# Momentum and Newton's Law of Restitution

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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.

( 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.

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#2

(Original post by

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.

**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.

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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(Original post by

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.

**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.

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.

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#4

(Original post by

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.

**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.

Last edited by mqb2766; 2 weeks ago

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(Original post by

The conservation of momentum perpendicular to the wall means e=1. That is why it's not applicable here.

**mqb2766**)The conservation of momentum perpendicular to the wall means e=1. That is why it's not applicable here.

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#6

(Original post by

So for conservation of momentum parallel to the wall, e is not equal to 1?

**Tesla3**)So for conservation of momentum parallel to the wall, e is not equal to 1?

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(Original post by

e only applies perpendicular to the wall. So parallel to the wall, motion is unaffected.

**mqb2766**)e only applies perpendicular to the wall. So parallel to the wall, motion is unaffected.

Last edited by Tesla3; 2 weeks ago

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#8

(Original post by

In the second part, e = 1/3, why is it not e = 1. ?

**Tesla3**)In the second part, e = 1/3, why is it not e = 1. ?

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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(Original post by

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?

**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?

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#11

(Original post by

mqb2766 you are saying that momentum is not conserved perpendicular to the barrier. Why is it not conserved?

**Tesla3**)mqb2766 you are saying that momentum is not conserved perpendicular to the barrier. Why is it not conserved?

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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(Original post by

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.

**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.

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#13

(Original post by

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.

**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.

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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(Original post by

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.

**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.

Is it wrong?

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#15

(Original post by

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?

**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?

I think this is where your misunderstanding is.

Last edited by mqb2766; 2 weeks ago

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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....

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....

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#17

(Original post by

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....

**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 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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(Original post by

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.

**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.

That got cleared

But about the 2nd confusion:

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.....

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#19

(Original post by

I see so the velocity changes direction so does the momentum, but horizontal speed remains the same.

That got cleared

But about the 2nd confusion:

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.....

**Tesla3**)I see so the velocity changes direction so does the momentum, but horizontal speed remains the same.

That got cleared

But about the 2nd confusion:

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.....

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(Original post by

The value of e is irrelevant for how you solve the problem, however obviously the value determines the actual solution.

**mqb2766**)The value of e is irrelevant for how you solve the problem, however obviously the value determines the actual solution.

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