# 2D Momentum Problem

1. May 16, 2012

### Probie1

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

Car 1 weighing 1000 kg crashes into the rear of parked car 2 weighing 1100 kg and stops. Car2 moves ahead 2 ms2. What is the velocity of car1 at impact?

2. Relevant equations

(m1v1+m2v2)=(m1v3+m2v4)

3. The attempt at a solution

(m2V4)/m1
(1100*2)/1000
2.2ms2= 2200/1000

Okay, now I know this is right, but what I don't know is how do they arrive at the equation (m2v4)/m1 ????

2. May 16, 2012

### Steely Dan

If you examine your original conservation of momentum equation, let $v_1, v_3$ be the initial and final speeds of car 1, respectively, and let $v_2,v_4$ be the initial and final speeds of car 2, respectively. From the context of the problem, what must be true about speeds $v_2$ and $v_3$?

3. May 17, 2012

### Probie1

(1000*2.2+1100*0)=(1000*0+1100*2)

V2 and v3 have no velocity so they have no momentum...correct?

4. May 17, 2012

### Steely Dan

Correct.

5. May 17, 2012

### Probie1

So because they have no momentum they cancel out which means you rewrite the equation to solve for velocity1

(1000*2.2+1100*0)=(1000*0+1100*2)

(1000)=(1100*2)

(m1)=(m2v4)

(m2V4)/m1

(1100*2)/1000

2.2ms2= 2200/1000

Is that how it is done?

6. May 17, 2012

### Steely Dan

Yes, that's the idea. Conservation of momentum dictates $m_1 v_1 = m_2 v_4$, solving for the initial speed is just an algebra problem.

7. May 17, 2012

### Probie1

Thanks Steely Dan... that wasn't as hard as I thought it would be... mind you that was an easy momentum problem for you, but difficult for me. I imagine it would be harder if you had 1 vehcile t-bone another and they both have pre and post impact velocities.

8. May 17, 2012