Coefficient of restitution

In summary, we have a problem where small object A with mass M collides with small object B with mass m on a horizontal stand fixed to a floor. Both objects then shoot off the stand and fall to the floor. The horizontal distance D that A lands from the edge of the stand is half of the distance d that B lands. We are asked to find the coefficient of restitution between A and B. After going through calculations, it is determined that the final velocity of A and B cannot be determined without knowing the initial velocities before the collision. However, it is noted that object B will fall further and have a higher velocity after the collision if it reaches twice the distance.
  • #1
Yoruichi
17
0

Homework Statement


As shown in the figure below, small object A (mass: M) collides with small object B (mass: m), which is initially at rest, on top of a horizontal stand fixed to a horizontal floor. Both A and B proceed to shoot horizontally off the stand and fall to the floor. Horizontal distance D from the edge of the stand to the point where A lands is 1/2 of the horizontal distance d to the point where B lands. Friction between the two objects and the stand is negligible.

https://scontent-kul1-1.xx.fbcdn.net/hphotos-xfp1/v/t34.0-12/12767193_1257518357595899_654751098_n.jpg?oh=5ee0294835905cc57ce2d8b8b2eed855&oe=56CC2F3E

What is the coefficient of restitution between A and B?

Homework Equations



a899eacd21a1ea3a57908b2beefff863.png


The Attempt at a Solution


If I simply substitute M1U1 + M2U2 = M1V1 + M2V2 and the equation above, I will come out with (M - m)/ 2M but that's wrong since I didn't put in the distance factor..

I'm not sure how to bring in the distance factor into the restitution formula. Should I use the v^2 = u^2 + 2as formula?

Also, is it that both A and B are in rest when they land on the floor? If that's the case, won't the final velocity of the object be 0?
 
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  • #2
What do you think which object falls further?
You have a collision and a horizontal projection following it.

Yoruichi said:
Also, is it that both A and B are in rest when they land on the floor? If that's the case, won't the final velocity of the object be 0?
No, you do not know what happens after the objects reach the floor. They can rebound, slide further, but you are interested in the velocity just before impact.
Push a ball from a table, and see what happens when it lands on the floor.
 
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  • #3
Yoruichi said:
If I simply substitute M1U1 + M2U2 = M1V1 + M2V2 and the equation above, I will come out with (M - m)/ 2M but that's wrong since I didn't put in the distance factor..

well i failed to understand - where you are wrong and what is correct according to you ?
try to explain !
Yoruichi said:
Also, is it that both A and B are in rest when they land on the floor? If that's the case, won't the final velocity of the object be 0?
do you think you are correctly describing the physical situation! check by doing a simple exercise of throwing a ball horizontally from a height h?
 
  • #4
Thanks for the replies!

ehild said:
How did you get that result?
drvrm said:
well i failed to understand - where you are wrong and what is correct according to you ?
try to explain !

Below is my attempt according to the two formulas.

M1U1 + M2U2 = M1V1 + M2V2
Since U2 = 0, M1U1 = M1V1 + M2V2

U1 = (M1V1 + M2V2) /M1
V1 = (M1U1 - M2V2) / M1
V2 = (M1U1 - M1V1) / M2

Using the formula
a899eacd21a1ea3a57908b2beefff863.png

Cr = [ (M1U1 - M1V1) / M2) - (M1U1 - M2V2) / M1] / [(M1V1 + M2V2) / M1]
= {[(M1M1U1 - M1M1V1) / M1M2 ] - [M1M2U1 - M2M2V2 / M1M2]} x M1 / (M1V1 + M2V2)
= { [M1M1U1 - M1M1V1 - M1M2U1+ M2M2V2] / M1M2 } x (M1V1 + M2V2)
= [ M1M1U1 - M1M1V1 - M1M2U1 + M2M2V2 ] / (M1M2V1 + M2M2V2)

*edited
sorry I'm kinda messed up with this..

My bad, after this I realize that this should be the right answer for my attempt.
But still, I doubt that this is the wrong answer. As this formula doesn't require the horizontal distance to come out with the answer... (and I guess this is a bit over complicated?)

ehild said:
No, you do not know what happens after the objects reach the floor. They can rebound, slide further, but you are interested in the velocity just before impact.
Push a ball from a table, and see what happens when it lands on the floor.

drvrm said:
do you think you are correctly describing the physical situation! check by doing a simple exercise of throwing a ball horizontally from a height h?

