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Energy [Car collisions]

  • #1

Homework Statement

Car A has a mass of 1900 kg and crashes into parked Car B with a mass of 1100 kg. Car A claims to have hit the brakes 15 metres before crashing into Car B. After the collision, Car A slid an additional 18 metres while pushing Car B 30 metres from its parked position. The coefficient of friction is .6 between the locked tires and the road. Prove that Car A was speeding. Legal speed limit of the road is 90 km/h.


Homework Equations


F=ma
W=[tex]\Delta[/tex]KE
KE=.5 x mv[tex]^{2}[/tex]
F[tex]_{net}[/tex]=F[tex]_{horizontal}[/tex]-F[tex]_{friction}[/tex]
F[tex]_{friction}[/tex]=[tex]\mu[/tex]mg


The Attempt at a Solution


I set force friction equal to the force of motion, and got that to be 11400 Newtons. Then I plugged that into the equation for work, and used 18m as the displacement, and 0 for the final velocity, and solved for the initial velocity. I got 14.697 m/s as that answer. Then I used the equation for work again, but this time I used 14.697 m/s as the final velocity, and found the initial velocity to be about 6 m/s. However, I don't think it's right, since I didn't use anything about Car B. I'm not sure exactly what to do with it.
 

Answers and Replies

  • #2
ideasrule
Homework Helper
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I set force friction equal to the force of motion, and got that to be 11400 Newtons. Then I plugged that into the equation for work, and used 18m as the displacement, and 0 for the final velocity, and solved for the initial velocity. I got 14.697 m/s as that answer. Then I used the equation for work again, but this time I used 14.697 m/s as the final velocity, and found the initial velocity to be about 6 m/s. However, I don't think it's right, since I didn't use anything about Car B. I'm not sure exactly what to do with it.
What's the "force of motion" and how did you calculate it? What equation for work are you talking about; is it W=Fd? I really don't understand what you did.
 
  • #3
On a similar problem my teacher posted, he set Force Friction equal to Force Motion. I'm not entirely sure why he did that, and as I can't ask him, I was just going with it.

And sorry, I use S as displacement, which is the same as distance. In my physics class we use different variables.
 
  • #4
ideasrule
Homework Helper
2,266
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OK, so what's the "force motion"?

I would approach the problem by calculating the kinetic energy of the two cars right after the collision. This is equal to the energy lost by friction before the cars came to rest. Then, I'd add on the energy lost by A as it was braking. This will give the initial kinetic energy of A.
 
  • #5
For the Force Motion, I set it equal to [tex]\mu[/tex] times Force Weight, and got 11400 for Car A, and 6600 for Car B.

Would I use the final velocity that I calculated for Car A as the velocity in the Kinetic Energy equation?
 

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