Solve 2D Momentum Problem: Car1 Velocity at Impact

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Homework Help Overview

The problem involves a collision between two cars, where car 1 crashes into a stationary car 2, resulting in car 2 moving forward. The objective is to determine the velocity of car 1 at the moment of impact using the conservation of momentum principle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the conservation of momentum equation and question how certain terms relate to the velocities of the cars involved. There is an exploration of the implications of zero velocity for car 2 before the impact and for car 1 after the impact.

Discussion Status

Participants have engaged in clarifying the momentum equation and its components. Some have confirmed their understanding of how to manipulate the equation to isolate the velocity of car 1. There is acknowledgment of the reasoning behind the cancellation of terms related to zero momentum.

Contextual Notes

Participants note that the problem may become more complex with different types of collisions, such as T-bone scenarios, which could involve additional velocities and momentum considerations.

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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 [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]?
 
(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 [itex]m_1 v_1 = m_2 v_4[/itex], 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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