Confused about conservation of momentum

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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?
 

Answers and Replies

  • #2
atyy
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Take M2 large but finite, then take M2 larger and larger.
 
  • #3
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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
Doc Al
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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
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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
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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
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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
atyy
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??

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
Doc Al
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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
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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
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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 transfered 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
Doc Al
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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
Doc Al
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V2 will be infinitely small, so result of momentum will be transfered 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
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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
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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!)

(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!
 

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