• SayedD
In summary, OP's method leaves the original momentum unchanged, but introduces an external energy that is added to the existing momentum.

#### SayedD

Homework Statement
A particle that is moving with velocity ##v## has kinetic energy ##K##. Suddenly the particle gained extra energy externally with magnitude ##3k/16## that causes the particle to split into two pieces. One piece moves at the speed of ##v## relative to the other piece with the direction of motion parallel with the motion before splitting. If the mass of the smaller piece is 1 kg, determine the mass of the other piece.
Relevant Equations
Momentum and Energy
My method is silly but here is my attempt

Initially we know that the kinetic energy is k = ##1/2mv^2## and the final kinetic energy at the moment before splitting is equal to ##(k + 3k/16)mv^2_1 = (19k/16)mv^2_1##. Substituting the value k to the second equation gives us ##v_1 = (\sqrt{19}/4)v##

Since the equation only tells us the relative motion between the pieces, I think we should deal the momentum in the reference frame of that piece. And also notice that ##M = m_1 + m_2## with ##m_1 = 1## This gives us the equation. And there is two cases since we do not know which piece is stationary or moving in this reference frame. So we have to check both case.

Case one, ##m_2## is moving with v relative to other piece.
##(m_1 + m_2)(v_1 - v) = m_2 v
\implies v(1 + m_2)(\sqrt{19}/4 - 1) = m_2 v
\implies m_2 = 0.097 ## which is smaller thus wrong

Case two ##m_1## is moving with v relative to other piece.
##(m_1 + m_2)(v_1 - v) = m_1 v
\implies v(1 + m_2)(\sqrt{19}/4 - 1) = m_1 v
\implies m_2 = 10.23 ## which is bigger thus could right

These all argument could be crap but I wanted to see your solutions.

Last edited by a moderator:
There is no reason to assume that one piece is stationary after explosion. The final kinetic energy may be split between the two pieces.

SayedD said:
Initially we know that the kinetic energy is k = ##1/2mv^2## and the final kinetic energy at the moment before splitting is equal to ##(k + 3k/16)mv^2_1 ##.
No, the energy after splitting is just ##k + 3k/16##.

nasu
Looking at OP's only previous thread, you can see that @SayedD (OP) never replied to any of the 23 postings made by other PF members.

berkeman
SammyS said:
Looking at OP's only previous thread, you can see that @SayedD (OP) never replied to any of the 23 postings made by other PF members.

Thanks for that heads-up, Sammy. OP has been given a not-so-gentle reminder from me to reply to the threads that he starts. Clock is ticking...

SammyS
nasu said:
There is no reason to assume that one piece is stationary after explosion. The final kinetic energy may be split between the two pieces.
In a certain reference frame it is right?

What reference frame are you using to solve the problem? In what reference frame is the velocity v and the kinetic energy k?

I have a question of interpretation. It is clear that the energy "gained externally" is added to the existing ##K##. However, I see no definitive indication that this external energy is added in way that leaves the initial momentum unchanged. I would feel more comfortable about it if there were an internal explosion that added the energy instead of an external force.

SammyS and nasu

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