2D Momentum Problem


by Probie1
Tags: momentum
Probie1
Probie1 is offline
#1
May16-12, 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 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 ????
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Steely Dan
Steely Dan is offline
#2
May16-12, 06:53 PM
P: 317
Quote Quote by Probie1 View Post
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 ????
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
Probie1 is offline
#3
May17-12, 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?

Steely Dan
Steely Dan is offline
#4
May17-12, 04:35 PM
P: 317

2D Momentum Problem


Quote Quote by Probie1 View Post
V2 and v3 have no velocity so they have no momentum...correct?
Correct.
Probie1
Probie1 is offline
#5
May17-12, 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?
Steely Dan
Steely Dan is offline
#6
May17-12, 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.
Probie1
Probie1 is offline
#7
May17-12, 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 t-bone another and they both have pre and post impact velocities.
Probie1
Probie1 is offline
#8
May17-12, 05:18 PM
P: 38
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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