Confused about conservation of momentum

In summary, the conservation of momentum states that the total momentum before a collision is equal to the total momentum after the collision. In the case of a head-on collision between two masses, M1 and M2, where M2 is infinitely large, the final velocity of M2, V2final, will be 0. This means that the final velocity of M1, V1final, will be equal to its initial velocity, V1initial, but in the opposite direction.
  • #1
stewdonym
4
0
I’m confused about conservation of momentum. I remember learning that the momentum before a collision is the same as after it:

Pinitial = Pfinal .

So I do that for a mass M1 colliding with another mass M2 that’s initially at rest,

M1 x V1initial + 0 = M1 x V1final + M2 x V2final

and I get an equation for the velocity V2final ,

V2final = (M1/M2) x (V1initial - V1final) .

Now let’s say that M2 is so large that in the limit it becomes infinite. In that case, V2final = 0, so that

V1final = V1initial .

But since M1 bounces off of M2 , shouldn't V1final and V1initial be pointing in opposite directions?
 
Physics news on Phys.org
  • #2
Take M2 large but finite, then take M2 larger and larger.
 
  • #3
atyy said:
Take M2 large but finite, then take M2 larger and larger.

??

In the limit, unless V1initial - V1final is also "large but finite," and grows that way with increasing M2 , V2final = 0 .
 
  • #4
stewdonym said:
I’m confused about conservation of momentum. I remember learning that the momentum before a collision is the same as after it:

Pinitial = Pfinal .

So I do that for a mass M1 colliding with another mass M2 that’s initially at rest,

M1 x V1initial + 0 = M1 x V1final + M2 x V2final

and I get an equation for the velocity V2final ,

V2final = (M1/M2) x (V1initial - V1final) .
OK.

Now let’s say that M2 is so large that in the limit it becomes infinite. In that case, V2final = 0,
OK.
so that

V1final = V1initial .
That doesn't follow. You cannot deduce anything about the difference between the initial and final speeds of M1, since you are essentially multiplying that difference by zero.

But since M1 bounces off of M2 , shouldn't V1final and V1initial be pointing in opposite directions?
Yes, if the collision is elastic. Use conservation of energy to deduce that the initial and final speeds are the same. And the fact that M1 doesn't pass through M2..
 
  • #5
I assume that what you are doing is to put the result [itex]v_{2final}=0[/itex] back into your first equation and then set [itex]m_2 v_{2final}=0[/itex] but that is not correct since you have taken the mass to be infinitely large you have something of the form [itex]\infty \times 0[/itex] which is not necessarily zero. In fact if you keep the expression for [itex]v_{2final}[/itex] and plug it into your first equation you get [itex]v_{1init}=v_{1init}[/itex] which is not very helpful :)

The problem is that you have 2 unknown variables [itex]v_{1final}[/itex] and [itex]v_{2final}[/itex] and only one equation. You need to also consider conservation of energy in order to solve the equations.EDIT: Ooops Doc Al was faster...
 
  • #6
stewdonym said:
But since M1 bounces off of M2 , shouldn't V1final and V1initial be pointing in opposite directions?
Not necessarily. From the initial conditions there are two unknowns, the two final velocities. So you have one equation and two unknowns. You need another piece of information in order to have a fully-determined system. As Doc Al mentioned, the usual additional piece of information is either a specification of the collision as perfectly elastic (KE also conserved) or perfectly plastic (final velocities of each object are equal).
 
  • #7
You need to conserve momentum separately in every degree of freedom, so Px, Py, and Pz are separately conserved. Visualize an asteroid bouncing off the Earth. More than half the off-center asteroid impacts will result in a forward final asteroid velocity.
 
  • #8
stewdonym said:
??

In the limit, unless V1initial - V1final is also "large but finite," and grows that way with increasing M2 , V2final = 0 .

magnitude of V2final=some smaller and smaller number, not zero.

Actually, I'm not sure if it's a strict limit, or just an approximation. But you can't divide by infinity to get zero if you attempt to do it "strictly". So solve everything with finite M2, and only take the limit/approx (?) M2 >> M1 in the final step.
 
  • #9
Bob S said:
You need to conserve momentum separately in every degree of freedom, so Px, Py, and Pz are separately conserved. Visualize an asteroid bouncing off the Earth. More than half the off-center asteroid impacts will result in a forward final asteroid velocity.
True, but I suspect that the OP was thinking of a head-on collision in one dimension.
 
  • #10
stewdonym said:
I’m confused about conservation of momentum. I remember learning that the momentum before a collision is the same as after it:

Pinitial = Pfinal .

