Alexander F. answered • 07/04/14

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Hey L,

Since the two cars are sticking together after the collision, this would be an inelastic case, which is given by the equation:

m1v1 + m2v2 = (m1 + m2)vf

In this case, lets treat m2 as the car at rest, this means that v2 = 0. If you plug in v1=10m/s, and we solve for the final velocity of the two cars stuck together, you will see that it has to be:

vf = (10 *m1)/(m1 + m2)

which is less than 10m/s. The answer is b.

For the second question, if the velocity changes from 30 to 60 km/h, then we would have a change in kinetic energy, not potential energy. So answer a. is out of the question. For work to be done on the car, we need to apply a force on the car in the same direction as motion, but we aren't applying any forces on the car itself, so no work was done on the car. Answer b looks correct! But just to be sure, we can use kinematics to disprove answer c.:

vf1 = vi1 + at1 and vf2 = vi2 + at2

Solving both equations for a, we can set the two equations equal to one another (since we are assuming acceleration will remain constant for both speeds). Keep in mind, the final velocity in either equation is zero, since we are coming to a complete stop!

a = -vi1/t1 and a = -vi2/t2

-vi1/t1 = -vi2/t2

-30/t1 = -60/t2

30t2 = 60t1

**t2 = 2t1**

So the time it would take the car to stop when its going 60m/s is twice as long as when the speed is 30m/s. Answer c. cannot be right, and answer d. by default can't be correct if any of the answers are wrong. Answer b. is the final!

Michael A.

07/04/14