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Analyzing Elastic Collisions w/ Conservation of Energy and Momentum

  • #1

Homework Statement


There is a 4 kg mass that has a speed of 6 m/sec on a horizontal frictionless surface. The mass collides head-on and elastically with an identical 4 kg mass initially at rest.

The final speed of the first 4 kg mass is:
(a) 0 m/s (b) 2 m/s (c) 3 m/s (d) 6 m/s

Homework Equations


p = mv
KE = (1/2)mv2

The Attempt at a Solution


To start off, one way of solving this problem seems to be knowing that the two masses have equal mass. Since the collision is elastic,

pi = pf
m1v1init + m2v2init = m1v1final + m2v2final

And since v2init = 0, this simplifies to

m1v1init = m1v1final + m2v2final

Since the two masses have equal mass, we can assume that the (absolute values of the) two final momentums (m1v1final and m2v2final) are equal. Therefore, if I just change m2v2final to be m1v1final, the equation should still yield us a correct answer:

m1v1init = m1v1final + m1v1final
m1v1init = 2 * m1v1final

Now I isolate the equation for v1final:

v1final = v1init / 2

Substituting variables for their values yields me the correct answer:

v1final = (6m/s) / 2
v1final = 3 m/s

(Technically, if the final momentums of the two masses are equal and opposite, they should add up to zero! This is also something I am confused with. Is this even correct to do? But that's a later question.)

------------------------

Now, the problem is, I can't reproduce this same answer when solving for the same v1final through just Conservation of Energy and Conservation of Momentum, with no other assumptions made... Here are the steps I tried:

First, use Conservation of Momentum to find an expression equal to v2final:

m1v1init + m2v2init = m1v1final + m2v2final

And again, v2init = 0, so

m1v1init = m1v1final + m2v2final
m2v2final = m1v1init - m1v1final
v2final = (m1v1init - m1v1final) / m2
v2final = (4kg * (6m/s) - 4kg * v1final) / 4kg
v2final = 6m/s - v1final

We can use this later to substitute v2final into our next equation, from Conservation of Energy:

(1/2)m1v1init2 + (1/2)m2v2init2 = (1/2)m1v1final2 + (1/2)m2v2final2
(1/2)m1v1init2 + 0 = (1/2)m1v1final2 + (1/2)m2v2final2
m1v1init2 = m1v1final2 + m2v2final2
4kg * 6m/s2 = 4kg * v1final2 + 4kg * (6m/s - v1final)2

(Omitting units because it gets messy)

4 * 62 = 4 * v1final2 + 4 * (6 - v1final)2
144 = 4 * v1final2 + 4 * (36 - 12v1final + v1final2)
36 = v1final2 + (36 - 12v1final + v1final2)
36 = 2v1final2 - 12v1final + 36
0 = 2v1final2 - 12v1final
0 = 2v1final(v1final - 6)

Therefore:

v1final = 0 or 6m/s

But the answer found earlier was 3m/s. How come the same number is not produced?

Sorry for the long post, and thank you for your help.
 
Last edited:

Answers and Replies

  • #2
Merlin3189
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In your first method you say, "Since the two masses have equal mass, we can assume that the (absolute values of the) two final momentums (m1v1final and m2v2final) are equal." I can't see any reason for that.
 
  • #3
Orodruin
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Since the two masses have equal mass, we can assume that the (absolute values of the) two final momentums (m1v1final and m2v2final) are equal
This is not correct. Your assumption is equivalent with giving both masses the same velocity post-collision, i.e., a completely inelastic collision.
 
  • #4
Merlin3189
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(Technically, if the final momentums of the two masses are equal and opposite, they should add up to zero! This is also something I am confused with. Is this even correct to do? But that's a later question.)
But why should the final momenta be equal and opposite?

Are you maybe getting confused with relative velocities?
 
  • #5
I see, the assumption about the equal but opposite momenta was incorrect... That should be true for only perfectly inelastic collisions with equal masses.

So, does that mean the answer from the second method, 6m/s, is correct? I found that if I plugged that number into the Conservation of Energy equation again and solved for the other velocity, v2final, I get zero:

(1/2)m1v1init2 + 0 = (1/2)m1v1final2 + (1/2)m2v2final2
(1/2)(4kg)(6m/s)2 = (1/2)(4kg)(6m/s)2 + (1/2)(4kg)v2final2
0 = (1/2)(4kg)v2final2
v2final = 0

Is this correct?
 
