# Car Collision Impulse Problem

1. Apr 21, 2014

### Bgerst103

1. The problem statement, all variables and given/known data

A 1343 kg car experiences an impulse of 30,000 N during a collision with a wall. If the collision takes 0.43 s, what was the velocity of the car just before the collision?

A) 22 m/s
B) 9 m/s
C) 51 m/s
D) 18 m/s

2. Relevant equations

F=dp/dt
p=mv

3. The attempt at a solution

I first tried to work backwards by doing 30,000 x .43 = dp which comes out to be 12900. Since I'm assuming the final momentum is 0, I plugged 12900 into p=mv along with 1343 and got v=9.6. I don't think 9 m/s is the answer, I'm pretty sure it's either A or D but I can't seem to come out with answer close to either. Any help is appreciated.

2. Apr 21, 2014

### AlephZero

Did you read the question correctly? The units for impulse are not Newtons.

3. Apr 21, 2014

### Bgerst103

30,000 N/s is what it says.

4. Apr 21, 2014

### haruspex

Are you sure? The units for impulse are as for momentum. Could be kg m/s or, equivalently, Ns (Newton-seconds).
You tried multiplying by the duration, giving you units of kg m (or, Ns2), neither of which relates to any physical quantity I can think of.
Given a time, a momentum and a mass, you want a velocity. What function of the given variables will do that?

5. Apr 22, 2014

### rude man

Replace 'impulse' with force in the problem statement and you get your answer which I think is correct, the given choices notwithstanding.

6. Apr 22, 2014

### haruspex

On the other hand, leave it as impulse, with units of Newton-seconds, and you do get one of the offered answers.

7. Apr 22, 2014

### Bgerst103

I ended up getting 22 m/s by just dividing 30,000 N-s by 1343. Not completely sure why that worked but it gave me one of the two answers that I thought it could potentially be.

8. Apr 22, 2014

### rude man

Good point, haruspex and OP. In which case I fell for the red herring of the 0.43 s.
Perhaps we'll never know the real story ...

9. Apr 22, 2014

### rude man

It worked because the basic equation is impulse = force F times time t = change in momentum Δp = mass m time change in velocity Δv, or

∫Fdt = Δp = mΔv + vΔm but in this case Δm = 0.

The equation follows directly from good old F = ma = m dv/dt so just move the differentials around and integrate both sides of the equation.