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Re: [Phys-L] elastic collisions



I actually like the "brute force" method! There is something worth seeing
there. I let my students choose the numbers for the initial conditions.
Then, we set up the momentum equation, substitute into the energy equation
and tidy up. The quadratic equation you land on always factors, to the
amazement of the students. Of course, it is easier to factor a quadratic
expression when you already know one of the factors -- one way for momentum
and energy to be conserved is that the objects continue with their initial
velocities.

While this is not the quickest solution, it provides another chance to talk
about the meaning of the "other" root when a quadratic equation shows up in
physics. I believe this was discussed in TPT a while back.



On Wed, Dec 7, 2016 at 8:25 PM, Donald Polvani <dgpolvani@verizon.net>
wrote:

Not having looked into collisions for some time, I decided to see how much
work was really involved in directly solving the momentum and kinetic
energy
equations for elastic collisions. The old (and new) textbooks that I have
shy away from such a "brute force" approach. Turns out, in my opinion, it
is too much algebra for the average beginning physics student, but,
otherwise, although tedious, it's quite straightforward. I started with
the conservation of momentum and kinetic energy equations for elastic
collisions:

m_1*v_1i + m_2*v_2i = m_1*v_1f + m_2*v_2f
(1)

(1/2)*m_1*v_1i^2 + (1/2)*m_2*v_2i^2 = (1/2)*m_1*v_1f^2 + (1/2)*m_2*v_2f^2
(2)

Where m_1 and m_2 are known point masses with known initial speeds v_1i and
v_2i. We want to solve Equations (1) and (2) for the final speeds v_1f and
v_2f.
The solution sequence was:

1) Solve for v_2f in terms of v_1f from Equation (1).
2) Substitute for v_2f in Equation 2 and solve for v_1f (This required
solving a quadratic equation and determining that the minus square root
option is the correct choice).
3) Use the now known v_1f to evaluate v_2f using the results of Step 1.

The final results (after considerable algebraic manipulation) for v_1f and
v_2f are:

v_1f = (2*v_2i - (1 - r)*v_1i)/(1 + r) (3)

v_2f = (2*r**v_1i + (1 - r)*v_2i)/(1 + r) (4)

Where r = mass ratio = m_1/m_2.

Although Equations (3) and (4) were tedious to derive, once determined,
they
rather simply give the well-known relationship for the final and initial
relative speeds:

v_1f - v_2f = -(v_1i - v_2i) (5)

They also easily yield the relationships Philip Keller mentioned in his
post
on 11/30/16 at 8:57 PM:

v_1f = 2*v_cm - v_1i (6)

v_2f = 2*v_cm - v_2i (7)

Where the center of mass speed is given by:

v_cm = (m_1*v_1i + m_2*v_2i)/(m_1 + m_2) = (r*v_1i + v_2i)/(1 + r) (8)

As a further check on the algebra, I tried several popular test cases and
got the expected results. For example, for two equal masses (r = 1) with
equal and opposite initial speeds, Equations (3) and (4) yield v_1f = -
v_1i
and v_2f = -v_2i, and for a small mass (arbitrary v_1i) colliding with an
arbitrarily large stationary mass, they yield v_1f arbitrarily close to -
v_1i and v_2f arbitrarily close to 0.

The only requirements for Equations (3) and (4) are point masses and
elastic
collisions.

Don Polvani
Adjunct Faculty, Physics, Retired
Anne Arundel Community College
Arnold, MD 21012


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