  • #6
haruspex
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I see, the assumption about the equal but opposite momenta was incorrect... That should be true for only perfectly inelastic collisions with equal masses.

So, does that mean the answer from the second method, 6m/s, is correct? I found that if I plugged that number into the Conservation of Energy equation again and solved for the other velocity, v2final, I get zero:

(1/2)m1v1init2 + 0 = (1/2)m1v1final2 + (1/2)m2v2final2
(1/2)(4kg)(6m/s)2 = (1/2)(4kg)(6m/s)2 + (1/2)(4kg)v2final2
0 = (1/2)(4kg)v2final2
v2final = 0

Is this correct?
Yes.
 
  • #7
Orodruin
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I see, the assumption about the equal but opposite momenta was incorrect... That should be true for only perfectly inelastic collisions with equal masses.
No, and it is not what your assumption was based on the equation you wrote down:
Therefore, if I just change m2v2final to be m1v1final, the equation should still yield us a correct answer:

m1v1init = m1v1final + m1v1final
That assumption was clearly equal momenta in the same direction.
 
  • #8
Merlin3189
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There is a 4 kg mass that has a speed of 6 m/sec on a horizontal frictionless surface. The mass collides head-on and elastically with an identical 4 kg mass initially at rest.

Since the collision is elastic,
pi = pf True irrespective of elastic collision.
Total momentum before = total momentum after

m1v1init = m1v1final + m2v2final

Since the two masses have equal mass, then just get rid of it !
v1init = v1final + v2final
6 = v1final + v2final

Now it all goes wrong

Now I isolate the equation for v1final:

v1final = v1init / 2

Substituting variables for their values yields me the correct answer:

v1final = (6m/s) / 2
v1final = 3 m/s So do the energy check: (m/2) ( 62 ) = (m/2) ( 32 + 32 ) ⇒ 36=18
So KE is not conserved. The collision is inelastic. Contradicting the assumptions of the Q.

So instead of making unnecessary assumptions about momentum, just use, elastic collision ⇒ KE conservation.


Now, the problem is, I can't reproduce this same answer when solving for the same v1final through just Conservation of Energy and Conservation of Momentum, with no other assumptions made... That's because those assumptions were unjustified. Why didn't you use momentum and KE to start with, as suggested in the question?

First, use Conservation of Momentum

m1v1init + m2v2init = m1v1final + m2v2final
always 0,
And again, v2init = 0, so and again all masses equal, so if you don't put them in, you don't need to take them out !

v1init = v1final + v2final I might just stop here until I look at the energy. (Result M )
v2final = v1init - ]v1final

v2final = 6m/s - v1final

We can use this later to substitute v2final into our next equation, from Conservation of Energy:
Again, notice the factor of m/2 is going to be in every term. Either leave it out altogether or at least give up on the subscripts.

(1/2)m1v1init2 + (1/2)m2v2init2 = (1/2)m1v1final2 + (1/2)m2v2final2
v1init2 + 0 = v1final2 + v2final2
Now substitute Result M in here
(v1final + v2final )2 = v1final2 + v2final2
Remove the common terms on each side
Cf. ## (A+B)^2 = A^2 +B^2 \ \ \ gives \ \ A^2 +2AB + B^2 = A^2 + B^2 \ \ \ or \ \ 2AB = 0 ##
So 2v1finalv2final = 0
Then whatever the final velocities are, one must be zero.
If v2final = 0 , then the second mass never moves and the first mass must have missed it and carried on at the same velocity.
So v1final = 0 and the other mass simply took its place, moving at the same velocity. KE and momentum preserved.

4kg * 6m/s2 = 4kg * v1final2 + 4kg * (6m/s - v1final)2

(Omitting units because it gets messy) Didn't it drive you crazy to keep writing those 4 kg 's?

4 * 62 = 4 * v1final2 + 4 * (6 - v1final)2
144 = 4 * v1final2 + 4 * (36 - 12v1final + v1final2)
36 = v1final2 + (36 - 12v1final + v1final2)
36 = 2v1final2 - 12v1final + 36
0 = 2v1final2 - 12v1final
0 = 2v1final(v1final - 6)

Therefore:

v1final = 0 or 6m/s

But the answer found earlier was 3m/s. How come the same number is not produced? Because this time you did it right !!!

