Maximum angle for 2D collision

[Luca 123]

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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[Dr. Courtney]

I picture would help a lot, as well as distinguishing between vectors and scalars in your notation.

 

[jbriggs444]

ΘΘΘΘ

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?

A moment's thought shows that the given answer cannot be right. By inspection, if m1 < m2 then any angle at all up to 180 degrees is attainable. Like bouncing a ping pong ball from a basketball.

In addition, both the given solution and your solution could be dramatically simplified. Start by taking the cosine of both sides.

cos θ_max = sqrt(1-(m₁/m₂)²)

Square both sides

cos² θ_max = 1 - (m₁/m₂)²

Apply a simple trig identity

sin² θ_max = m₁/m₂²

Take the square root

sin θ_max = m₁/m₂

 

[haruspex]

A moment's thought shows that the given answer cannot be right. By inspection, if m1 < m2 then any angle at all up to 180 degrees is attainable. Like bouncing a ping pong ball from a basketball.

I suspect that the original problem either stated or assumed m1>m2. It is then clear that m1 and m2 are swapped in the given answer.
Luca, I agree with your answer for the case where m1>m2, but you may still be puzzled by the result it gives when m1<=m2. Consider the inequality you had just before "And hence". What do you get with m1<m2 in there?