2D Conservation of Momentum Example using Air Hockey Discs


– Good morning. Let’s do a two-dimensional conservation of momentum problem using air hockey discs,
like on an air hockey table. ♫ Flipping Physics ♫ Billy, could you please read the problem and Bobby, could you please translate? – A 28.8 gram yellow air hockey disc elastically strikes a 26.9 gram stationary red air hockey disc. – The mass of the yellow
disc is 28.8 grams. The mass of the red disc is 26.9 grams. The initial velocity of
the red disc is zero. – I think it would be confusing
if we use y for yellow and y for y direction. – Maybe we should label
them disc number one and disc number two. – Sure. Let’s make the
yellow disc number one and the red disc number two. – Billy, please read on. – If the velocity of the yellow disc before the collision is 33.6 centimeters per second in the x direction and after the collision it is
10.7 centimeters per second at an angle of 63.4 degrees south of east what is the velocity of the red disc after the collision? – Velocity initial of the
yellow disc or disc one is 33.6 centimeters per
second in the x direction so the initial velocity
of disc one is zero in the y direction. The final velocity of disc
one after the collision is 10.7 centimeters per second at an angle of 63.4 degrees south of east and the final velocity of
the red disc or disc two equals question mark. – Let’s take a quick look
at the demonstration. (click)
(click) Bo, how would you like
to start the problem? – The final velocity of disc one is neither directly in
the x or the y direction so let’s break that into its components. Sin theta equals opposite over hypotenuse or velocity one final y
over velocity one final. Multiply by velocity one final
to get velocity one final y equals velocity one final times sin theta or 10.7 times sin of 63.4 degrees which is 9.5675 centimeters per second. – Negative, because it’s going south. – Yeah, negative because it’s going south. We can do the same thing with cosine. Cosine theta equals
adjacent over hypotenuse or velocity one final x
over velocity one final. Which means velocity one final x equals velocity one
final times cosine theta or 10.7 times cosine of 63.4 which equals 4.7910 centimeters per second. And it’s positive because it is east. – Bobby, what should we do now? – It’s a collision so we
know momentum is conserved. The sum of the initial
momenta of the system equals the sum of the final
momenta of the system. – Momentum is a vector. – Yeah, so we need to do
conservation of momentum in both the x and y directions. Let’s start with conservation of momentum in the x direction. That means the mass of disc one times the velocity of disc
one initial in the x direction plus the mass of disc two times the velocity of disc two
initial in the x direction equals the mass of disc one times the velocity of disc
one final in the x direction plus the mass of disc two times the velocity of disc
two final in the x direction. Because disc two is initially at rest we know the velocity of disc two initial in the x direction is zero. That whole term cancels out. We can subtract the mass of disc one times the velocity of disc one final in the x direction from both sides. Then we can divide both
sides by the mass of disc two to get the velocity of disc
two final in the x direction equals the mass of disc
one times the velocity of disc one initial in the x direction minus the mass of disc one times the velocity of disc
one final in the x direction all divided by the mass of disc two. Now we just substitute in numbers. The velocity of disc two
final in the x direction equals 28.8 times 33.6 minus 28.8 times the 4.7910 all divided by 26.9 which equals 30.8438
centimeters per second. – Notice how because in this
instance grams cancel out we do not need to convert our
known values to base SI units and we get the final velocity for disc two in the x direction in
centimeters per second. Now we need it in the y direction. Bo, could you please do that? – We start with the same
conservation of momentum equation only now it is in the y direction instead of the x direction. The initial velocity in the y direction of both discs is zero so the whole left hand side
of the equation is zero. Subtract mass one times velocity one final in the y direction from both sides and then divide the whole
equation by mass two. We now have the velocity two final y equals negative mass one
times velocity one final y divided by mass two or negative 28.8 times
negative 9.5675 divided by 26.9 which equals 10.2432
centimeters per second. – Now that we have its components Billy, could you please solve for the final velocity of disc two? – Let’s start by drawing
the vector diagram. Velocity final of disc
two in the x direction plus velocity final of
