# Momentum and Newton's Law of Restitution

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#1
Hi, I am unable to understand what I have done wrong in this question. I have used Newton's Law of Restitution to calculate an equation:

( v * cos(alpha) = e * u * cos(alpha) (1st Equation)

Also using the fact that the velocity component does not change in the direction parallel to the barrier I get this equation

( v * sin(alpha) = u * cos(alpha) ). (2nd Equation)

So far it works well, and dividing 1st equation by 2nd equation I get that:

tan^2(alpha) = 1 / e

But the thing that I cannot seem to understand is that when I apply Conservation of Momentum in the horizontal direction I get:

m * u * sin(alpha) = m * v * cos(alpha)

u * sin(alpha) = v * cos(alpha)

and when I plug in this value of ( v * cos(Alpha) ) in the 1st equation, I get e = 1 which shouldn't be the case as in the second part the question is asking for the value of e and it is not 1. I would really appreciate if anyone can help. Thanks a lot. 0
2 weeks ago
#2
(Original post by Tesla3)
Hi, I am unable to understand what I have done wrong in this question. I have used Newton's Law of Restitution to calculate an equation:

( v * cos(alpha) = e * u * cos(alpha) (1st Equation)

Also using the fact that the velocity component does not change in the direction parallel to the barrier I get this equation

( v * sin(alpha) = u * cos(alpha) ). (2nd Equation)

So far it works well, and dividing 1st equation by 2nd equation I get that:

tan^2(alpha) = 1 / e

But the thing that I cannot seem to understand is that when I apply Conservation of Momentum in the horizontal direction I get:

m * u * sin(alpha) = m * v * cos(alpha)

u * sin(alpha) = v * cos(alpha)

and when I plug in this value of ( v * cos(Alpha) ) in the 1st equation, I get e = 1 which shouldn't be the case as in the second part the question is asking for the value of e and it is not 1. I would really appreciate if anyone can help. Thanks a lot. For horizontal shouldn't it be
u*cos(alpha) = v*sin(alpha)

Looks like you did conservation perp to the wall which is e=1.
Last edited by mqb2766; 2 weeks ago
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#3
(Original post by mqb2766)
For horizontal shouldn't it be
u*cos(alpha) = v*sin(alpha)

Looks like you did conservation perp to the wall which is e=1.
I applied conservation of momentum perpendicular to the wall for which:
u * sin(alpha) = v * cos(alpha)

But why is it leading to e=1 as in the next part e is not equal to 1.
0
2 weeks ago
#4
(Original post by Tesla3)
I applied conservation of momentum perpendicular to the wall for which:
u * sin(alpha) = v * cos(alpha)

But why is it leading to e=1 as in the next part e is not equal to 1.
The conservation of momentum perpendicular to the wall means e=1. That is why it's not applicable here.
Last edited by mqb2766; 2 weeks ago
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#5
(Original post by mqb2766)
The conservation of momentum perpendicular to the wall means e=1. That is why it's not applicable here.
So for conservation of momentum parallel to the wall, e is not equal to 1?
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2 weeks ago
#6
(Original post by Tesla3)
So for conservation of momentum parallel to the wall, e is not equal to 1?
e only applies perpendicular to the wall. So parallel to the wall, motion is unaffected.
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#7
(Original post by mqb2766)
e only applies perpendicular to the wall. So parallel to the wall, motion is unaffected.
In the second part, e = 1/3, why is it not e = 1. ? Why doesn't conservation of momentum apply over here?
Last edited by Tesla3; 2 weeks ago
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2 weeks ago
#8
(Original post by Tesla3)
In the second part, e = 1/3, why is it not e = 1. ?
Your "proof" that e=1 is because you assumed momentum must apply perpendicular to the wall, which does indeed give e=1. The original assumption is wrong however. The normal force the wall exerts on the ball is in the perpendicular direction only.

Your diagram is a bit complex. Try redrawing with alpha and the reflection clearly marked and the ball as a point. Your first two equations should be something like
e*u*sin() = v*cos() .... Perpendicular
u*cos() = v*sin() ... Parallel
You have the same trig function on the rhs of the two equations which suggests resolving motion has not been done correctly.

There are a few (quick google) YouTube vids on this, maybe give one a look?
Last edited by mqb2766; 2 weeks ago
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#9
(Original post by mqb2766)
Your "proof" that e=1 is because you assumed momentum must apply perpendicular to the wall, which does indeed give e=1. The original assumption is wrong however. The normal force the wall exerts on the ball is in the perpendicular direction only.

Your diagram is a bit complex. Try redrawing with alpha and the reflection clearly marked and the ball as a point. Your first two equations should be something like
e*u*sin() = v*cos() .... Perpendicular
u*cos() = v*sin() ... Parallel
You have the same trig function on the rhs of the two equations which suggests resolving motion has not been done correctly.

There are a few (quick google) YouTube vids on this, maybe give one a look?
Is the diagram correct now? Also why is my assumption incorrect that the momentum is conserved perpendicular to the wall?
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#10
mqb2766 you are saying that momentum is not conserved perpendicular to the barrier. Why is it not conserved?
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2 weeks ago
#11
(Original post by Tesla3)
mqb2766 you are saying that momentum is not conserved perpendicular to the barrier. Why is it not conserved?
Yes, if e <1, then momentum is not conserved perpendicular to the wall. That is what restitution represents. Energy gets lost in the impact. Think of a bouncing ball where the height on successive bounces follows a geometric sequence with r=e. Note that even with e=1, the velocity switches direction, so momentum isn't strictly conserved as it's sign would switch.

