# Inelastic collision

1. Jan 20, 2010

### squintyeyes

A 1900 kg car moving at 20.0 m/s crashes into a brick wall and is stopped. The driver's mass is 70.0 kg, and although the driver is not wearing his seatbelt, the air bag thankfully slows him to a stop in 0.100 s. What is the magnitude of the average force that the air bag exerts on the driver?
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(Side Note: Air bags and seat belts act to slow you to a rest in a longer time. By increasing the time, the force on your body decreases. An average force exceeding roughly 2.00 x 105 N will probably kill you.)

How would you do this problem?

2. Jan 20, 2010

### pgardn

Have you done impulse and change in momentum? The driver has an impulse supplied by the air bag that causes him to change momentum.

3. Jan 20, 2010

### squintyeyes

so what would the equation look like? how would you solve it?

4. Jan 21, 2010

### pgardn

Have you studied F*t...?

5. Jan 24, 2010

### squintyeyes

So i read up on this and did this. It was marked wrong can someone help me figure out why?

m*v = f*t
(1900 + 70)(20) = F * (0.1)
39400= 0.1F
F=394000

6. Jan 24, 2010

### pgardn

I believe they were asking about the force on the driver (70 kg) and he was moving at 20 m/s and then 0 m/s (stopped). So the driver's change in momentum is due to a force acting over a 0.1 second time period.

7. Jan 24, 2010

### squintyeyes

so would it be like this then?

m*v = f*t
(70)(20) = F * (0.1)
1400= 0.1F
F=14000

8. Jan 24, 2010

### pgardn

yeps...

I am assuming on the left hand side of your equation you had final momentum being zero. So really you end up with a force that is (-). Which basically means the driver was accelerated in the opposite direction as his initial motion.