Collisions in Two Dimensions

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  • #1
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After a completely inelastic collision, two objects of the same mass and same initial speed are found to move away together at 1/4 their initial speed. Find the angle between the initial velocities of the objects.

I figured.. 2m(v/4)
but then i just get lost... can someone help! thanks!
 

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  • #2
nrqed
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mb85 said:
After a completely inelastic collision, two objects of the same mass and same initial speed are found to move away together at 1/4 their initial speed. Find the angle between the initial velocities of the objects.

I figured.. 2m(v/4)
but then i just get lost... can someone help! thanks!


Pick a direction for the first object, let's say East. Let's say the second object is coming from an angle [itex] \theta [/itex] North of East. Then Decompose their momenta into x and y components, apply conservation of momentum along x and y directions. You will find the final x and y velocities along x and y in terms of the initial speed v and the angle [itex] \theta [/itex]. Impose that [itex] {\sqrt { v_{x final}^2 + v_{y final}^2}} = v/4 [/itex] and that will give you a single equation for [itex] \theta [/itex].


Patrick
 
  • #3
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OK, so the collision is in 2 dimensions, so you know you're going to be using vectors. So let me start you off with what you know.

[tex]m(\underline{v}_1 + \underline{v}_2 ) = 2m \underline{v}_{final}[/tex]

Do you know how to make vectors loose their directional components?

(Hint: Multiply both sides by a vector you know)

Hope this Helps, Sam
 
  • #4
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OK, pick the vector carefully.

I have just done this question in 5 lines (instead of 10) by picking another vector. If you don't choose your vector well, you'll have to use some trig. identities (which I don't like if I can avoid it).

Draw a diagram and note what the angles have in relation to each other due to symmetry. If you want, I'll put up a diagram of what I'm trying to say... just ask if you want it.

Regards,
Sam
 
  • #5
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hey thanks. this is what i did so far. but im still have trouble.

M1V1i + M2V2i = M1V1f + M2V2f

initial for 1
X = mv cos Theta
y = mv sin theta

final for 1
X = mv
y=0

Initial for 2
X = mv cos theta
Y = mv sin theta

final for 2
X =mv
y = 0


mv cos theta + mv sin theta = MVi
Mv (cos theta + sin theta) = MVi

???? then i get lost.
 
  • #6
335
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The components of their initial velocities perpendicular to the direction of their final velocity adds to zero. And the components along this direction must thus be same and add upto v/4.
 
  • #7
335
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Sorry. The second sentence is - the component of their initial velocities along the direction of final velocity must thus be same and add upto v/4
 
  • #8
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im still confused.

can someone show me the steps?
 
  • #9
335
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Let a be the angle made with their initial velocities along the direction of final veloctitis.
Now,
v*cosa + v*cosa = v/4.
Thus cosa = 1/8
 

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