What Are the Velocities of Two Blocks After an Elastic Collision?

[Carlitos]

Homework Statement

Two blocks collide on a perfectly elastic collision, after sliding on a frictionless surface.
va = 1m/s
vb = 1m/s
ma = 0.05kg
mb = 0.03kb

I need to find the speed of both blocks after the crash.

Homework Equations

Conservation of momentum: mava+mbvb = mava' + mbvb'
Conservation of energy: 1/2mava^2 + 1/2mbvb^2 = 1/2mava'^2 + 1/2mbvb'^2

General solution for these two equations: -(vb−va)=vb'-va'

Refer to this thread for cool formatting of these equations (i don't really know how to do that):

https://www.physicsforums.com/threads/elastic-collision.753780/

I don't really understand where that last equation comes from or how to get there.

The Attempt at a Solution

I solved for the conservation of momentum:

mava+mbvb = mava' + mbvb'
0.05kg.1m/s + 0.03kg.1m/s = 0.05kg.va' + 0.03kg.vb'
0.08ms = 0.05va' + 0.03vb' (*)

Then on the third equation:

-(vb−va)=vb'-va'
-(-1m/s - 1m/s) = vb'-va'
2m/s = vb'-va'
2m/s +va' = vb'

Then i plug this on (*) and i get

0.08m/s = 0.05va' + 0.03(2m/s + va')
0.08m/s = 0.05va' + 0.06m/s + 0.03va'
0.02m/s = 0.08va'
0.25m/s = va'

then vb' = 2.25m/s

The expected solution is 0.5m/s and 2m/s .

So my two questions are, am i doing something wrong here? And how do i get to the third equation?

 

[ehild]

Homework Statement

Two blocks collide on a perfectly elastic collision, after sliding on a frictionless surface.
va = 1m/s
vb = 1m/s
ma = 0.05kg
mb = 0.03kb

I need to find the speed of both blocks after the crash.

If both blocks have the same velocity they never collide. Check the problem text.

Homework Equations

Conservation of momentum: mava+mbvb = mava' + mbvb'
Conservation of energy: 1/2mava^2 + 1/2mbvb^2 = 1/2mava'^2 + 1/2mbvb'^2

General solution for these two equations: -(vb−va)=vb'-va'

It is not true.

Refer to this thread for cool formatting of these equations (i don't really know how to do that):

https://www.physicsforums.com/threads/elastic-collision.753780/

I don't really understand where that last equation comes from or how to get there.

That third equation is wrong. You get the correct one by arranging both the first and second equations that quantities labelled by "a" are on one side and those labelled by "b" are on the other side.

$m_a (v_a-v_a' ) = - m_b(v_b-v_b')$ *
$m_a(v_a^2-v_a'^2)= - m_b(v_b^2-v_b'^2)$ **
Factorize the second equation and divide it by the first one: you get a third equation
v_a+v_a' =v_b+v_b'

 

[Carlitos]

Thanks for the answer, my bad on that veocity, vb should be -1 m/s.

Then technically va + va' = vb + vb'
is the same as -(vb − va)=vb' - va'.

I see how to find that equation now.

 

[ehild]

Thanks for the answer, my bad on that veocity, vb should be -1 m/s.

Then technically va + va' = vb + vb'
is the same as -(vb − va)=vb' - va'.

Sorry, I misread your equation, they are the same