- #1

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I know that when m> m1 , angle <90'

when m<m1 , angle >90'

but there is no mathematical proof to show this result....

....

does anyone help me solve this problem?

- Thread starter mysqlpress
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- #1

- 83

- 1

I know that when m> m1 , angle <90'

when m<m1 , angle >90'

but there is no mathematical proof to show this result....

....

does anyone help me solve this problem?

- #2

Doc Al

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Can you prove the first result? (For equal masses.)

- #3

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mv=m1v1cosx+mv2cosy- (1)

v=v1cosx+v2cosy

0=m1v1sinx-mv2siny - (2)

0=v1sinx-v2siny

(1)^2+(2)^2

v^2 = (v1)^2+(v2)^2 +2v1v(cosxcosy-sinysinx)

KE conserved

mv^2 =m1v1^2+mv2^2

v^2=v1^2+v2^2

cos(x+y) = 90'

- #4

Doc Al

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- #5

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Thanks ! I've proven that. :)

- #6

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- #7

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I understand everything except what is in bold. First, you are finding the momentum along the X. Then finding the momentum along the Y. Then somehow combine them in the equation for the conservation of energy. I have all of that already, but I don't understand the combining itself and don't understand the bolded areas

mv=m1v1cosx+mv2cosy- (1)

v=v1cosx+v2cosy

0=m1v1sinx-mv2siny - (2)

0=v1sinx-v2siny

(1)^2+(2)^2

v^2 = (v1)^2+(v2)^2 +2v1v(cosxcosy-sinysinx)

KE conserved

mv^2 =m1v1^2+mv2^2

v^2=v1^2+v2^2

cos(x+y) = 90'

- #8

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- 0

Ø1 + ø2 = 90,

so it is safe to assume that :

cos(Ø1 + ø2) = 0

since cos(90) = 0. Now, for the actual proof, start by using the formula for conservation of kinetic energy:

K1i + K2i = K1f + K2f.

After simplifying, you will get:

V1i^2 = V1f^2 + V2f^2.

According to law of cosines,

c^2 = a^2 + b^2 -2(ab)cosC.

CosC is The same value as cos(ø1 + ø2), so

c^2 = a^2 + b^2 -2(ab)cos

- #9

Doc Al

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Good! (Sorry I didn't have time to respond earlier.)NvM. I got it. Law of cosines and sum difference formulas.

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