Elastic Collision Problem: Finding Velocities of Two Bodies

[toesockshoe]

Homework Statement

A 3 kg body (mass1) moving at 4 m/s makes an elastic collision with a stationary body(mass2) of mass 2 kg. Find the velocity of each body after the collision

Homework Equations

pi=pf
w=delta e

The Attempt at a Solution

so because it is elastic collision, it means that kinetic energy is conserved... we can do the following:
$W=\Delta E$
$0= \Delta KE_1 + \Delta KE_2$
$0= \frac{1}{2}m_1v_{f1}^2-\frac{1}{2}m_1v_{i1}^2+\frac{1}{2}m_2v_{2f}^2-\frac{1}{2}m_2v_{2i}^2$
$v_{f1} = \sqrt{\frac{m_1v_{i1}^2-m_2v_{2f}^2}{m_1}}$
ugh latex won't work:
vf1=((m1(v1f)^2-m2(v2f)^2)/m1)^1/2
now we can use momentum conservation to say

pi=pf
$m_1v_{i1}=-m_1v_{f1}+m_2v_{f2}$
we know vf1 from the energy part so we have the following:
m_1v_{i1}=m_1\sqrt{\frac{m_1v_{i1}^2-m_2v_{f2}^2}{m_1}+m_2v_{f2} [/itex]

so i have one unknown ($v_{f2}$) and one equation, but that seems awfully horrible to eliminate the final velcity of mass 2. am i even doing it correctly?

 

[RaulTheUCSCSlug]

Your equation should be m_1V_1=m_1V_1'+ m_2V_2

Think about it, why would the second m_1V_1 be negative?

Also I think you are confused on this equation:

0=12m1v2f1−12m1v2i1+12m2v22f=12m2v22i 0= \frac{1}{2}m_1v_{f1}^2-\frac{1}{2}m_1v_{i1}^2+\frac{1}{2}m_2v_{2f}^2=\frac{1}{2}m_2v_{2i}^2

the second mass initially has a velocity of zero.

 

[toesockshoe]

Your equation should be m_1V_1=m_1V_1+ m_2V_2

Think about it, why would the second m_1V_1 be negative?

Also I think you are confused on this equation:the second mass initially has a velocity of zero.

i thought it would be negative becasue it is going in the other direction...
i know v2i is 0... that's why i just took it out by the next step (its not in the answer inside my square root).

 

[haruspex]

equation should be m_1V_1=m_1V_1+ m_2V_2

You don't mean that, I hope.
I assume your point is that it is better to keep a constant definition of which direction is positive. I agree.

 

[RaulTheUCSCSlug]

You don't mean that, I hope.
I assume your point is that it is better to keep a constant definition of which direction is positive. I agree.

I meant that the second V_1 should be (V_1)' (or V_1 final)

I always thought that both would continue to go in the same direction after collision? Since the first one has no momentum, they will both share a momentum in the same direction won't they?

 

[toesockshoe]

I meant that the second V_1 should be (V_1)' (or V_1 final)

I always thought that both would continue to go in the same direction after collision? Since the first one has no momentum, they will both share a momentum in the same direction won't they?

actually, yes youre right. because the mass of the object being hit is smaller, they will go in the same direction. i should have changed that to a plus sign.

 

[grzz]

A simpler notation may help.

One can call the final velociies u and w and substitute for the masses right away.

Then conservation of momentum and conservation of KE will give two equations in u and w which can be solved.

 

[haruspex]

I meant that the second V_1 should be (V_1)' (or V_1 final)

I always thought that both would continue to go in the same direction after collision? Since the first one has no momentum, they will both share a momentum in the same direction won't they?

Whether they will continue in the same direction depends on the relative masses. In this case, as toesockshoe says, they will

 

[theodoros.mihos]

Make terms by the same masses and divide by part. You make a 1st class equation system.