Conservation of Linear Momentum

[Zulu2018]

Homework Statement

This question was on a recent AP Physics 1 exam as a multiple choice; "Three air track gliders, shown to the right all have the same mass M. Gliders 2 and 3 are initially at rest. Glider 1 is moving to the right with speed v. Glider 1 collides with glider 2 and sticks to it. This combination collides elastically with glider 3. Which of the following is most nearly the final speed of Glider 3?"

Homework Equations

p = mv, p1 + p2 = p1' + p2'

The Attempt at a Solution

initial p = mv, but inelastic results in 2mv/2, but second collision is elastic so 2m/v2 = (2mv)' + (1/3mv)', solve for the 1/3m's v'

 

[haruspex]

but second collision is elastic so 2m/v2 = (2mv)' + (1/3mv)', solve for the 1/3m's v'

This is most unclear. Please write out in more detail.
Is the 2m/v2 a typo for 2mv/2?
What is your final answer? Is the official answer known?

 

[Zulu2018]

Apologies, yes it is a typo, should read 2mv/2. The final answer is .67v is what is given. Thank you!

 

[haruspex]

= (2mv)' + (1/3mv)',

I still don't know what that means or how you arrive at it.

The final answer is .67v is what is given

That is the answer you got, or it is the official answer, or both? If both, what is your question?

 

[Zulu2018]

I am requesting assistance as to what the correct answer is. Thank you for your help

 

[haruspex]

I am requesting assistance as to what the correct answer is. Thank you for your help

No need for that, just be clear in what information you are giving.
So the 0.67v is the answer you got, is that true, yes or no?

Please write out in detail how you got your answer, explaining your steps. What you have posted so far is mysterious.

 

[Zulu2018]

Yes. .67v is my answer.
pi = m1v1 inelastically collides with identical mass, now momentum shows velocity cut in half 2mv/2. This double-mass collides elastically with another identical mass, that must have a higher velocity than the double mass AFTER elastistic collision. Therefore the double mass velocity must decrease, while the last mass' velocity must increase. It made sense that the last mass would have 1/3 higher velocity and the double-mass would have just 1/3 velocity in order to conserve momentum.

 

[jbriggs444]

"It made sense" is not a calculation. Please show the calculation.

 

[Zulu2018]

Precisely why I am asking for help, Thank you in advance for any assistance.

 

[haruspex]

Precisely why I am asking for help, Thank you in advance for any assistance.

After the first collision you have, as you wrote, a mass 2m moving at speed v/2.
Write out the two conservation equations for the second collision.
Please make it clear what each variable represents.

 

[neilparker62]

Momentum transfer - elastic collision: $$Δp=2μΔv$$
where μ is the reduced mass of the colliding objects:$$μ=\frac{m_1m_2}{m_1+m_2}$$
and Δv their relative velocity along the collision line.