How to back calculate vehicle speed after crash?

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To estimate the speed of Car 2 (C2) before the crash, the discussion highlights the use of momentum conservation principles. Car 1 (C1) weighs 3,000 kg and was at rest, while C2 weighs 1,200 kg and impacted C1, resulting in both vehicles moving together for 4 meters in 1 second. The momentum before the crash can be set equal to the momentum after the crash, allowing for the calculation of C2's initial speed. The coefficient of friction on the dry road is noted as 0.7, which may need to be factored into the calculations for a more accurate estimate. The conversation suggests that understanding the collision type—elastic versus inelastic—can also influence the approach to solving the problem.
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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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