I see, this is my misunderstanding..

ehild said:
What do you think which object falls further?

Object B falls further?
 
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  • #5
Yoruichi said:
Thanks for the replies!

Below is my attempt according to the two formulas.

M1U1 + M2U2 = M1V1 + M2V2
Since U2 = 0, M1U1 = M1V1 + M2V2

U1 = (M1V1 + M2V2) /M1
V1 = (M1U1 - M2V2) / M1
V2 = (M1U1 - M1V1) / M2

Using the formula
a899eacd21a1ea3a57908b2beefff863.png

= [ M1M1U1 - M1M1V1 - M1M2U1 + M2M2V2 ] / (M1M2V1 + M2M2V2)
My bad, after this I realize that this should be the right answer for my attempt.
But still, I doubt that this is the wrong answer. As this formula doesn't require the horizontal distance to come out with the answer... (and I guess this is a bit over complicated?)
It is not an answer. You need to give CR with known quantities, the masses of the objects M and m and the ratio of the horizontal distances they reach. You do not know V1 and V2. It has no sense to make the original formula more complicated by still having the same unknowns.

Yoruichi said:
Object B falls further?

Yes, what does it mean for its velocity after the collision?
 
  • #6
ehild said:
It is not an answer. You need to give CR with known quantities, the masses of the objects M and m and the ratio of the horizontal distances they reach. You do not know V1 and V2. It has no sense to make the original formula more complicated by still having the same unknowns.
Yes, what does it mean for its velocity after the collision?

I think its velocity after the collision will be faster.
 
  • #7
Yoruichi said:
I think its velocity after the collision will be faster.
How much faster if it reaches twice the distance then A?
 
  • #8
ehild said:
How much faster if it reaches twice the distance then A?
The velocity will also be 2 times faster than A, so the ratio of both distance and velocity will be 1:2. Am I right?
So the next step would be substitute the ratio into the formula?
 
  • #9
Yoruichi said:
The velocity will also be 2 times faster than A, so the ratio of both distance and velocity will be 1:2. Am I right?
So the next step would be substitute the ratio into the formula?
yes, into both formula. First find VA in terms of the masses and UA, then substitute both VA and VB=2VA into the formula for CR.
 
  • #10
ehild said:
yes, into both formula. First find VA in terms of the masses and UA, then substitute both VA and VB=2VA into the formula for CR.

Let V2 = 2V1

M1U1 = M1V1 + M2V2
M1U1 = M1V1 + 2M2V1
V1(M1+2M2) = M1U1
V1 = M1U1 / (M1 + 2M2)

CR = (2V1 - V1) / U1
CR = V1 / U1

Sub V1 = M1U1 / (M1 + 2M2)
CR = V1 = M1U1 / (M1 + 2M2) / U1
CR = V1 = M1U1 / (M1 + 2M2) x (1 / U1)
CR = M1 / (M1+2M2)

This is the correct answer, thanks a lot! :D
 
  • #11
You are welcome :smile:
 

What is the coefficient of restitution?

The coefficient of restitution, also known as COR or e, is a measure of the bounciness or elasticity of a collision between two objects. It is defined as the ratio of the relative velocity of the objects after the collision to the relative velocity before the collision.

How is the coefficient of restitution calculated?

The coefficient of restitution can be calculated using the equation e = (vf - vi) / (ui - uf), where vi and vf are the initial and final velocities of one object, and ui and uf are the initial and final velocities of the other object.

What factors affect the coefficient of restitution?

The coefficient of restitution can be affected by various factors, such as the material and surface properties of the objects involved, the angle and speed of the collision, and the presence of external forces like friction or air resistance.

Why is the coefficient of restitution important?

The coefficient of restitution is important in various fields of study, such as physics, engineering, and sports. It helps to understand the behavior of objects in collisions and can be used to optimize designs for maximum efficiency and safety.

Can the coefficient of restitution be greater than 1?

Yes, the coefficient of restitution can be greater than 1. This means that the objects involved in the collision will bounce back with a greater speed than before the collision. This is known as a superelastic collision and is often observed in sports like tennis and basketball.

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