Also, try to draw the ball as a point. Its simpler and that's generally the assumption in the equations.
Last edited by mqb2766; 2 weeks ago
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#12
(Original post by mqb2766)
Yes, if e <1, then momentum is not conserved perpendicular to the wall. That is what restitution represents. Energy gets lost in the impact. Think of a bouncing ball where the height on successive bounces follows a geometric sequence with r=e

Also, try to draw the ball as a point. Its simpler and that's generally the assumption in the equations.
Sorry for bothering bro, but can I say although momentum is not conserved perpendicular to the wall, momentum is conserved if we include both parallel and perpendicular sides to the wall. As I heard, that momentum is conserved regardless of the value of e.
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2 weeks ago
#13
(Original post by Tesla3)
Sorry for bothering bro, but can I say although momentum is not conserved perpendicular to the wall, momentum is conserved if we include both parallel and perpendicular sides to the wall. As I heard, that momentum is conserved regardless of the value of e.
No. Where did you hear that?
If e=0, what happens? What is the momentum before/after the collision? You apply momentum to a closed system where the objects exert equal and opposing forces on each other during the collision. However, the wall is stationary, so other forces are acting on it during the collision, so you can't apply momentum to the ball/wall collision.
Even if e=1, the perpendicular to the wall velocity switches sign, as would momentum.
The model solution considers either parallel/perpendicular velocity or energy, but not momentum.
If the two objects were a closed system (wall would move), then momentum would apply and the value of e would be irrelevant.
Last edited by mqb2766; 2 weeks ago
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#14
(Original post by mqb2766)
No. Where did you hear that?
If e=0, what happens? What is the momentum before/after the collision?
Even if e=1, the perpendicular to the wall velocity switches sign, as would momentum.
As the wall is stationary here, momentum isn't the way to go for this question.
The model solution considers either velocity or energy, but not momentum.
I think so when e = 1, no kinetic energy is lost and momentum is conserved (i.e. initial momentum = final momentum) but when e = 0, the maximum amount of kinetic energy is lost, the objects coalesce and momentum is conserved (i.e. initial momentum = final momentum).

Is it wrong?
0
2 weeks ago
#15
(Original post by Tesla3)
I think so when e = 1, no kinetic energy is lost and momentum is conserved (i.e. initial momentum = final momentum) but when e = 0, the maximum amount of kinetic energy is lost, the objects coalesce and momentum is conserved (i.e. initial momentum = final momentum).

Is it wrong?
See the "closed system/stationary wall" part of the (edited) previous post. Note that even with e=1, both the velocity and momentum flip sign in the perpendicular direction, so momentum is not conserved.
I think this is where your misunderstanding is.
Last edited by mqb2766; 2 weeks ago
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#16
So, basically,
As this is not a closed system (where no external forces are acting). It is an open system (where external forces are acting on the wall as the wall is stationary), therefore conservation of momentum does not apply over here.

This concept got cleared.

Confusion 1: "Even if e=1, the perpendicular to the wall velocity switches sign, as would momentum."

If we take e =1, then v * cos(alpha) = u * sin(alpha), this is kind of abnormal, It does not look possible, as it shows both the horizontal velocities are the same before and after the impact.

Confusion 2: "If the two objects were a closed system (wall would move), then momentum would apply and the value of e would be irrelevant."

Why would e become irrelavant here?

Thanks a lot btw for your time....
0
2 weeks ago
#17
(Original post by Tesla3)
So, basically,
As this is not a closed system (where no external forces are acting). It is an open system (where external forces are acting on the wall as the wall is stationary), therefore conservation of momentum does not apply over here.

This concept got cleared.

Confusion 1: "Even if e=1, the perpendicular to the wall velocity switches sign, as would momentum."

If we take e =1, then v * cos(alpha) = u * sin(alpha), this is kind of abnormal, It does not look possible, as it shows both the horizontal velocities are the same before and after the impact.

Confusion 2: "If the two objects were a closed system (wall would move), then momentum would apply and the value of e would be irrelevant."

Why would e become irrelavant here?

Thanks a lot btw for your time....
Conclusion 1) that's not velocity, it's speed. The velocity will switch direction/sign (it bounces), so will total momentum.
Conclusion 2) I mean that the wall would have mass and move. e is not irrelevant, but the value does not affect how you'd approach the problem.
Last edited by mqb2766; 2 weeks ago
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#18
(Original post by mqb2766)
Conclusion 1) that's not velocity, it's speed. The velocity will switch direction/sign (it bounces), so will total momentum.
Conclusion 2) I mean that the wall would have mass and move. e is not irrelevant, but the value does not affect how you'd approach the problem.
I see so the velocity changes direction so does the momentum, but horizontal speed remains the same.

That got cleared

Confusion 2: "If the two objects were a closed system (wall would move), then momentum would apply and the value of e would be irrelevant."

"I mean that the wall would have mass and move. e is not irrelevant, but the value does not affect how you'd approach the problem."

I think you accidentally typed the value of e will be irrelevant the first time.. Is that right, or.....
0
2 weeks ago
#19
(Original post by Tesla3)
I see so the velocity changes direction so does the momentum, but horizontal speed remains the same.

That got cleared

Confusion 2: "If the two objects were a closed system (wall would move), then momentum would apply and the value of e would be irrelevant."

"I mean that the wall would have mass and move. e is not irrelevant, but the value does not affect how you'd approach the problem."

I think you accidentally typed the value of e will be irrelevant the first time.. Is that right, or.....
The value of e is irrelevant for how you solve the problem, however obviously the value determines the actual solution.
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#20
(Original post by mqb2766)
The value of e is irrelevant for how you solve the problem, however obviously the value determines the actual solution.
Thanks a lot, I can't thank you enough, you saved the day. Further Maths can be tough sometimes, but just don't want that thing to make me give up. God bless you. 1
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