Find the coefficient of restitution

AI Thread Summary
The discussion revolves around calculating the coefficient of restitution for a scenario involving three equal balls, where a third ball strikes two stationary balls simultaneously. The initial approach used conservation of momentum and expressions for the coefficient of restitution but led to an incorrect conclusion. Participants highlighted the need to reconsider the expressions and check for errors in momentum conservation, noting that the collision is inelastic. The correct deduction reveals that the angles of impact are equal, leading to the conclusion that the coefficient of restitution is 2/3. This solution emphasizes the importance of accurately applying the definitions and principles of momentum and restitution in collision problems.
utkarshakash
Gold Member
Messages
852
Reaction score
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.

\frac{v_{2}}{vcosθ_{2}} = e

Also,
\frac{v_{1}}{vcosθ_{1}} = e

Applying Conservation of Momentum along X-axis

v=v_{1}cosθ_{1}+v_{2}cosθ_{2}
Along Y- axis
v_{1}sinθ_{1}=v_{2}sinθ_{2}

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

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

θ_{1}=θ_{2} and θ_{1} = \frac{∏}{4}

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

1=e(cos^{2}θ_{1}+cos^{2}θ_{2})
e=\frac{1}{cos^{2}θ_{1}+cos^{2}θ_{2}}
e=\frac{1}{2cos^{2}θ_{1}}
e=1

What's wrong here? :confused:
 

Attachments

  • PF1.gif
    PF1.gif
    19.4 KB · Views: 666
Physics news on Phys.org
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.
 
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.
 
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.
 
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,
\theta_1=\theta_2=\frac{\pi}{6}\\<br /> and\\<br /> v_1=v_2
Then your answer is therefore e=\frac{2}{3}
 

Attachments

  • line of impact.png
    line of impact.png
    3.4 KB · Views: 548
Last edited:
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,
\theta_1=\theta_2=\frac{\pi}{6}\\<br /> and\\<br /> v_1=v_2
Then your answer is therefore e=\frac{2}{3}

Hey, why didn't I think it earlier! Thank You.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top