Energy dissipated in 3-cars collision

[geoffrey159]

Homework Statement

Cars B and C are at rest with their brakes off. Car A plows into B at high speed, pushing B into C. If the collisions are completely inelastic, what fraction of the initial energy is dissipated when car
C is struck? The cars are identical initially.

Homework Equations

Collisions

The Attempt at a Solution

First collision:
Momentum is conserved and the collision is inelastic so $mv_1 = 2mv_2$. The energy dissipated after the first collision
is $Q_1 = \triangle K = \frac{1}{2} mv_1 ^2 - m v_2^2 = \frac{1}{4}mv_1^2 = \frac{1}{2} K_i$

Second collision:
Momentum is conserved and the collision is inelastic so $2mv_2 = 3mv_3$. The energy dissipated after the second collision is $Q_2 = \triangle K = m v_2 ^2 - \frac{3}{2}m v_3^2 = \frac{1}{3}mv_2^2 = \frac{1}{12}mv_1^2 = \frac{1}{6} K_i$

Total energy lost is $Q = Q_1 + Q_2 = \frac{2}{3} K_i$ so the fraction of initial energy dissipated after C is struck is 2/3.

Is that correct?

 

[Bystander]

Looks good. Anything else?

 

[geoffrey159]

Thank you ! What do you mean by 'anything else ?' ?

 

[PeroK]

You've not done anything wrong but can you see a simpler way? What if there were 10 cars?

 

[geoffrey159]

Yes, there is a recursive way to do it:

With momentum conservation and inelasticity: $mnv_n = m(n+1) v_{n+1}$

So $v_n =\frac{v_1}{n}$ and total energy lost for n-car collision is $Q = K_i - K_n = (1 - \frac{1}{n}) K_i$

 

[PeroK]

Yes, you need only consider energy and momentum at the beginning and end. Even if the cars were different masses.

 

[geoffrey159]

I see, it would have been simpler to consider the general case ! Thanks !