# Elastic Collision Momentum problem help

1. Oct 20, 2009

### dynoma5

1. The problem statement, all variables and given/known data
A block of mass m1 = 5.00 kg is released from A. It makes a head on elastic collision at B with a block of mass m2=10.0kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision

2. Relevant equations

Pi = Pf
Ef= Ei + other work

3. The attempt at a solution
1/2 m1v1^2 = m1gh
v1 = square root of 2gh = 10 m/s
Pi = Pf
m1v1 +m2v2 = m1c1 + m2c2
m1v1 = m1c1 + m2c2
m1(v1 - c1) = m2c2
1/2 m1v1^2 = 1/2 m1c1^2 + 1/2 m2c2^2
m1(v1^2 - c1^2) = m2c2^2
then i used that and divided by the Pi = Pf and i got
(v1+c1) = c2
all right I'm stuck around here and have no clue what to do from here. Someone help me please?

2. Oct 20, 2009

### ehild

You have got two equations for c1 and c2.

m1(v1 - c1) = m2c2

and

(v1+c1) = c2

Insert the second one for c2 into the first equation, solve for c1, then calculate c2.

ehild

3. Oct 20, 2009

### dynoma5

kk so then i'd get m1(v1 - c1) = m2(v1 +c1)
m1v1 - m2v1 = m2c1 + m1 c1
so c1= (m1v1 - m2v1)/(m2+m1)
but that would give me a negative which means that m1 would travel back after collision?

4. Oct 20, 2009

### ehild

"but that would give me a negative which means that m1 would travel back after collision?"

yes, m1 starts to travel backwards and rises uphill, and you can calculate its maximum height now.

ehild

backwards and rises to the

5. Oct 20, 2009

### dynoma5

omg i see it now. Even the question was implying that the m1 was gonna go backward after collision but I was thinking that m1 was suppose to go forward too after the collision so I thought I did something wrong. kk so for c1 i got -3.33 m/s and c2 i got 6.67 m/s. using conservation of energy it will be Ef = Ei which is also PEf + KEf = KEi + PEi. m1gh = 1/2 m1c1^2. Reducing it down will be h = c1^2/2g = .55m ?I feel so stupid but thanks for the help. Ive seen ppl use the formula 2m1v1/(m1+m2) + (m2-m1)/(m1+m2)v2 = c2 when solving for a elastic collision but how did they get that formula?

Last edited: Oct 20, 2009
6. Oct 21, 2009

### ehild

They use the same method you applied. In general,

m1v1+m2v2=m1c1+m2c2--->m1(v1-c1)=-m2(v2-c2)
m1v12+m2v22=m1c12+m2c22--->
m1(v1-c1)(v1+c1)=-m2(v2-c2)(v2+c2)

divide the equations :v1+c1=v2+c2.

You have two linear equations for c1 and c2:

v1+c1=v2+c2 (a)

m1(v1-c1)=-m2(v2-c2) (b)

Multiply eq. (a) by m1, and add to (b). c1 cancels. Solve for c2.

If multiplying (a) with m2 and subtracting (b), c2 will cancel.

About negative velocity after collision: push a light coin against a heavy one, what happens? Never tried?

ehild

7. Oct 21, 2009

### dynoma5

ahh i see if you put it that way then i can see waht you mean. Again thanks for the clarification!