Question

A teenage driver backs into a bush at 2 mph. The impact stops the car. The car is not damaged. Later that day, they back into a fencepost at 2 mph. The impact stops the car. The bumper is dented by the fencepost. Use physics terms to explain whey the bumper is not dented when the car backs into the bush.

a. The bush gives and the fencepost doesn't. Since the bush gives, it doesn't apply as much force to the car as the car does to it. The fencepost doesn't give, so it can apply the same force to the car that the car applies to it.

b. The fencepost has more mass than the bush, so the fencepost is able to apply a larger force to the car than the fencepost can. (I know this is wrong)

c. The bush gives, so the collision with it lasts longer than the collision with the fencepost. The longer the collision time, the lower the force applied.

d. The bush gives, so momentum is conserved in the collision with the bush. The fencepost does not move, so momentum is not conserved when the car hits the fencepost.

Answer 1

If we see carefully the the time taken to make the car stop in the case of the bust is more than when it is the fencepost.

Now, initially the total momentum of the car, p = m*v

where m = mass of the car

v = speed of the car = 2 mph

After stopping the final momentum is 0 <--- as velocity is 0

So, the change in momentum = mv - 0 = mv

Now, the time taken to stop is 1st case( when stopped by bush) is t1 and when stopped by fencepost be t2

Now, as observed, t1 > t2

Now, the force exerted by the collision, F = p/t = change in momentum/time

So, F1 = p/t1 and F2 = p/t2

As t1 > t2 , so, F1 < F2

This means the force exerted in the case of bush is less than that of the fencepost.

So, the car is undamaged in 1st case and damaged in 2nd <--------answer

SO, as explained :

the answer should be :

c. The bush gives, so the collision with it lasts longer than the collision with the fencepost. The longer the collision time, the lower the force applied.