Kinetic Energy / Momentum Problem

[dimeking]

Homework Statement

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.

Homework 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.

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) / (m_{1 - m2).}

 

[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).

 

[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.

 

[Pi-Bond]

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

 

[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.

 

[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.

 

[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?

 

[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.