What is the Speed of the Incoming Sphere After a Perfectly Elastic Collision?

[Nguyenfa.t]

Homework Statement

Example 6: Two spheres
shown in the figure below. They both hang under the influence of
gravity on 2.0 m long strings and start at rest. The 2.0 kg sphere
starts at an angle  to the vertical such that
2.0 kg sphere is released. (
assume that g = 10.0
1. What is the speed of the 2.0 kg sphere just before it
collides with the 4.0 kg sphere
2. Assuming a perfectly elastic collision
the 2.0 kg sphere immediately after the collision
3. After the collision, what is the maximum possible angle
achieved by the 2.0 kg sphere
swings back up)?
4. If both spheres had a mass of 2.0 kg, what would the speed
of the incoming sphere have been after the perfectly
elastic collision (

I just need help with number 4 I did the rest. Thanks guys

Homework Equations

The Attempt at a Solution

 

[LowlyPion]

Sorry you need to supply more information about the layout of your problem, and more importantly where it is you are stuck in your solution.

 

[Nguyenfa.t]

i have attached the problem and the work that i have done. if someone could show me how to do number 4 i would really appreciate it. the work and the solution would be great.

 

[Nguyenfa.t]

can anyone please help??

 

[Nguyenfa.t]

can somebody please do number 4. I would really appreciate it.

 

[Nguyenfa.t]

can somebody please solve number 4 for me by midnight. thanks

 

[majin_andrew]

Hey, I'm not sure which time zone you are in, so I don't know if this is before midnight, but the worked example in this link seems very similar to Question 4. http://farside.ph.utexas.edu/teaching/301/lectures/node82.html

 

[Nguyenfa.t]

im in pacific time zone.

is there any way you can work it out for me please? I've been trying to do this problem for a while now and can't seem to figure it out

 

[Nguyenfa.t]

after using that link i got the answer of 2 m/s is that correct?

 

[queenofbabes]

The question asks about two equal masses in a perfectly elastic collision, so the answer should be zero. It's the only way to conserve both energy and momentum.

 

[Nguyenfa.t]

is there any way you can show the work for that please?

 

[Nguyenfa.t]

but the masses arent equal one is 2 kg while the other is 4kg

 

[queenofbabes]

"4. If both spheres had a mass of 2.0 kg..."

If you could do 1-3, how can you not do 4? It's the same thing with a different value of mass!

 

[Nguyenfa.t]

oh my bad misread the problem. i did 1-3 in class and have no idea how to do 4

 

[Nguyenfa.t]

can you please just show me how its done. thanks

 

[Nguyenfa.t]

i use the mv=mv+mv formula right then what?

 

[queenofbabes]

That was to conserve momentum. Now conserve energy, KE of sphere before collision = KE of both spheres after collision. It's exactly the same as question 2.

 

[Nguyenfa.t]

i don't see how you get 0 can you explain please by showing the work. it would be greatly appreciated.

 

[queenofbabes]

Conservation of momentum:
$$m_1u = m_1 v_1 + m_2 v_2$$

Conservation of energy:

$$\frac{1}{2}m_1 u^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$$


You know m1, m2, and you can work out u. Solve for v1.

 

[LowlyPion]

i don't see how you get 0 can you explain please by showing the work. it would be greatly appreciated.

You will note that to conserve both energy and momentum after impact with twice the mass involved, something has to give.
You at once have Va = Va' + Vb' and Va² = Va'² + Vb'².
Since Vb' must be non-zero, there is only one condition that can eliminate the extra middle term 2Va'Vb', namely that Va' must be 0.

Va'² + 2Va'Vb' + Vb'² = Va'² + Vb'²