Correct statement about elastic collision

In summary: So "##u_1+u_2##" gives the difference in the two velocities (the closure rate) and not the sum of the two velocities (which would be relevant for momentum conservation).Similarly, since the arrows on the right hand side in the OP both point in the same direction, the sign conventions for ##v_1## and ##v_2## are identical. So the difference of the two figures gives a recession rate rather than the velocity sum (which would be relevant for...
  • #1
2,264
316
Homework Statement
Please see below
Relevant Equations
Conservation of momentum

Conservation of KE
1661993937298.png


From conservation of momentum:
m1u1 + m2u2 = m1v1 + m2
u1 - u2 = v1 + v2 (u2 is negative because the object moves to the left)

From conservation of KE, I got answer (C)

So there are two correct answers, (B) and (C)?

Thanks
 
Physics news on Phys.org
  • #2
songoku said:
Homework Statement:: Please see below
Relevant Equations:: Conservation of momentum

Conservation of KE

So there are two correct answers, (B) and (C)?
Yes, both are correct. In fact, since the masses are the same and the collision is elastic:
v2=u1, v1=-u2.

If inelastic, none would be correct (other than by chance). Instead, u1+u2=v2-v1.
 
  • Like
  • Skeptical
Likes jbriggs444, songoku, PeroK and 1 other person
  • #3
I agree that (C) is correct. I am not sure that (B) is also correct. Given the statement of the problem, the case can be made that (A) is also correct instead of (B). Here is why.

The problem clearly states that ##u_1## and ##u_2## are the speeds before the collision and that ##v_1## and ##v_2## are the speeds after the impact. We know that when two equal masses collide head-on elastically, they exchange velocities. This also means that they exchange speeds. Using the given symbols for the speeds it follows that
##v_2=u_1##
##v_1=u_2##
If we subtract the bottom equation from the top, we get choice (A).

Frankly, I don't like this problem because it muddles the distinction between components and magnitudes in one dimension, but it is what it is.
 
  • Like
Likes songoku, robphy, topsquark and 1 other person
  • #4
The diagram picturing after the collision is misleading for the case of equal masses and elastic collision. If the two spheres travel in opposite directions before the collision, they must also travel in opposite directions after the collision. As pointed out by @kuruman, the spheres just exchange velocities.
 
  • Like
Likes songoku, robphy, topsquark and 1 other person
  • #5
TSny said:
The diagram picturing after the collision is misleading for the case of equal masses and elastic collision. If the two spheres travel in opposite directions before the collision, they must also travel in opposite directions after the collision.
That is exactly right. I let that go by thinking that presumably the arrows after the collision were drawn generically in order not to "give anything away." As I already said, I don't like this problem.
 
  • Like
Likes songoku, topsquark, Lnewqban and 1 other person
  • #6
kuruman said:
That is exactly right. I let that go by thinking that presumably the arrows after the collision were drawn generically in order not to "give anything away." As I already said, I don't like this problem.
I see a lot of teaching websites that refer to speed when what they mean is the scalar value of a velocity component in a known vectoral direction. I've never come across a good terminology for describing such.
 
  • Like
Likes songoku
  • #7
TSny said:
As pointed out by @kuruman, the spheres just exchange velocities.
Which is equivalent for identical spheres to their not colliding at all and each continuing with its original velocity.
 
  • Like
Likes songoku, Lnewqban, TSny and 1 other person
  • #8
Please give the source of the question.
As others have pointed out, the question is poorly posed.
 
  • Like
Likes songoku
  • #9
haruspex said:
I see a lot of teaching websites that refer to speed when what they mean is the scalar value of a velocity component in a known vectoral direction. I've never come across a good terminology for describing such.
At the expense of extra notation (and necessary use in forthcoming 2D problems), I use component-subscripts ##v_{1,x}## etc.

Update:
an alternative notation is to use vectors everywhere:
for velocity use ##\vec v##, and for speed use ##|\vec v|##.

I think this highlights a problem with many notations.
Some notations aren't clearly defined [for the reader] and
many definitions aren't read or understood.
 