So I do that for a mass M1 colliding with another mass M2 that’s initially at rest,

M1 x V1initial + 0 = M1 x V1final + M2 x V2final

and I get an equation for the velocity V2final ,

V2final = (M1/M2) x (V1initial - V1final) .

Now let’s say that M2 is so large that in the limit it becomes infinite. In that case, V2final = 0, so that

V1final = V1initial .

But since M1 bounces off of M2 , shouldn't V1final and V1initial be pointing in opposite directions?

It must be
P(initial)+P(final)=o
P(initial)=-P(final)
M1.V(1initial)+M2.V(2initial)= -(M1.V(1final)+M2.V2(final))
M1.V(1initial)+o = -(M1.V(1final)+M2.V(2final))
M1.V(1initial)+M1.V(1final)= -M2.V(2final)
M1/M2(V(1initial)+V(1final))= -V(2final)
if M2 is infinitally large, V(2final)=o
that means V(initial)+V(1final)=0
V(initial) = -V(1final)
Velocity in opposite direction
 
Last edited:
  • #11
Now let’s say that M2 is so large that in the limit it becomes infinite. In that case, V2final = 0, so that

V1final = V1initial

V2 will be infinitely small, so result of momentum will be transferred to m1 in the opposite direction.

V2final = (M1/M2) x (V1initial - V1final) .

If m2 becomes infinite in this formula, then v2 final will be 0.

And v1 final will be -

v2final (M2/M1) - v1inital = -v1final

That way v1final = infinite :P
 
  • #12
coverme said:
It must be
P(initial)+P(final)=o
That's incorrect. Conservation of momentum says:
P(initial) = P(final)

P(initial)=-P(final)
M1.V(1initial)+M2.V(2initial)= -(M1.V(1final)+M2.V2(final))
M1.V(1initial)+o = -(M1.V(1final)+M2.V(2final))
M1.V(1initial)+M1.V(1final)= -M2.V(2final)
M1/M2(V(1initial)+V(1final))= -V(2final)
if M2 is infinitally large, V(2final)=o
that means V(initial)+V(1final)=0
This is an incorrect conclusion, as already pointed out several times in this thread.
 
  • #13
dE_logics said:
V2 will be infinitely small, so result of momentum will be transferred to m1 in the opposite direction.

V2final = (M1/M2) x (V1initial - V1final) .

If m2 becomes infinite in this formula, then v2 final will be 0.

And v1 final will be -

v2final (M2/M1) - v1inital = -v1final

That way v1final = infinite :P
Wise guy. :smile: :tongue:
 
  • #14
Doc Al said:
That's incorrect. Conservation of momentum says:
P(initial) = P(final)


This is an incorrect conclusion, as already pointed out several times in this thread.

Thanks for telling my math errors,I appologise.It was math deleimma in my mind at instant
 
  • #15
cool Doc Al..you are superb
 
  • #16
Doc Al said:
You cannot deduce anything about the difference between the initial and final speeds of M1, since you are essentially multiplying that difference by zero.

How correct of you...and dumb of me. (Thanks!)

Doc Al said:
(regarding whether V1final and V1initial point in opposite directions):
Yes, if the collision is elastic. Use conservation of energy to deduce that the initial and final speeds are the same. And the fact that M1 doesn't pass through M2..

The "pass through" implication never occurred to me. Very nice!
 

1. What is the conservation of momentum?

The conservation of momentum is a fundamental law of physics that states that the total momentum of a closed system remains constant over time. This means that the total amount of momentum before and after an interaction or event must be equal.

2. Why is conservation of momentum important?

Conservation of momentum is important because it helps us understand and predict the behavior of objects in motion. It is also a key principle in many real-world applications, such as rocket propulsion, collisions, and the movement of fluids.

3. What are the conditions for conservation of momentum?

The conditions for conservation of momentum are that the system must be closed, meaning there are no external forces acting on it, and that there is no loss or gain of momentum within the system. This means that the total momentum of all objects before and after an event must be the same.

4. How is conservation of momentum related to Newton's Third Law?

Conservation of momentum is related to Newton's Third Law, which states that for every action, there is an equal and opposite reaction. This means that in a closed system, the total momentum before and after an event must be equal due to the equal and opposite forces acting on each object involved.

5. Can conservation of momentum be violated?

No, conservation of momentum is a fundamental law of physics and cannot be violated. However, it may appear to be violated in some situations, such as when friction or external forces are present, but in reality, the law is still being obeyed.

Similar threads

Replies
30
Views
1K
Replies
11
Views
2K
Replies
3
Views
963
Replies
11
Views
2K
Replies
3
Views
1K
Replies
40
Views
2K
Replies
4
Views
1K
Back
Top