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Luca 123
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Homework Statement
A mass m1, with initial velocity u, collides elastically with mass m2, which is initially at rest. After collision, m1 deflects by angle θ. Find the maximum value of θ. The answer given is θmax=acos(sqrt(1-(m1/m2)^2)). Does this mean that the maximum angle cannot exist if m1>m2?
Homework Equations
Let m2 deflect by angle α from the initial direction of m1, and v1, v2 be the velocity of m1, m2 respectively after collision. Then
1.(m1)(u)= (m1)(v1)(cosθ)+(m2)(v2)(cosα)
2. (m1)(v1)(sinθ)=(m2)(v2)(sinα)
3. (m1)(u)^2=(m1)(v1)^2 + (m2)(v2)^2
The Attempt at a Solution
I got θmax =acos(sqrt(1-(m2/m1)^2)) instead.
By 1. and 2.
[(m1)(u)-(m1)(v1)(cosθ)]^2 +[(m1)(v1)(sinθ)]^2 = [(m2)(v2)(cosα)]^2 +[(m1)(v1)(sinα)]^2
Hence
(m1u)^2 +(m1v1)^2 - 2(m1)^2(v1u cosθ)^2 = (m2v2)^2
By 3.
(m2v2)^2 = (m2m1)(u^2-v1^2)
Thus
m1(u^2) +m1(v1^2) - 2m1(v1u cosθ)^2 = (m2)(u^2-v1^2)
Dividing by v1^2 , we get:
(m1-m2)(u/v1)^2 -(2m1cosθ)(u/v1)+(m1+m2)=0
Hence the equation (m1-m2)x^2-(2m1cosθ)x+(m1+m2)=0 has a solution. Since it has a solution, we must have b^2-4ac≥0. Thus:
(2m1cosθ)^2 ≥4(m1-m2)(m2+m1)
And hence
cosθ≥sqrt(1-(m2/m1)^2). Since the maximum θ of means the minimum of cosθ, θmax =acos(sqrt(1-(m2/m1)^2)).
What went wrong?
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