Elastic collision -- Energy & Momentum

[Deadawake]

Homework Statement

Please see the attached photo. (down)
H_minitial= 1.5R
M = 2/3m
Perfectly elastic collision

What is the velocity of object m immidiatly after the collision? (by m,g,R)

Homework Equations

Conservation of energy
Conservation of momentum

The Attempt at a Solution

I assumed that because of the elastic collision the M object bounce back, so I considered positive velocity to the right and negative to the left (when we're looking exactly on the collision moments)

First I found the "before the collision" velocity of the falling ball M (with conservation of energy) which is √(3Rg)

Now I used two equation to find the velocities after the collision :

Conservation of momentum: MV_i + 0 = MV_f +mv_f
Conservation of energy: V₀ + V_f = v₀ + v_f

After some work I get :
V_f = ⅓⋅√(3Rg)
v_f = 4⋅√(3Rg) / 3

It seems make sense when the speed of the small ball got higher and the big ball, which bounced backward, slower .

But in the other hand I know that the M velocity after the collision has to be negative. because it bounced back!, perfectly elastic collision...
But I got 2 positive answers!

when I try to change the direction/sign (to minus) of the variable MV_f on the first equation I get different and not logical answer (V_f = √(3Rg) ) , like the bigger ball didn't lost energy at all while it was giving energy to the smaller ball...

Please some help :)

Thanks.

 

[Doc Al]

But in the other hand I know that the M velocity after the collision has to be negative. because it bounced back!, perfectly elastic collision...

That's not true. They have to "bounce apart" (not stick together) and conserve kinetic energy, but the first mass needn't reverse direction. Imagine that first mass being 100 times more massive. Is it likely that hitting a small mass would make it move backwards?

 

[Deadawake]

That's not true. They have to "bounce apart" (not stick together) and conserve kinetic energy, but the first mass needn't reverse direction. Imagine that first mass being 100 times more massive. Is it likely that hitting a small mass would make it move backwards?

You right. So basically the answers tell me that the bigger mass in this case didn't move backwards, It continued forward with third of its velocity?

 

[Doc Al]

You right. So basically the answers tell me that the bigger mass in this case didn't move backwards, It continue forward with third of its velocity?

Right!

 

[Deadawake]

Right!

Thank you very much.
For some reason I thought it needs to bounce backwards.

 

[haruspex]

Thank you very much.
For some reason I thought it needs to bounce backwards.

In a perfectly elastic head-on collision with a stationary object, the incomng object will bounce back if, and only if, it is lighter than the other. If they are equal it will come to rest.

 

[ehild]

Conservation of momentum: MV_i + 0 = MV_f +mv_f
Conservation of energy: V₀ + V_f = v₀ + v_f

After some work I get :
V_f = ⅓⋅√(3Rg)
v_f = 4⋅√(3Rg) / 3

Your result is not correct.

 

[Deadawake]

Your result is not correct.

What am I missing?

 

[ehild]

What am I missing?

I do not know without seeing your work in detail.
The numerical values do not fulfill the equations for conservation of momentum and energy.

 

[Doc Al]

Yikes, ehild is right!

Conservation of momentum: MVi + 0 = MVf +mvf
Conservation of energy: V0 + Vf = v0 + vf

That equation for conservation of energy is incorrect.

After some work I get :
Vf = ⅓⋅√(3Rg)
vf = 4⋅√(3Rg) / 3

Thus these results are incorrect.

Fix your conservation of energy equation and resolve.

Thanks to ehild for paying attention!

(Sorry for not checking your work earlier!)

 

[ehild]

That equation for conservation of energy is incorrect.

It is correct, if Vi stands for V0 and vi stands for v0 in
V0 + Vf = v0 + vf

Capital letters refer to the big ball, small case letters mean the velocities of the small ball.

 

[Doc Al]

It is correct, if Vi stands for V0 and vi stands for v0 in
V0 + Vf = v0 + vf

Capital letters refer to the big ball, small case letters mean the velocities of the small ball.

Once again, ehild is correct (as always).

(That equation isn't a statement of energy conservation only, but is a result of energy conservation plus momentum conservation.)

@Deadawake: You must have made an algebra/computational error; try it once more.

 

[ehild]

Once again, ehild is correct (as always).

This is not true

 

[Deadawake]

Once again, ehild is correct (as always).

(That equation isn't a statement of energy conservation only, but is a result of energy conservation plus momentum conservation.)

@Deadawake: You must have made an algebra/computational error; try it once more.

Correct. this statement is result of conservation of energy and momentum equations.
V_{before collision} = √(3Rg)

1)
MV₀ + 0 = MV_f+mv_f

⇒ 1.5mV₀ = 1.5mV_f +mv_f
⇒ v_f = 1.5⋅√(3Rg) - 1.5V_f

2)
V₀ + V_f = v₀ + v_f
⇒ √(3Rg) +V_f = 0 +v_f
⇒v_f = √(3Rg) +V_f

⇒⇒⇒√(3Rg) +V_f = 1.5⋅√(3Rg) - 1.5V_f
⇒⇒⇒2.5V_f = 0.5⋅√(3Rg) →here was my error. I wrote 1.5 instead of 2.5
⇒⇒⇒V_f = 0.2√(3Rg)
⇒⇒⇒v_f = √(3Rg) +0.2√(3Rg) = 1.2√(3Rg)

 

[Doc Al]

⇒⇒⇒Vf = 0.2√(3Rg)
⇒⇒⇒vf = √(3Rg) +0.2√(3Rg) = 1.2√(3Rg)

Good. My answers agree with yours.

 

[Deadawake]

Some other question is - What height does the small mass ball reach after the collision? (h_f)
(Just a reminder - we are talking about vertical standing circle/ring...)
So I have the initial speed of the small ball (after the collision) and I can use conservation of energy
½mv₀² = mgh_f
½m⋅(1.2√3gh)² = mgh_f
½⋅1.44⋅(3Rg) = g⋅h_f
h_f = 2.16⋅R

We know that the radius of the circle is R , so the maximum possible height is 2R.
Then the answer tells me that the small mass has enough energy to reach the top of the circle and a bit more.
The cunclusion is that the small ball will get to the maximum height which is 2R.
Am I right?

 

[Doc Al]

Sounds good to me.