Elastic Collision - but no Final Velocities given

[PhysicsPhil]

[SOLVED] Elastic Collision - but no Final Velocities given!

Homework Statement

An object of mass 2.0kg is moving at 4.0 m/s to the right. It collides elastically with an object of mass 1.0kg moving at 12.0 m/s to the left. Calculate the velocities of the objects after the collision.

Homework Equations

$$m_1=2.0kg v_1 =4.0m/s m_2=1.0kg v_2=-12m/s$$

The Attempt at a Solution

I know the equation for an elastic collision:
$$m_1*v_1,i + m_2*v_2,i = m_1*v_1,f + m_2*v_2,f$$
(the i = initial, f = final)

I've been given this challenge problem, and all my samples in the textbook have one of the unknowns given to us ... I'll show you what I mean, when I plug-in:

$$(2.0kg)(4.0m/s) + (1.0kg)(-12m/s) = (2.0kg)v_1f + (1.0kg)v_2f$$

How can I solve for v1f and v2f if I don't have one of them given to me? In the examples from the book one of those is always given, but not in this challenge problem, so I'm not sure how to go about solving it. I tried a system of equations where I solved for v2f in terms of v1f, but then when I plugged back into the same equation I got 0 (which of course I expected since I need two equations to solve a system of equations) ... I'm really at a loss ... any help?

 

[fliinghier]

what else is conserved?

 

[mrdummy]

this is easy
in any physics equation
especially these kinds
u have to use the equation p = mv
right = pos...left = -
momentum = mass * velocity

starting momentum = 2 * 4 = 8 and p = 1 * -12
total P = 8 - 12
p = -4
this has to be the ending momentum too
due to the law of conservation of momentum

- 4 = ( 2 + 1) kg * v
v = - 4 / 3

 

[fliinghier]

actually it is an elastic collision. the objects rebound off of one another, they don't stick together.

 

[mrdummy]

so r u saying
momentum is not conserved

 

[fliinghier]

no, it is. hoever there are an infinite number of possibilities for the velocities of both objects. they are not necessarily equal. something else is conserved, and using it will get you an answer.

 

[PhysicsPhil]

Kinetic Energy is conserved in any elastic collision. I'll try to see if I can piece something together from that, but wouldn't KE depend on velocity as well?

 

[PhysicsPhil]

I got this sucker. Conservation of Momentum. Conservation of KE. Solve system of equations! :)

 

[clipperdude21]

just to see if i did this right... were ur answers 9.33 m/s to the right for the larger mass and -6.66 m/s for the smaller one.

 

[PhysicsPhil]

just to see if i did this right... were ur answers 9.33 m/s to the right for the larger mass and -6.66 m/s for the smaller one.

Yes... :)

 

[Ajwrighter]

PEi = PEf -----> m1V1 + m2V2 = m1V1' + m2V2' then (using algebra) --------------------> m1(V1-V1') = -m2(V2 - V2')

KEi = KEf ------> $$\frac{1}{2}$$(m1)(V1^2) + $$\frac{1}{2}$$(m2)(V2^2) = $$\frac{1}{2}$$(m1)(V1^2') + $$\frac{1}{2}$$(m2)(V2^2')
then -----> m1(V1^2-V1^2') = -m2(V2^2 - V2^2')

once your equations are set up start with PEi = PEf ---> replace your known variables and move everything over except the V2' usually looks something like V2' = $$\frac{m1V1 -V1}{m2}$$

now you start on your KEi = KEf again fill in all the blanks and replace your V2' with your new equation from PEi = PEf , when solving for it it should be something like this ($$\frac{m1V1 -V1}{m2}$$) ( $$\frac{m1V1 -V1}{m2}$$) one of your m2 should cancle out the m2 next to the V2' then you multiply the remainder with all the other varialbes. Then you add your V1'2 and subtract your numbers on the left with the remaining numbers on the right and solve for Vi' you will get two answers (Vi' = -x) Vi' = +x) just look at it logicly and you should be able to determine which one is correct. Once you have your answer you can easily solve for V2' by filling in V1' on PEi = PEf.

 

[Ajwrighter]

hmm my ( ' ) marks didnt all come through .. oh well you get the general idea