How to back calculate vehicle speed after crash?

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SUMMARY

The discussion focuses on calculating the estimated speed of Car 2 (C2) before a crash involving Car 1 (C1) and C2. C1 weighs 3,000 kg and was at rest, while C2 weighs 1,200 kg and impacted C1, resulting in both vehicles moving 4 meters in 1 second post-collision. The user considers using momentum conservation and kinetic energy equations, noting the coefficient of friction on a dry road is 0.7. The conclusion emphasizes that the momentum before and after the crash can be equated to find C2's initial speed.

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  • Understanding of momentum conservation principles
  • Basic knowledge of kinetic energy calculations
  • Familiarity with vehicle dynamics and crash analysis
  • Knowledge of friction coefficients and their impact on motion
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  • Study the principles of momentum conservation in inelastic collisions
  • Learn how to apply kinetic energy formulas in crash analysis
  • Research the effects of friction on vehicle motion post-collision
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This discussion is beneficial for automotive engineers, accident reconstruction specialists, and students studying vehicle dynamics or physics related to collisions.

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I am struggling on solving a car crash question. Could anyone help me out?

Statement:
Car 1 (C1) weighs 3,000kg, C2 weighs 12,00kg. C1 was at rest and hit by C2 from side. Both vehicles moved 4m in 1 second before total rest. In other words, C1 was pushed away from side for 4m and C2 traveled forward 10ft.

Question: Assuming the fraction coefficient of the dry road is 0.7. What is the estimated speed of C2 before crash?

I am not sure if I should solve this question by using Momentum (M)
M(before)=(Mass of C1 x Velocity of C1) + (Mass of C2 x Velocity of C2)
= (3,000 x 0) + (1,200 x V2)
=1,200V2

Upon crash, in the 1-second time, cars traveled 4m... so their speed after crash is 4m/s
M (After)=(Mass of C1 x 4m/s) + (mass of C2 x 4m/s)
=(3,000 x 4) + (1,200 x 4)

Now if I assume M (before) = M (after)... then I can figure out initial speed V2
But I did not use the friction here. Do I need any other factors for the calculation?

Or should I approach this question from kinetic energy approach
C2 was moving with kinetic energy: F2 = 0.5x (mass of C2) x (Velocity of C2)2
C1 was at rest, so no kinetic energy.
But I do not know how to proceed from here

Any help is appreciated!
 
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mcta said:
I am struggling on solving a car crash question. Could anyone help me out?

Statement:
Car 1 (C1) weighs 3,000kg, C2 weighs 12,00kg. C1 was at rest and hit by C2 from side. Both vehicles moved 4m in 1 second before total rest. In other words, C1 was pushed away from side for 4m and C2 traveled forward 10ft.

Question: Assuming the fraction coefficient of the dry road is 0.7. What is the estimated speed of C2 before crash?

I am not sure if I should solve this question by using Momentum (M)
M(before)=(Mass of C1 x Velocity of C1) + (Mass of C2 x Velocity of C2)
= (3,000 x 0) + (1,200 x V2)
=1,200V2

Upon crash, in the 1-second time, cars traveled 4m... so their speed after crash is 4m/s
M (After)=(Mass of C1 x 4m/s) + (mass of C2 x 4m/s)
=(3,000 x 4) + (1,200 x 4)

Now if I assume M (before) = M (after)... then I can figure out initial speed V2
But I did not use the friction here. Do I need any other factors for the calculation?

Or should I approach this question from kinetic energy approach
C2 was moving with kinetic energy: F2 = 0.5x (mass of C2) x (Velocity of C2)2
C1 was at rest, so no kinetic energy.
But I do not know how to proceed from here

Any help is appreciated!
Since elastic collisions are almost ideal considering an inelastic collision the answer is worth easy since the cars have a common velocity after collision
 

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