Last edited:
  • Like
Likes songoku
  • #10
robphy said:
Please give the source of the question.
As others have pointed out, the question is poorly posed.
From my senior and he said he got it from the teacher.

I just skip this question because it is not clear.

Thank you very much for all the help and explanation haruspex, kuruman, Tsny, Perok, robphy
 
  • #11
robphy said:
At the expense of extra notation (and necessary use in forthcoming 2D problems), I use component-subscripts ##v_{1,x}## etc.

Update:
an alternative notation is to use vectors everywhere:
for velocity use ##\vec v##, and for speed use ##|\vec v|##.

I think this highlights a problem with many notations.
Some notations aren't clearly defined [for the reader] and
many definitions aren't read or understood.
It's not the notation that bothers me, it's the terminology. It's a scalar, so it's not speed and it's not a velocity, which is a vector. It is isomorphic to a vector in a 1-D space.
 
  • #12
haruspex said:
u1+u2=v2-v1.
If I read the arrows in the OP as denoting a sign convention (the least strained interpretation I can come up with) then the sign convention for ##u_1## and ##u_2## are opposite.

So "##u_1+u_2##" gives the difference in the two velocities (the closure rate) and not the sum of the two velocities (which would be relevant for momentum conservation).

Similarly, since the arrows on the right hand side in the OP both point in the same direction, the sign conventions for ##v_1## and ##v_2## are identical. So the difference of the two figures gives a recession rate rather than the velocity sum (which would be relevant for momentum conservation).'

This is presumably why option B under the OP reads "##u_1-u_2 = v_2+v_1##". It expresses momentum conservation under the sign conventions indicated by the arrows.
 
  • Like
Likes songoku
  • #13
jbriggs444 said:
If I read the arrows in the OP as denoting a sign convention (the least strained interpretation I can come up with) then the sign convention for ##u_1## and ##u_2## are opposite.
OMG, that is strained beyond suspension of disbelief. Are you saying that "to the right" is positive for one mass and negative for the other? When I see arrows with labels of the kind shown in the OP or in FBDs, I interpret the combined arrow-label to be a graphic depiction of a vector where the label is the magnitude and the arrow the direction. I am not sure what you mean by "opposite" sign convention. To me, the algebraic expressions corresponding to the "before" and "after" total momenta shown in the drawing would be
$$\begin{align} & \mathbf{P}_{\text{before}}= m u_1~(\mathbf{\hat x} )+m u_2~(-\mathbf{\hat x}) \nonumber \\ & \mathbf{P}_{\text{after}}= m v_1~(\mathbf{\hat x} )+m v_2~(\mathbf{\hat x}). \nonumber \end{align}$$
 
  • Like
Likes songoku
  • #14
kuruman said:
OMG, that is strained beyond suspension of disbelief. Are you saying that "to the right" is positive for one mass and negative for the other?
In a word, yes. It is stupid, but it is what it is.

As you yourself seem to agree with your formulation for ##P_\text{before}##. The two "u" coordinates are given according to two different bases (two different sign conventions). Meanwhile, the two "v" coordinates are given according to the same vector basis.

Unless one is trying to test on comprehension of sign conventions, using silly conventions is just making things difficult and obscuring the physics.
 
Last edited:
  • Like
Likes songoku and kuruman
  • #15
jbriggs444 said:
In a word, yes. It is stupid, but it is what it is.
That's one more reason for me not to like this problem.
 
  • Like
Likes jbriggs444
  • #16
jbriggs444 said:
So "##u_1+u_2##" gives the difference in the two velocities (the closure rate) and not the sum of the two velocities (which would be relevant for momentum conservation).
I was applying Newton's Experimental Law for the special case of perfect elasticity. The closure rate before equals the separation rate after. This works even if the masses differ.
 
  • #17
haruspex said:
I was applying Newton's Experimental Law for the special case of perfect elasticity. The closure rate before equals the separation rate after. This works even if the masses differ.
The statement to which I responded was:
haruspex said:
If inelastic, none would be correct (other than by chance). Instead, u1+u2=v2-v1.
I read that to indicate that u1+u2=v2-v1 applied to the inelastic case.
 
