Collisions in two Dimensions

  • Thread starter mike_24
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  • #1
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Hi, I am having problems with a particular question it says:
Two 30kg children, each with a speed of 4.0m/s, are sliding on a frictionless frozen pond when they collide and stick together because they have Velcro straps on their jackets. The two children then collide and stick to a 75kg man who is sliding at 2.0m/s. After the collision, the three-person composite is stationary. What is the angle between the initial velocity vectors of the two children?
So what i did was you have the mass and velocity of both kids, they are both the same, and i found the total velocity of them using V=m1v1+m2v2/m1+m2.
Then i used the mass and velocity and mass of the man and I used the new velocity I found and the combined mass (doubled the original mass) and this time solved for the angle using Tan(theta)= m2v2/m1v1 and took the inverse of that to find the angle.
I am not sure if that was the correct angle to find or not, or if I even went about this problem the right way, if anyone could help me out it would be much appreciated. Thanks in advance

Mike
 

Answers and Replies

  • #2
Doc Al
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mike_24 said:
So what i did was you have the mass and velocity of both kids, they are both the same, and i found the total velocity of them using V=m1v1+m2v2/m1+m2.
Careful. The kids have the same speed, but they move in different directions. You can't just add their momenta together, you need to add them like vectors.

Here's a hint: Since, after the kids collide with the man, the total momentum is zero, what must be the combined momentum of the kids? Use that to figure out the angle between the kid's initial velocities.
 
  • #3
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But you don't know the direction they are comming from, how would you figure that out?, because you find the angles at the end of the problem, that is what the question wants....
 
  • #4
Doc Al
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You'll find that you don't need to know where they are coming from to find the angle the kids make with each other. Try this: Pretend the man is sliding along in the -x direction. What's his momentum? What must be the total momentum of the kids? (Before they collide.)
 
  • #5
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So when finding the new velocity when the two kids collide, do I find the center of mass velocities in the x and y components and then find the magnitude of that to find their new velocity?, or do I need to use cosine and sine in my equations and find the velocity that way. If i did use cosine and sine I am not really sure how to go about that.
 
  • #6
radou
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mike_24 said:
So when finding the new velocity when the two kids collide, do I find the center of mass velocities in the x and y components and then find the magnitude of that to find their new velocity?, or do I need to use cosine and sine in my equations and find the velocity that way. If i did use cosine and sine I am not really sure how to go about that.

What center of mass do you mean? Isn't it stated that the two kids stick together when they collide? Or am I missing something?
 
  • #7
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No that is right, they do stick together. I just don't really know what to do with this problem and I was just throwing out some ideas I came up with. Im pretty confused at the moment with it though.
 
  • #8
Doc Al
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First things first: Answer the two (hopefully easy) questions that I asked in post #4. I'll repeat them here:

(1) Pretend the man is sliding along in the -x direction. What's his momentum?

(2) Given the above, what must be the momentum of the two kids (together) just before they collide with the man?​

Once you answer these questions, you'll be in a better position to solve the problem.
 
  • #9
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Well
1. The mans momentum would be his mass(velocity)= -150 because he is moving along the negative x axis.
2. The momentum of the 2 kids must have a momentum of 150, which will make them all stop once they collide.
 
  • #10
Doc Al
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Right! Now you can get to work on the real problem:

(1) What was the momentum of each kid (magnitude only) before they collided?

(2) Since the total momentum of the two kids together has an x-component of 150, what must be the x-component of the momentum of each kid separately?

(3) What angle from the x-axis must each kid have had to make their x-component equal to that value? (One kid will be above the x-axis, the other below it.) This requires finding the component using a little trig.

(4) Now that you know the angle each kid made with the x-axis, what's the angle between them?
 
  • #11
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1. Each kid would have a momentum on 120 each ( 30kg*4m/s)
2. Would each kid have equal momentum?, meaning 75 each?
 
  • #12
Doc Al
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1. Each kid would have a momentum on 120 each ( 30kg*4m/s)
Right.
2. Would each kid have equal momentum?, meaning 75 each?
Exactly. Each kid would have an x-component of momentum equal to 75 kg-m/s.

Keep going.
 

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