disc two in the y direction equals the velocity final of disc two. We can use the Pythagorean theorem a squared plus b squared equals c squared. Substituting in our velocities and taking the square root of both sides means that the velocity final of disc two equals the square root of the quantity of the velocity final of disc
two in the x direction squared plus the velocity final of disc two in the y direction squared. Substituting in values means that we have the
square root of the quantity 30.8438 squared plus 10.2432 squared which is 32.500 or 32.5
centimeters per second. Velocity is a vector, so we
need to find the direction. Tangent theta equals
opposite over adjacent so the tangent of the
angle for disc two final equals the final velocity of
disc two in the y direction over the final velocity of
disc two in the x direction. We can take the inverse
tangent of both sides and substitute in our values to get the theta for disc two final equals the inverse tangent of 10.2432 divided by 30.8438 which is 18.3713 or 18.4 degrees. The final velocity of disc two is 32.5 centimeters per second at an angle of 18.4 degrees north of east. – This is our predicted
final velocity for disc two. Our measured value is 32.0
centimeters per second at an angle of 13 degrees north of east. I’d say we were pretty close. But what I’d like to know is is this really an elastic collision? Bobby, what needs to be true for this to be an elastic collision? – In an elastic collision
the kinetic energy needs to be conserved. – [Mr. P] Bobby, you are correct. Now let’s determine if the total kinetic energy was conserved. Please, Bobby, start by determining the kinetic energy initial of the system. – Do we need to first
convert our known values to kilograms and meters? – Yes, when determining energy, all of our units should be in
base SI units, so go ahead. – Convert both of our masses
from grams to kilograms by multiplying by one
kilogram over 1000 grams. Convert the magnitudes of all our velocities to meters per second by multiplying by one
meter over 100 centimeters and now the sum of the
initial kinetic energies is the sum of the initial
kinetic energy for both discs. However, the initial
velocity of disc two is zero so it has no initial kinetic energy therefore the initial
kinetic energy of the system is just the initial
kinetic energy of disc one. One half times the mass of disc one times the initial velocity
of disc one squared or one half times 0.0288
times 0.336 squared which is 0.0016257 joules. – Which we can convert to millijoules by multiplying by 1000
millijoules over one joule and we get 1.6257 millijoules. – Hold up, Mr. P.
– Yes, Bo? – Why did you convert to
millijoules and do we have to? – Two reasons: one, 1.6257 millijoules makes more sense to me
than 0.0016257 joules and two, in my experience students need as much practice as
possible doing conversions. But no, we don’t actually have
to convert to millijoules. – Okay, thanks. – [Mr. P] We also need to determine the net kinetic energy final. The sum of the final kinetic energies is the final kinetic energy of disc one plus the final kinetic energy of disc two. With numbers that works out to be one half times 0.0288 times 0.107 squared plus one half times
0.0269 times 0.32 squared. That works out to be 1.5421 millijoules. In other words, before the collision we measured that the system had
1.6257 millijoules of energy and after the collision we measured that the system had 1.5421
millijoules of energy. Now we can use the relative error formula to determine what percentage
of the kinetic energy was dissipated during the collision. Using the measured net
kinetic energy final as our observed value and the measured net
kinetic energy initial as our accepted value we get a change in the
measured total kinetic energy of the system to be negative 5.14% with
three significant digits. In other words, 5.14% of
the initial kinetic energy was converted to heat and sound during the collision. I would say that’s pretty
close to an elastic collision. Thank you very much for
learning with me today. I enjoyed learning with you. – Hold up, Mr. P.
– Yes, Bo? – I think we need to
discuss Movie Character Day. (air horn)
– Movie Character Day. – [Mr. P] Sure, go ahead. Who did you all dress up as? – I’m Barry B. Benson from the Bee Movie. – I’m Zoolander.
– I’m Fix-It Felix. – Fix-It Felix doesn’t
wear a hat or overalls. – Zoolander never wore a
silver outfit or gold cape. – I think he did wear a silver outfit and I love this cape. – Mr. P, who are you dressed up as? – That is correct. – What?
– What are you saying? – I am dressed up as the
movie character Mr. P. – You are not a movie character. – Really? You do not think
I am a movie character? How much time have you spent
watching me on this screen?

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