Last edited:
  • #18
jbriggs444 said:
The statement to which I responded was:

I read that to indicate that u1+u2=v2-v1 applied to the inelatic case.
Ah, yes. That was a mistake.
 
  • #19
jbriggs444 said:
In a word, yes. It is stupid, but it is what it is.

As you yourself seem to agree with your formulation for ##P_\text{before}##. The two "u" coordinates are given according to two different bases (two different sign conventions). Meanwhile, the two "v" coordinates are given according to the same vector basis.

Unless one is trying to test on comprehension of sign conventions, using silly conventions is just making things difficult and obscuring the physics.
If I understand correctly what you're saying, there are two kinds of unit vectors one for the mass 1 on the left and an opposite one for mass 2 on the right in the before picture. In terms of the standard Cartesian unit vectors, one would write ##\mathbf{\hat u}_1= \mathbf{\hat x}## and ##\mathbf{\hat u}_2= -\mathbf{\hat x}##. Then according to the picture on the left, $$\mathbf{P}_{\text{before}}=m u_1~\mathbf{\hat u}_1 +m u_2~\mathbf{\hat u}_2 =m u_1~(\mathbf{\hat x} )+m u_2~(-\mathbf{\hat x}).\tag{1}$$According to the picture on the right, $$\mathbf{P}_{\text{after}}=m v_1~\mathbf{\hat u}_1 +m v_2~(-\mathbf{\hat u}_2) =m v_1~(\mathbf{\hat x} )+m v_2~(\mathbf{\hat x}).\tag{2}$$I note that in equations (1) and (2) the ##u_i## and ##v_i## represent positive magnitudes. Momentum conservation requires that $$\mathbf{P}_{\text{before}}=\mathbf{P}_{\text{after}} \implies u_1-u_2=v_1+v_2.\tag{3}$$Equation (3) is indeed choice (B) in the OP. However there is a big BUT because it leads to an absurd result.

We all agree that the energy conservation equation, choice (C) is also correct, ##u_1^2+u_2^2= v_1^2+v_2^2##. If we square both sides of equation (3), and substitute the energy conservation equation, we get $$\begin{align} u_1^2+2u_1u_2+u_2^2 & = v_1^2-2v_1v_2+v_2^2 \nonumber \\ \cancel{u_1^2}+2u_1u_2+\cancel{u_2^2} & = \cancel{v_1^2}-2v_1v_2+\cancel{v_2^2} \nonumber \\ 2u_1u_2 &=- 2v_1v_2 \nonumber \end{align}$$ This result is absurd because we have already stipulated that the ##u_i## and ##v_i## represent positive magnitudes. The only way to reconcile the result with physical reality is to assert what we know to be the case, ##\mathbf{v}_1## is in a direction opposite to what is shown in the picture. Then, $$\mathbf{P}_{\text{after}}=m v_1~(-\mathbf{\hat x} )+m v_2~(\mathbf{\hat x}).$$ Now momentum conservation requires that $$u_1-u_2=-v_1+v_2$$ which is choice (A) as I indicated in post #3 using another line of reasoning.
 
Last edited:
  • #20
kuruman said:
If I understand correctly what you're saying, there are two kinds of unit vectors one for the mass 1 on the left and an opposite one for mass 2 on the right in the before picture.
The basis vectors for the left hand side are right-positive for u1 and left-positive for u2.
The basis vectors for the right hand side are right-positive for both v1 and v2.
kuruman said:
This result is absurd because we have already stipulated that the ##u_i## and ##v_i## represent positive magnitudes.
No such requirement exists. ##v_1## will be almost certainly be negative since ##\vec{v_1}## is leftward and the sign convention for its scalar multiple (##v_1##) is right-positive.

We are not totting up unsigned magnitudes. We are totting up signed scalars.
 
Last edited:
  • #21
jbriggs444 said:
We are not totting up unsigned magnitudes. We are totting up signed scalars.
I now see what you mean. It's a twisted way of looking at things. This kind of thinking is likely to prompt students to substitute ##g = -9.8## m/s2 when they see a "down" arrow labeled ##mg## in a free body diagram.
 
  • Like
Likes jbriggs444

Suggested for: Correct statement about elastic collision

Back
Top