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Car Crash Problem.

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**Not a homework problem but I was directed here**

Hello, over the weekend I was involved in a car accident and received a careless driving ticket because the officer believed that I was going an insane amount of speed. My first instinct to react to this would to be to sit down and do the physics behind it. Now I know that the physics I have used here is not going to be 100% precise as I have not factored in the crumple zone of the car, or how much hitting the curb had slowed me down. However, I am just trying to get a general idea of how fast I truly hit the box. I have not taken physics for over a year and a half now so I am a bit rusty so please bear with me.

A 1590kg car is travelling at an unknown speed. It hits a 1140kg metal box sitting on concreate which is not bolted down. If the box moves 1m, how fast was the car travelling right before impact. Based on my research I have found that the coefficient of friction of steel on concrete varies between 0.45 and 0.7, lets use 0.7 for the benefit of the doubt. Correct me if I'm wrong on this but this would be how i would go about this equation:

Normal force of box=Frictional force=MA=(1140kg)(9.81m/s^2)=11,183.4N
W=FDµ (11,183.4N)(1m)(0.7)= 7,828.38J

Ekinitial=Ekfinal (0.5)(M)(V)^2=7,828.38J V=√((2)(7828.38J))/(1590kg)

The speed of the car before the crash is 3.14m/s
1m/s=3.6km/h (3.14m/s)(3.6)= 11.3km/h
 

rude man

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Use enrgy conservation.
The box moved 1m. How much frictional energy dissipation does that represent, assuming mu_k = 0.7 and mass = 1140 kg?

Then equate that to the k.e. of the car just before impact.
 
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Sorry what? I've only just finished physics 30 and don't understand how to factor in frictional energy dissipation. As for relating that to the kinetic energy of the car-box that is what I had done.
 
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An interesting situation, I'm not sure any physics will do you well here but anyways....

You need to consider both kinetic and static friction, but assuming just kinetic for now (actually about 0.4 from what I hear)

[itex] F_{fk} = μ_{k} F_{N} = μ_{k} mg = 4473.36\;N[/itex]

[itex] W = Fxcosθ = Fx = 4476.36\;J[/itex]

By the work energy theorem and final speed is zero:

[itex] W = ΔK = \frac{1}{2}m(v_{f}-v_{i}) = \frac{1}{2}mv_{i} [/itex]

[itex] v_{i} = \frac{2W}{m} = 5.63 \; \frac{m}{s} = 12.6 \frac {mi}{hr}[/itex]


From what I think, you could account for static friction by conservation of momentum, but that would require an estimate of the speed of the concrete block immediately after you hit it.
 

haruspex

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You need to consider it in two stages. First, there's the collision with the box. Your car is designed to crumple to absorb the impact, so this is almost entirely inelastic. So for this part use conservation of momentum:
(M+m)vf = Mvi
From there, the energy equation is mμgs = (M+m)vf2/2
Combining these:
vi2 = 2(M+m)mμgs/M2
which gives vi = 4.1m/s.
However, this assumes you made no attempt to brake, which might also be considered careless driving. Did you? If so, approximately when?
 

rude man

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Sentin31 reflects what I tried to say. That's energy conservation.

Momentum conservation does not work since upon impact some of the momentum of your car was transferred not only to the box but to the Earth due to static friction.
 
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haruspex

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Sentin31 reflects what I tried to say. That's energy conservation.

Momentum conservation does not work since upon impact some of the momentum of your car was transferred not only to the box but to the Earth due to static friction.
No, you have to consider two separate phases: the collision itself, where momentum is conserved but there's an unknown loss of kinetic energy, and the subsequent slide. Since you know the distance and force in the slide you can use energy for that phase.
 

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