Sorry for the long post, and thank you for your help At least your long post enabled us to see what you did. I wish more people would follow your example..
 
  • #9
Aha, now it makes much more sense, and I see why the assumption in the first part was wrong. Thank you all for your helpful replies!
 
  • #10
ehild
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There is a very simple solution. The masses are the same, so you have the equations
v1i=v1f+v2f from conservation of momentum, and
v21i=v21f+v22f from conservation of energy.
Take the square of the first equation and subtract the second one. You get 2v1fv2f=0. That can be if either v1f=0 or v2f=0. The second case means that the balls avoid each other, there is no collision. In the other case, v1f=0, v2f=v1i,
 
  • #11
Orodruin
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Should be mentioned that, regardless of the masses, the relative speed between the bodies remains the same.
 
  • #12
Merlin3189
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Should be mentioned that, regardless of the masses, the relative speed between the bodies remains the same.
Interesting. That's what I remember using in these situations,under the name of Newton's experimental law.
I didn't mention it here, as the OP thought of momentum and KE, which does the job. (Though this question is a do it in your head job, when you use it.)

But I did look up to see whether this rule is still ok, as I've not seen people here use it. The references I found said things like, this used to be taught in high school, as if it were a bit of dubious fudge and not rigorous. The very name, experimental law, suggests it is not based in maths (and therefore unacceptable to PFists?)

To me it is like a definition of a coefficient of restitution. So for fully elastic collision it turns out to be 1, for fully inelastic collision to be 0 and for others, somewhere between to be determined by experiment. If a question did specify a coefficient of restitution, it would be like saying, I've done the experiment on these bodies and I know the relative velocity after collision is this fraction of the initial relative velocity.
But it does imply that this coefficient will exist and be the same irrespective of the collision velocities. I wonder if that is the weak link and the reason it is not used in serious physics?
 
  • #13
Orodruin
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Interesting. That's what I remember using in these situations,under the name of Newton's experimental law.
I didn't mention it here, as the OP thought of momentum and KE, which does the job. (Though this question is a do it in your head job, when you use it.)

But I did look up to see whether this rule is still ok, as I've not seen people here use it. The references I found said things like, this used to be taught in high school, as if it were a bit of dubious fudge and not rigorous. The very name, experimental law, suggests it is not based in maths (and therefore unacceptable to PFists?)
It is a direct consequence of conservation of momentum and kinetic energy easiest seen in the center of mass frame, but it can be easily derived in any frame.

Edit: To be more specific. Let ##M = m_1 + m_2## be the total mass and ##\mu = m_1m_2/M## be the reduced mass of the system. The total momentum is ##\vec p = m_1 \vec v_1 + m_2 \vec v_2## and is conserved, therefore the total momentum squared is also conserved and hence
$$
M^2 v^2 = m_1^2 v_1^2 + m_2^2 v_2^2 + 2 m_1 m_2 \vec v_1 \cdot \vec v_2
$$
is a conserved quantity (where ##\vec v## is the center of mass velocity). Now, consider the difference ##\vec u = \vec v_1 - \vec v_2##. Squaring it gives
$$
u^2 = v_1^2 + v_2^2 - 2\vec v_1 \cdot \vec v_2
$$
from which we can solve for the inner product ##\vec v_1 \cdot \vec v_2## and insert it in the expression for the squared momentum to obtain
$$
P^2 = M^2 v^2 = M(m_1 v_1^2 + m_2 v_2^2) - m_1 m_2 u^2.
$$
It follows that
$$
u^2 = \frac{2M E_k - P^2}{m_1 m_2},
$$
where ##E_k## is the kinetic energy of the system and ##P## its total momentum. Since everything on the right-hand side is conserved, ##u^2## (and therefore ##u##) must be conserved as well.
 
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  • #14
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Alternate solution (which I seem to be repeating monotonously!).

Collision impulse (elastic collision) Δp = 2μΔv were μ is the reduced mass [m1*m2/(m1+m2)] and Δv the relative velocity.

Thus Δp (moving mass) = -2 x 4 x 4 / (4 + 4) x 6 = -24Ns. Therefore p + Δp = 24 + (-24) = 0. The initially stationary mass will have momentum 0 + 24 = 24Ns.
 

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