- #1

- 44

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I figured.. 2m(v/4)

but then i just get lost... can someone help! thanks!

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- Thread starter mb85
- Start date

- #1

- 44

- 0

I figured.. 2m(v/4)

but then i just get lost... can someone help! thanks!

- #2

nrqed

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mb85 said:

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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[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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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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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

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- #7

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- #8

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im still confused.

can someone show me the steps?

can someone show me the steps?

- #9

- 335

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Now,

v*cosa + v*cosa = v/4.

Thus cosa = 1/8

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