2 swinging balls (pendulum) problem

[awiest82]

Homework Statement

A pendulum of mass m_1 is released at an angle theta_1 and strikes the pendulum of mass m_2, which rises to angle theta_2. The coefficient of restitution e for the two masses is given. Derive an expression for the angle theta_2.

Homework Equations

KE_1+PE_1=KE_2+PE_2

e=(v₂'-v₁')/(v₁-v₂)

The Attempt at a Solution

Solving for v₁ at time of impact using conservation of energy I get v₁=Sqrt(2gl(1-cos(theta_1)))

I solve the same equation for v'₂=sqrt(2gl(1-cos(theta_2)))

I then plug in the coefficient of restitution equation and solve for theta_2

theta_2= arccos(1-((e*Sqrt(2gl(1-cos(theta_1)))+v'₁)^2)/2gl

My issue is determining whether or not this is a complete elastic collision. If so I assume the v'₁ would be omitted, correct?

 

[kuruman]

To determine if the collision is perfectly elastic, you need to look at the coefficient of restitution. If e =1, then the collision is perfectly elastic. If e = 0, the collision is perfectly inelastic (the masses stick together). If 0 < e < 1, the collision is inelastic. Unless a number for e is given, that's all you can say.

Why do you say that v'₁ = 0 for an elastic collision? This would be the case only if m₁=m₂.

 

[awiest82]

The problem is purely symbolic and since its not specified otherwise I assume that m1=m2. If that was the case then v'₁ would be zero correct? I leave the coefficient of restitution in since the problem states that's it's a known.

Besides those questions you had, does it look like I am on the right track or answered correctly?

 

[kuruman]

The problem is purely symbolic and since its not specified otherwise I assume that m1=m2.

If it is not specified otherwise, then you must assume the most general case, and that is that the two masses are not the same.

If that was the case then v'₁ would be zero correct?

Correct, but like I said above you may not assume that the masses are equal in which case v'₁ is not zero. You need to find what it is. I suggest that you use the momentum conservation equation to do so.

I leave the coefficient of restitution in since the problem states that's it's a known.

That is good.

Besides those questions you had, does it look like I am on the right track or answered correctly?

Address the issues that I raised, and you will be on the right track.

 

[awiest82]

I have taken your suggestion and used the conservation of momentum equation to solve for v'₁ and the resultant equation I get is:

v'₁=v₁-(m₂)/(m₁)*v'₂

plugging this into the coefficient of restitution equation along with the other solved variables leaves me with:

e=(Sqrt(2gl(1-cos(theta_2))-Sqrt(2gl(1-cos(theta_1))-(m₂/m₁)*Sqrt(2gl(1-cos(theta_2)))/(Sqrt(2gl(1-cos(theta_1)))

Before I begin the "painful" exercise of simplifying and solving this, is this looking a bit better?

 

[kuruman]

I have taken your suggestion and used the conservation of momentum equation to solve for v'₁ and the resultant equation I get is:

v'₁=v₁-(m₂)/(m₁)*v'₂

plugging this into the coefficient of restitution equation along with the other solved variables leaves me with:

e=(Sqrt(2gl(1-cos(theta_2))-Sqrt(2gl(1-cos(theta_1))-(m₂/m₁)*Sqrt(2gl(1-cos(theta_2)))/(Sqrt(2gl(1-cos(theta_1)))

Before I begin the "painful" exercise of simplifying and solving this, is this looking a bit better?

Much better. The solution is not all that painful. Note that you have a common factor of sqrt(2gl) in both numerator and denominator. Once you get rid of that, solve for sqrt(1-cos(θ₂)) in terms of sqrt(1-cos(θ₁)) and you are almost done.

 

[awiest82]

Ok let's see how this looks for a final answer

Theta₂ = arccos(1-((e+1)²*m₂²*(1-cos(theta₁)))/(m₂-m₁)²)

The only issue I see with this answer is that if m₁ and m₂ are identical, we then have a 0 in the denominator. I have worked it out twice and gotten the same answer both times, so I think this is right.

 

[kuruman]

After looking at your expression for e more carefully, I see that you have a wrong sign in front of the mass ratio. If
e = |v'₂ - v'₁|/v₁
and
v'₁ = v₁ - v'₂*m₂/m₁
then
e = |v'₂ - (v₁ - v'₂*m₂/m₁)|/v₁
and
e = |v'₂*(1+m₂/m₁) - v₁|/v₁

You can take it from here.

 

[awiest82]

Aha! I forgot to distribute that minus sign!

Here is what I get now and no longer have a 0 in denominator.

Theta₂ = arccos(1-((e+1)²*(m₂)²*(1-cos(theta₁)))/(m₂+m₁)²)

Look better?

 

[kuruman]

I haven't done the algebra, but it looks OK. Why don't you test your expression? What should θ₂ be when m₁=m₂ and you have a perfectly elastic collision, i.e. e = 1? Does your expression give you what you expect?