Change in kinetic energy in a 2-car collision

[clope023]

[SOLVED] change in kinetic energy in a 2-car collision

Homework Statement

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, a 1850kg car traveling to the right at 1.60m/s collides with a 1450kg car going to the left at 1.10m/s . Measurements show that the heavier car's speed just after the collision was 0.270m/s in its original direction. You can ignore any road friction during the collision.

a) What was the speed of the lighter car just after the collision?

got this to be .597m/s

b) (where I'm stuck) Calculate the change in the combined kinetic energy of the two-car system during this collision.

Homework Equations

\DeltaK = K2-K1?

The Attempt at a Solution

I'm pretty sure I'm missing something obvious, but I'v tried like 5 different combinations.

(1/2(ma + mb)(v2a+v2b)^2)-(1/2(ma+mb)(v1a+v1b)^2) = 190J wrong

with other combos I've gotten as high as 1490J which was also wrong.

any help is greatly appreciated.

 

[Doc Al]

You know the speed of each car before and after the collision. So just calculate the total kinetic energy before and after and compare. Calculate the KE of each car separately, then add them up.

 

[clope023]

okay, so:

K1a = 1/2(1850)(1.60)^2 = 2368J
K1b = 1/2(1450)(1.10)^2 = 877J

K2a = 1/2(1850)(.270)^2 = 67J
K2b = 1/2(1450)(.597)^2 = 258J

deltaK = (2368+877)-(67+258) = 2920J?

 

[Doc Al]

Looks good to me. (Assuming you calculated the final speed of the lighter car correctly; I didn't check.)

Edit: I just confirmed your speed calculation.

 

[clope023]

Looks good to me. (Assuming you calculated the final speed of the lighter car correctly; I didn't check.)

Edit: I just confirmed your speed calculation.

thanks, yeah I know part A is right for sure (masteringphysics)

thanks again Doc Al!

 

[samelliz]

As i understood it, the two cars collided with an elastic form of collision. Therefore momentum and kinetic energy is conserved (same before and after the collision). How come that in the computation of the two kinetic energy before and after (the one you presented with us) are not equal (3245 J before collision and 325 J after collision)? I thought they will be equal. This is confusing for me.

You just showed combined kinetic energy.
Please help me. Explain it in detail.

 

[Doc Al]

As i understood it, the two cars collided with an elastic form of collision. Therefore momentum and kinetic energy is conserved (same before and after the collision).

No, the collision is not elastic. Only momentum is conserved. You have enough information to solve for the final speeds using only conservation of momentum (that was part a).

(Realize that you are responding to a post that is over three years old.)