Calculating Final Speed in a Momentum Word Problem

AI Thread Summary
The discussion revolves around solving momentum-related problems involving collisions and movement on frictionless surfaces. In the first scenario, two carts collide, and the correct final speed is determined by accounting for their opposing directions, leading to the equation (4.3)(5.8) + (3.2)(-4.5) = (4.3+3.2)Vf. A second problem involves a girl walking on a plank, where conservation of momentum is necessary to find her velocity relative to the ice surface. Participants emphasize the importance of using momentum conservation principles rather than forces in these calculations. The thread highlights the collaborative nature of problem-solving in physics, with users seeking and providing guidance on complex concepts.
JassC
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Two carts with masses of 4.3 kg and 3.2 kg
move toward each other on a frictionless track
with speeds of 5.8 m/s and 4.5 m/s, respec-
tively. The carts stick together after colliding
head-on.
Find their final speed. Answer in units of
m/s.

I plugged in the numbers into this equation
4df879733089c570fbd48698e428fb34.png


(4.3)(5.8) + (3.2)(4.5) = (4.3+3.2)Vf
39.34 = 7.5Vf
5.2453 = Vf

That apparently isn't the correct answer =/
 
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Remember that momentum is a vector it has magnitude and direction, so you need to take this into account when you work the problem. Notice that the problem says the two carts ae heading towards each other, so they have to be moving in different directions, so take velocity in one direction as positive and the other as negative and you should be able to get the correct answer.
 
Okay cool, I got it . Thanks!
 
JassC said:
Okay cool, I got it . Thanks!

No problem, glad I could help.
 
I don't want to make a new thread so I'll ask this one.

A 37.9 kg girl is standing on a 98 kg plank.
The plank, originally at rest, is free to slide on
a frozen lake, which is a flat, frictionless sup-
porting surface. The girl begins to walk along
the plank at a constant speed of 1.54 m/s to
the right relative to the plank.
What is her velocity relative to the ice sur-
face? Answer in units of m/s.

Am I using conservation of momentum or force here? I don't know where to start.
 
JassC said:
I don't want to make a new thread so I'll ask this one.

A 37.9 kg girl is standing on a 98 kg plank.
The plank, originally at rest, is free to slide on
a frozen lake, which is a flat, frictionless sup-
porting surface. The girl begins to walk along
the plank at a constant speed of 1.54 m/s to
the right relative to the plank.
What is her velocity relative to the ice sur-
face? Answer in units of m/s.

Am I using conservation of momentum or force here? I don't know where to start.

Conservation of momentum, I'm pretty sure there is no such thing as conservation of force. Since she and the plank are initially at rest there is no initial momentum, so when she starts moving the plank must move in the other direction in order for momentum to be conserved.
 
I meant using conservation of momentum or just "FORCE"...
 
Yes, use coservstion of momentum .
Why did the doubt arise ?
 
Momentum

Hey i need help i have a web assignment due soon and i could really use some guidence.

A 10 metric ton train moves toward the south at 70 m/s. At what speed must it travel to have four times its original momentum? Answer in units of m/s.
 
  • #10
br0wneyes786 said:
Hey i need help i have a web assignment due soon and i could really use some guidence.

A 10 metric ton train moves toward the south at 70 m/s. At what speed must it travel to have four times its original momentum? Answer in units of m/s.

What do you think? What have you tried?

You know the equation that describes its current momentum. If you want it to be four times greater, what do you do to it?
 
  • #11
i've solved it already thanks
 
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