This course will help prep you to earn the CompTIA Healthcare IT Technician certification showing that you have the knowledge and skills needed to succeed in installing, managing, and troubleshooting IT systems in medical and clinical settings.

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!

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|

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?

m --->

From the diagram, the speeds are shown as B > A

The difference in velocity is

Using CM approach as noted in previous post..

For the

(1) 0 = m

(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) ]

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).

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.

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.

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.

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.

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.

Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.