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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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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|

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) ]

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.

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.

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