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oblique impact

How to do part d ? I have created an equation for tanY using the compound angle formulae with tanY= tan(A-B) where A is alpha and B is Beta . Do I just differentiate and solve for Y ?

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Reply 1
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Reply 2
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Original post by RLangdon569
Screenshot 2021-05-15 at 12.17.00.png

What did you get for c?
Reply 3
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Original post by mqb2766
What did you get for c?

tanY = (-2tanA)/(1+3(tanA)^2)
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Original post by mqb2766
What did you get for c?

hopefully that is right
Reply 5
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Original post by RLangdon569
tanY = (-2tanA)/(1+3(tanA)^2)

Not worked it through but if A is acute, tan(A) is positive so tan(Y) <= 0. The max would be at (0,0). That doesnt sound right..
(edited 2 years ago)
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Original post by mqb2766
Not worked it through but if A is acute, tan(A) is positive so tan(Y) <= 0. The max would be at (0,0). That doesnt sound right..

well is Y=A-B ?
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Original post by RLangdon569
well is Y=A-B ?

Can you post your working and give me 5 to go over it?
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Original post by RLangdon569
well is Y=A-B ?

B-A
I suspect which would make the numerator positive and make the answer more sensible..
(edited 2 years ago)
Original post by mqb2766
B-A
I suspect which would make the numerator positive and make the answer more sensible..

B is less than A ?
Original post by RLangdon569
B is less than A ?

No
tan(B) = 3 tan(A)
so ...

Note also, you could write the expression as
2x / (1+3x^2)
and differentiate with respect to x. Then use atan() to get the angle. tan() is monotonic so the max values will be equivalent.
(edited 2 years ago)
Original post by mqb2766
No
tan(B) = 3 tan(A)
so ...

Note also, you could write the expression as
2x / (1+3x^2)
and differentiate with respect to x. Then use atan() to get the angle. tan() is monotonic so the max values will be equivalent.

I am a bit confused because I get that tanY = tanA.
2-6x^2 = 0 , x^2=1/3. x= 1/rt3 . tanY = (2/rt3)/(1+3(1/3)) therefore tanY=1/rt3 . so tanY = tanA ?!?
Is the angle of deflection A+B ?
Original post by RLangdon569
2-6x^2 = 0 , x^2=1/3. x= 1/rt3 . tanY = (2/rt3)/(1+3(1/3)) therefore tanY=1/rt3 . so tanY = tanA ?!?

Assuming no errors, "yes". Yes in quotes because tan y = tan a for the specific value of a that maximizes the deflection in this question. (What the question wants is just the numerical value).
(edited 2 years ago)
Incidentally, I'm pretty sure there's a "cute" way of getting the result by minimizing 1/tan y instead of maximizing tan y and then using the AM-GM inequality.
Original post by RLangdon569
2-6x^2 = 0 , x^2=1/3. x= 1/rt3 . tanY = (2/rt3)/(1+3(1/3)) therefore tanY=1/rt3 . so tanY = tanA ?!?

Sory been asleep. As you say, the max occurs when alpha 30 is beta is 60 so alpha+beta = 90.
Gamma=30 as well.
(edited 2 years ago)
Original post by RLangdon569
Is the angle of deflection A+B ?

The diagram you should have is basically Q38 in

https://madasmaths.com/archive/maths_booklets/mechanics/m3_m4_oblique_collisions.pdf

So B-A
(edited 2 years ago)

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