Solving a Car Collision: Kinetic Energy & Force

[shanie]

Hello, I've tried solving the following question:

A car collides with a tree with a velocity of 75km/h. The car was compressed 1.1m after the collission and the car's mass is 1100kg (including the driver), and the driver's mass is 71kg.
a) What is the kinetic energy just before the collission?
b) What is the average force that works on the driver during the collission?

Now I calculated a) by the regular mv²/2 giving E_k=239 kJ

And then b) by v²-v₀²=2as, where s is the distance 1.1m. Converting 75km/h to 20.833m/s, and setting the equation 20.833²/(2*1.1)=197.29 m/s², this would then be the constant acceleration. According to F=ma I set the equation: 71kg*197.29=14 kN on average during the collission.
Is this calculation correct? I would appreciate some help, thanks!

 

[Chrisas]

In my opinion that is not correct. Your method assumes that the car accelerates at a constant value. Nothing in the problem says that this is the case.

Think about why you were asked for Kinetic energy in part a). It can be used for part b). There is a key physics word in the sentence of part b that gives you a hint of how to do it.

 

[Doc Al]

It's all good.

You can also do part b by setting the driver's KE equal to the work done on him by the force of the collision. (It's completely equivalent to what you've done.)

[Note: In the future, please post homework/coursework questions in the appropriate homework help forum (Intro Physics). Do not post such questions in the main forums.]

 

[Doc Al]

In my opinion that is not correct. Your method assumes that the car accelerates at a constant value. Nothing in the problem says that this is the case.

Since the problem asks for average force, this method works OK. (It actually finds the average acceleration.)

Using energy methods also assumes that the force is constant, which is OK for calculating the average force.

 

[Chrisas]

Ok, I can see that since F=ma and as you said Force is considered constant in the average, which implies that "a" can also be considered constant.

Thanks.