
Perhaps on reflection I ought to have awarded phoffric more, although the essential point of the Conservation of Linear Momentum was not brought out by his subsequent calculation, which, I still think, lacks explaination.
All of my "quizz" type questions have a theme,. so that there is some PAQ value to them and it is not just point passing to others. I should have phrased it was "prove" rather than "show" which would have required the justification in the steps chosen, and then judged the result on the simplicity of the proof.
I will of course accept any moderation which you think appropiate.
I only asked for the equation of the difference in velocity. I thought ozo ought to have all the points, since it was the first posted correct answer. The posting from phoffric ID: 38733578 doesn't provide a solution although it does suggest the correct procedure. That is why I awarded him a few points as an honorable mention.
Perhaps on reflection I ought to have awarded phoffric more, although the essential point of the Conservation of Linear Momentum was not brought out by his subsequent calculation, which, I still think, lacks explaination.
All of my "quizz" type questions have a theme,. so that there is some PAQ value to them and it is not just point passing to others. I should have phrased it was "prove" rather than "show" which would have required the justification in the steps chosen, and then judged the result on the simplicity of the proof.
I will of course accept any moderation which you think appropiate.
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