Proving that mass is an additive quantity

[Nirmal Padwal]

Homework Statement

Use a three-particle interaction to show that mass is an additive quantity.

Relevant Equations

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$.

I already have referred to the solution to this problem. But the way I originally solved the problem is completely different from how the available solution proceeds. I wish to know if my solution is right or wrong.

My Solution:
Consider three particles undergoing one-dimensional motion all moving in the same direction. Let the leftmost particle be $A$ whose mass and initial velocity are $m_1$ and $u_1$ respectively. Let the particle in between be $B$ whose mass and initial velocity are $m_2$ and $u_2$ respectively. Let the rightmost particle be $C$ whose mass and initial velocity are $m_3$ and $u_3$ respectively. Consider that $u_1 > u_2 > u_3$ and the distances between $A$, $B$ and $C$ are such that $A$ and $B$ collide first and the combined mass then collides with $C$. We assume inelastic collision.

By conservation of linear momentum (not even mentioned in the actual solution), momentum before and after collision between $A$ and $B$ are related by $$ m_1 u_1 + m_2 u_2 = m_1 u + m_2 u = (m_1 +m_2) u = m_c u $$ where u is the velocity of combined mass

Similarly, by conservation of linear momentum, momentum before and after collision between combined mass $m_c$ and $C$ are related by $$ m_c u + m_3 u_3 = m_c u' + m_3 u' = (m_c +m_3) u' = (m_1 + m_2+m_3) u' = m u' $$ where u' is the velocity of combined mass of all three particles.

Thus we have $m u' = (m_1 + m_2+m_3) u'$ or $m = m_1 +m_2 +m_3$. Hence mass is an additive quantity

Is this solution correct?

 

[kuruman]

Are you sure you are not begging the question, i.e. assume what you want to prove in order to prove it? Conservation of linear momentum relies on the constancy of the velocity of the center of mass which itself assumes that masses are additive. Can you post the proof that was given?

My proof would be to put mass $m_1$ on a scale and measure its weight; put mass $m_2$ on the same scale and measure its weight; finally put both masses on the sme scale, measure their combined weight and compare the result with the sum of previous two measurements..

 

[PeroK]

It seems a strange question, but I think it follows from $\vec F = m\vec a$ if forces are additive.

 

[jbriggs444]

Are you sure you are not begging the question, i.e. assume what you want to prove in order to prove it? Conservation of linear momentum relies on the constancy of the velocity of the center of mass which itself assumes that masses are additive. Can you post the proof that was given?

My proof would be to put mass $m_1$ on a scale and measure its weight; put mass $m_2$ on the same scale and measure its weight; finally put both masses on the sme scale, measure their combined weight and compare the result with the sum of previous two measurements..

No matter what you do, you are going to have to rely on something as an assumption. Any truth that you can deduce from nothing is pretty much vacuous.

Given foreknowledge of what special relativity brings to the table, one might want to specify:

1. The momentum of a single object is given by $p=mv$
2. The mass of an object does not change when that object is accelerated or acted upon by a force.
3. Momentum is an additive property.

You are still going to need more. Such as Newton's third law plus something like @PeroK's suggestion that a three body interaction can be decomposed into additive pairwise forces.

 

[PeroK]

I was thinking of something simpler:

1) Adjust forces $F_1, F_2$ on $m_1$ and $m_2$ until they have the same acceleration $a$.

2) Consider $m_1$ and $m_2$ as a system and apply force $F = F_1 + F_2$ to them. As the acceleration is still $a$ (the system is moving in unison) and the forces add, then the mass of the system must be $m = m_1 + m_2$.

 

[Nirmal Padwal]

@kuruman The actual solution has the same approach as that of @PeroK.

My logic was that, since momentum is additive, after inelastic collision as both the masses will be moving with the same velocity, I can take the velocity $u$ common thanks to vector algebra. And now if a second person enters the room and he has not witnessed the collision, he will say that the momentum of the moving particle is $m_c u$. Now if I equate his observation with the expression I got using conservation law and cancel out the $u$ assuming the particles don't come to rest after collision, I end up with, $m_c = m_1 + m_2$. So what exactly is wrong with this approach?

 

[anuttarasammyak]

Let me show you my attempt with Galilean transformation, which does not change mass, and conservation of momentum for two particle case, not three -particle case of Homework.

Conservation of momentum says,
$$m_1v_1 + m_2v_2= m_1' v'_1 + m'_2v'_2$$
where LHS is before and RHS is after collision.
In IFR where particle of m_1 is at still, the above formula is transferred to :
$$m_2(v_2-v_1)= m_1'( v'_1-v_1) + m'_2(v'_2 - v_1)$$
Doing the same for m_2, and substracting in the both sides and dividing by $v_2-v_1$ which is not zero so that the collision takes place,
$$m_1+m_2= m_1'+ m'_2$$
This is the equation of conservation of total mass.

I hope this will give a hint to Homework of three-particle case.