Solve 2D Momentum Problem: Car1 Velocity at Impact

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SUMMARY

The discussion centers on a 2D momentum problem involving two cars, where Car 1 (1000 kg) crashes into parked Car 2 (1100 kg), which then moves at 2 m/s after the impact. The conservation of momentum equation used is (m1v1 + m2v2) = (m1v3 + m2v4). The solution reveals that the initial velocity of Car 1 at impact can be calculated using the rearranged equation (m2v4)/m1, leading to a result of 2.2 m/s. The participants confirm the correctness of the approach and clarify the cancellation of terms due to zero initial velocities for Car 2.

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Homework Statement



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?


Homework Equations



(m1v1+m2v2)=(m1v3+m2v4)


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 ?
 
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Probie1 said:

Homework Statement



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?


Homework Equations



(m1v1+m2v2)=(m1v3+m2v4)


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 ?

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?
 
(1000*2.2+1100*0)=(1000*0+1100*2)

V2 and v3 have no velocity so they have no momentum...correct?
 
Probie1 said:
V2 and v3 have no velocity so they have no momentum...correct?

Correct.
 
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?
 
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.
 
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.
 
I was just re reading your posts...

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.

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

Thanks again Steely Dan
 

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