2D Momentum Problem

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 ????

Steely Dan

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

Probie1

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

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

Steely Dan

Correct.

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?

Steely Dan

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

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.

Probie1

I was just re reading your posts...

...when I re read this post the light started to flicker.

Thanks again Steely Dan