Did I do this right?

  • Thread starter cooper43
  • Start date
  • #1
cooper43
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Homework Statement



I was wondering if anybody might help me on this UNLESS i did this correct, but I'm not totally sure.

Well here's the question.

Jorge and Elizabeth are both going to the Antique Road show on PBS. Jorge pushes an antique desk weighing 300kg on a car to the right, while Elizabeth pushes a similar cart to the left with a dresser weighing 225kg. The mass of each cart is 500kg and Jorge pushes his with a speed of 3.0m/s and Elizabeth at a speed of 4.0m/s. If the cars collide and stick together then what is their final velocity and which direction are they headed?

Homework Equations



KE=1/2*m*v^2
P=mv

The Attempt at a Solution



Car 1
500kg+300kg=800kg
3.0 m/s to the left

KE=(.5)mv^2
KE=(.5)(800)(3.0)^2
KE=3600

Car 2
225kg+500kg=725kg
4 m/s to the right

KE=(.5)mv^2
KE=(.5)(725)(4.0m/s)^2
KE=5800

5800-3600=2200

KE=(.5)mv^2
2200=(.5)(mass of Car1 and 2, 1525kg)( v )^2

Solved using calculator

Answer
V=1.6986m/s to the right

Am I right?
 
Last edited:

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
11
Looks like you are using conservation of kinetic energy when you should be using conservation of momentum. Stick-together collisions are not elastic - kinetic energy is lost.
 
  • #3
cooper43
13
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So should I be using P=mv?
 
  • #4
jgens
Gold Member
1,593
50
You definitely should be using conservation of momentum. In collisions that aren't completely elastic (this one's not as the carts stick together) kinetic energy is not conserved as some energy is lost to heat.
 
  • #5
cooper43
13
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What is conservation of momentum?

formula please :P
 
  • #6
jgens
Gold Member
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50
The law of conservation of momentum states that if no net external force acts on a system that the initial and final momentum of the system must be the same; hence, Pi = Pf
 
  • #7
cooper43
13
0
So i should use these

e26532f7b06dba85c58474ef61fb42ab.png

807ef01bd2f00bae37644bc468c4b543.png


If so, which one?

i got 3.95082 with the first and
2.95082 with the second
 
  • #8
Delphi51
Homework Helper
3,407
11
I would start with "momentum before = momentum after".
Then put an mv on each side for each object that is moving.
mv + mv = mv (where all the m's and v's are different and perhaps should be numbered)
Put in the numbers and solve for the one unknown speed.
 
  • #9
cooper43
13
0
so is it
((M1*V1)+(M2*V2))/(M1+M2)
that way velocity is isolated?

Now I got -.000215

I'm moving on to the next question now
 
  • #10
jgens
Gold Member
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Cooper43, the equations you posted only apply in the case of a completely elastic collision. The scenario you've described appears to be completely inelastic.

Edit: I can't agree with your most recent answer.
 
Last edited:
  • #11
cooper43
13
0
I (think I) FOUNT IT!!!

cea19d6735094aa9af285bee2a873b47.png
 
Last edited by a moderator:
  • #12
jgens
Gold Member
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Yes, that equation will suffice for the scenario described in the problem.
 
  • #13
cooper43
13
0
(800) (3) + (725) (-4) = (1525)V

V = -0.328 m/s?

they move to the left?
 
  • #14
jgens
Gold Member
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That's what I got. They move left with a velocity of v = 0.33 m/s. Good job!
 
  • #15
cooper43
13
0
Yes, finally. Thanks jgens :)
 
  • #16
jgens
Gold Member
1,593
50
You're welcome!
 

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