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2D Momentum Problemby Probie1
Tags: momentum 
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#1
May1612, 06:08 PM

P: 38

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 ms^{2}. 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 2.2ms^{2}= 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
May1612, 06:53 PM

P: 317




#3
May1712, 04:10 PM

P: 38

(1000*2.2+1100*0)=(1000*0+1100*2)
V2 and v3 have no velocity so they have no momentum...correct? 


#4
May1712, 04:35 PM

P: 317

2D Momentum Problem



#5
May1712, 04:55 PM

P: 38

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
May1712, 04:56 PM

P: 317

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.



#7
May1712, 05:01 PM

P: 38

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 tbone another and they both have pre and post impact velocities.



#8
May1712, 05:18 PM

P: 38

I was just re reading your posts...
Thanks again Steely Dan 


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