Find the coefficient of restitution

In summary, the problem involves three equal balls in contact on a table and a third ball striking them simultaneously, resulting in the third ball remaining at rest after the impact. The goal is to show that the coefficient of restitution is 2/3. Using conservation of momentum, equations are derived for the coefficient of restitution and it is found that the value is actually 1, which indicates an elastic collision. However, it is later realized that the collision is in fact inelastic, and after considering the definition of the coefficient of restitution and the conditions of the problem, it is concluded that the correct value is 2/3.
  • #1
utkarshakash
Gold Member
854
13

Homework Statement


Two equal balls are in contact on a table. A third equal ball strikes them simultaneously and remains at rest after the impact. Show that the coefficient of restitution is 2/3.
I have attached an image for clarity of problem. Please open it.

Homework Equations


Conservation of Momentum

The Attempt at a Solution


Let the mass of balls be m and coefficient of restitution be e.

[itex]\frac{v_{2}}{vcosθ_{2}} = e[/itex]

Also,
[itex]\frac{v_{1}}{vcosθ_{1}} = e[/itex]

Applying Conservation of Momentum along X-axis

[itex]v=v_{1}cosθ_{1}+v_{2}cosθ_{2}[/itex]
Along Y- axis
[itex]v_{1}sinθ_{1}=v_{2}sinθ_{2}[/itex]

Substituting the value of v1 and v2 in the above equation
[itex]evcosθ_{1}sinθ_{1}=evcosθ_{2}sinθ_{2}[/itex]
[itex]cosθ_{1}sinθ_{1}=cosθ_{2}sinθ_{2}[/itex]

Multiplying 2 on both sides
[itex]sin2θ_{1}=sin2θ_{2}[/itex]
Rearranging and simplifying
[itex]2cos(θ_{1}+θ_{2})sin(θ_{1}-θ_{2})=0[/itex]
[itex]θ_{1}+θ_{2}=\frac{∏}{2}[/itex]
[itex]θ_{1}-θ_{2}=0[/itex]

[itex]θ_{1}=θ_{2} and θ_{1} = \frac{∏}{4}[/itex]

Now substituting the value of θ1 in equation of momentum along X-axis

[itex]1=e(cos^{2}θ_{1}+cos^{2}θ_{2})[/itex]
[itex]e=\frac{1}{cos^{2}θ_{1}+cos^{2}θ_{2}}[/itex]
[itex]e=\frac{1}{2cos^{2}θ_{1}}[/itex]
[itex]e=1[/itex]

What's wrong here? :confused:
 

Attachments

  • PF1.gif
    PF1.gif
    19.4 KB · Views: 577
Physics news on Phys.org
  • #2
I will say first to reconsider your expressions for the coefficients of restitution. Also check the momentum conservation. Furthermore it may help to consider energy conservation as well. As this gives a further restraint on the values in the problem.
 
  • #3
klawlor419 said:
I will say first to reconsider your expressions for the coefficients of restitution. Also check the momentum conservation. Furthermore it may help to consider energy conservation as well. As this gives a further restraint on the values in the problem.
energy conservation principle can't be applied here because the collision is not elastic. I also suspect something wrong in my expressions for coefficient of restitution. But I can't figure out what is it. Also I don't think that momentum conservation equations are wrong.
 
  • #4
Well if it is indeed an inelastic collision than you are correct. It was not specified in your problem statement. You're off by a minus sign in one of your momentum conservation equations. The definition of the coefficient of restitution is fractional value of the ratio of speeds before and after a collision. You should be able to find the error by just considering this definition and looking at your expressions for, e.
 
  • #5
If the 3 balls are identical and the collisions occur simultaneously. You can immediately deduce that.
The lines of impacts of both collisions pass through the centres and therefore,
[tex]\theta_1=\theta_2=\frac{\pi}{6}\\
and\\
v_1=v_2[/tex]
Then your answer is therefore [itex]e=\frac{2}{3}[/itex]
 

Attachments

  • line of impact.png
    line of impact.png
    3.4 KB · Views: 496
Last edited:
  • #6
shutdo said:
If the 3 balls are identical and the collisions occur simultaneously. You can immediately deduce that.
The lines of impacts of both collisions pass through the centres and therefore,
[tex]\theta_1=\theta_2=\frac{\pi}{6}\\
and\\
v_1=v_2[/tex]
Then your answer is therefore [itex]e=\frac{2}{3}[/itex]

Hey, why didn't I think it earlier! Thank You.
 

Related to Find the coefficient of restitution

1. What is the coefficient of restitution?

The coefficient of restitution is a measure of the elasticity of a collision between two objects. It is a dimensionless number that represents the ratio of the relative velocity of the objects after the collision to the relative velocity before the collision.

2. How is the coefficient of restitution calculated?

The coefficient of restitution is calculated by dividing the relative velocity of the objects after the collision by the relative velocity before the collision. This can be calculated using the initial and final velocities of the objects, as well as the mass and elasticity of the objects.

3. What factors affect the coefficient of restitution?

The coefficient of restitution is affected by the elasticity or stiffness of the objects involved in the collision, as well as the materials and surface properties of the objects. Other factors such as the angle and speed of the collision can also impact the coefficient of restitution.

4. Why is the coefficient of restitution important?

The coefficient of restitution is important in understanding the behavior of objects during collisions. It can help determine the amount of energy lost or conserved during a collision, as well as the potential damage or deformation that may occur. It is also an important concept in sports and engineering, as it affects the performance and safety of equipment.

5. How is the coefficient of restitution used in real-world applications?

The coefficient of restitution is used in a variety of real-world applications, such as designing sports equipment like balls and racquets, analyzing car crashes and designing safer cars, and understanding the behavior of materials in impact testing. It is also used in physics and engineering simulations to accurately model the behavior of objects in collisions.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
902
  • Introductory Physics Homework Help
Replies
15
Views
314
  • Introductory Physics Homework Help
Replies
1
Views
139
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
773
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
11K
  • Introductory Physics Homework Help
Replies
13
Views
779
  • Introductory Physics Homework Help
Replies
10
Views
922
Back
Top