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Energy dissipated in 3-cars collision

  • #1
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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?
 

Answers and Replies

  • #2
Bystander
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Looks good. Anything else?
 
  • #3
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Thank you ! What do you mean by 'anything else ?' ?
 
  • #4
PeroK
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You've not done anything wrong but can you see a simpler way? What if there were 10 cars?
 
  • #5
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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 ##
 
  • #6
PeroK
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Yes, you need only consider energy and momentum at the beginning and end. Even if the cars were different masses.
 
  • #7
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I see, it would have been simpler to consider the general case ! Thanks !
 

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