Physics frame of reference, is this possible?

1. Apr 4, 2013

x86

1. The problem statement, all variables and given/known data
Generic collision question.

For example, lets say you are asked the following question:

Object 1 is 10kg and is traveling at 10 m/s

Object 2 is 5kg and is traveling at 2 m/s

Find their final collisions after impact

2. Relevant equations
m1v1 + m2v2 = m1vf1 + m2vf2
Ek1i + Ek2i = Ek1f + Ek2f

Ek = (1/2)mv^2

3. The attempt at a solution

Now, instead of doing massive algebra and combining the two equations, can I include an object 3 into the scenario, and say it is going at 2 m/s
?

According to this new reference, object 1 will be going at 8 m/s and object 2 will be going at 0m/s and thereby be stationary.

Now, when I plugin values to the two equations, it gets a WHOLE lot easier because I can instantly cancel some values out and derive two new equations.

So, then I can say according to object 3, object 1 goes X m/s after collision, and object 2 goes Y m/s after collision.

Now, if i SUBTRACT the 2 m/s from object 1 and object 2, relative to the first reference I will get the final answers X+2 and Y+2 --> m/s

So, is this correct? Can I do this? Newtons laws state I should be able to, but my teacher said this is impossible and it doesn't work like that because I can't just add and subtract velocities. I said that all I'm doing is adding a reference point and removing it. I firmly think I can do this. If I cant do this, can someone explain why?​

2. Apr 4, 2013

SammyS

Staff Emeritus

What you are trying to do is valid. A couple of the details appear to be misstated, which may be what gives your teacher heartburn.​

3. Apr 4, 2013

x86

What details am I mistaking? I would like to know so that I can understand physics more.

Thank you.

4. Apr 5, 2013

ehild

You do have object 2 travelling at V=v2=2 m/s to the right. What you do is introducing a frame of reference, travelling with 2 m/s to the right.
Correct
You can say that object 1 travels with velocity X m/s and object 2 travels with Y m/s after the collision, with respect to the frame of reference travelling with 2 m/s.

If you subtract the 2 m/s of any velocities with respect to the ground, you get the velocities with respect to the new frame of reference.

You are correct, the final answers for the velocities with respect to the ground are X+2 and Y+2.
You can use any initial frame of reference, and all frames of references travelling with constant velocities are inertial.
-----------------------------------
Anyway, you can derive the formulae for velocities without much algebra without introducing the new frame of reference.

If the initial velocities are v1 and v2, the final ones are u1 and u2, conservation of momentum and energy yield the equations

m1v1+m2v2=m1u1+m2u2
m1v12+m2v22=m1u12+m2u22

Rearrange

m1(v1-u1)=-m2(v2-u2) (1)

m1(v12-u12)=-m2(v22-u22) (2)

Factorise the second equation

m1(v1-u1)(v1+u1)=-m2(v2-u2)(v2+u2)
and divide by (1)

v1+u1=v2+u2 (3)
Together with (1), you have two simple equations to be solved. Multiply (3) with m1 and add to (1), solve for u2: $$u_2=\frac{2m_1v_1+(m_2-m_1)v_2}{m_1+m_2}$$
Exchanging the subscripts, you get the analogous equation for u1.

ehild​