# Kinetic Energy / Momentum Problem

1. Jul 17, 2011

### dimeking

1. The problem statement, all variables and given/known data
Two Railway cars, m1 and m2, are moving along a track with velocities v1 and v2, respectively. The cars collide, and after the collision the velocities are v'1 and v'2. Show that the change in kinetic energy, K' - K, will be maximum if the cars couple together.
Hint: Set d(K' - K)/dv'1 = 0 and show that v'1 = v'2.

2. Relevant equations
Conservation of linear momentum: m1v1 + m2v2 = m1v'1 + m2v'2.
Kinetic energy K = 0.5mv^2
Difference in kinetic energy: K' - K = 0.5m1v1^2 + 0.5m2v2^2 - 0.5m1v'1^2 - 0.5m2v'2^2.

3. The attempt at a solution
I solved the conservation of momentum equation for v1 and substituted that into the K' - K equation. This yields v'2 = v2.
I then solved the conservation of momentum equation for v2 and substituted that into the K' - K equation. I got v'1 = (m1v1 - m2v'2) / (m1 - m2).

2. Jul 17, 2011

### Pi-Bond

Since the hint mentions finding the derivative of the kinetic energy difference with respect to $v'_{1}$, you should find the value of $v'_{2}$ in terms of $v'_{1}$. Substitute that into the energy difference and then use the hint (Note that $v_{1}$ and $v_{2}$ are constant in that expression).

Last edited: Jul 17, 2011
3. Jul 17, 2011

### dimeking

Thank you for responding, Pi-Bond. If I solve the momentum equation for v2-prime in terms of the other v's, and then substitute this into the energy equation, the energy equation wil be in terms of v1-prime, v1, and v2. I will not be able to show that v1-prime equals v2-prime.

4. Jul 17, 2011

### Pi-Bond

If you do that, what value for $v'_{1}$ do you get?

5. Jul 17, 2011

### Pi-Bond

Your two equations are sufficient; you can get the result by using the conservation equation to find an expression $m_{1} v_{1}$, which can be substituted into your second equation along with your first equation.

6. Jul 18, 2011

### dimeking

I still cannot solve this.

When I solve the conservation of momentum equation for m2v'2 and then substitute this into the K' - K equation and then take its derivative and set it equal to zero, I get v'1 = 0.5v'2.

When I solve the conservation of momentum equation for m1v1 and then substitute this into the K' - K equation and then take its derivative and set it equal to zero, I get v'1 = 0.5v1.

Finally, when I solve the conservation of momentum equation for m2v2 and then substitute this into the K' - K equation and then take its derivative and set it equal to zero, I get v'1 = 0.5v2.

7. Jul 19, 2011

### Pi-Bond

You got the equations:

$v'_{2}=v_{2}$
$v'_{1}=\frac{m_{1} v_{1} - m_{2} v'_{2} }{m_{1}-m_{2}}$

From conservation of momentum,

$m_{1} v_{1} = m_{1} v'_{1}+ m_{2} v'_{2} - m_{2} v_{2}$

Substitute that above to get:

$m_{1} v'_{1} - m_{2} v'_{1} = m_{1} v'_{1}+ m_{2} v'_{2} - m_{2} v_{2} - m_{2} v'_{2}$

Can you get the result now?

8. Jul 19, 2011

### dimeking

OK, I finally got it. It would have gone quicker if I had just followed the advice in your first post.

I solved the conservation of momentum equation for v'2, then substituted that into the kinetic energy equation, then set its derivative equal to zero and solved for v'1.

I then substituted this value for v'1 back into my equation for v'2 and showed that v'2 reduced to the same expression as v'1.

Pi-Bond, thank you for your help and patience.