here we've got two billiard balls colliding and i gave them both equal masses
and you should notice in this problem that there are three unknowns and that's the final speed
of one of the balls the final speed of the other and the angle of just one of them and
the only thing we know about the final state is the angle of one of the balls at 40 degrees
now conceptually this problem isn't so bad but mathematically it gets really ugly so conceptually
we just use conservation of momentum and then it's an elastic collision so we use
conservation of energy and we set up three equations which is why i have three unknowns
set up in the problem so i'm going to look at x momentum and write down an equation
in my initial state i have a mass of 0.5 kilograms moving at 2 meters per second
and in my final state i've got a mass of 0.5 kilograms moving at v1 cosine
40 degrees and another mass of 0.5 kilograms moving at v2 cosine of this unknown angle theta
of course all those masses are going to cancel out
and i end up with my first equation and it's 2. if i just disappear my units for simplicity
2 equals v1 cosine 40.
plus v2 cosine of this unknown angle theta next we look at py
and in the y direction my initial momentum is zero and so in the final state it must also be zero
so rather than writing down zero equals a difference of two things over here i'm just
going to say that the upward momentum's magnitude must be equal to the downward momentum's magnitude
it's just a little simpler to write down that way so i have a mass of 0.5 kilograms
moving upward at v1 sine 40 degrees and that should be equal to the momentum i get
downward from the second ball 0.5 kilograms
v2 sine of the unknown angle theta and the masses cancel out and i get a second
useful equation v1 sine 40 degrees equals v2 sine of theta finally i look at the kinetic energy
and in the initial state i have one half times 0.5 kilograms times this initial speed
squared and in the final state i have two things moving
the top ball has a speed of v1
and the bottom ball has a speed of v2 those are both unknowns
the one-halfs cancel out the masses cancel out and i get a third equation so 2 squared is 4 so
i get 4 equals v1 squared plus v2 squared okay so in principle the problem is done we applied
the physics concepts we got three equations we have three unknowns but i do want to illustrate
how we go about solving these equations because it's a weird system of equations it's non-linear
so let me summarize up here okay and the strategy i'm going to take with this is to eliminate theta
first and i'm going to use a trick for that i'm going to use the fact that sine squared theta plus
cosine squared theta is equal to one so i'm going to solve the second equation here for sine theta
i'm going to solve the first equation for cosine theta and then i'm going to add the squares of
those two equations and it'll come out to 1 and theta will be gone so let's get into it
sine theta is going to be v1 over v2 sine of 40 degrees
and at this point to simplify my life i'm going to actually move to decimal approximations so
the sine of 40 degrees i'm going to keep a lot of decimal points here so 0.6428
in equation number one i'm going to solve for cosine theta
and that's 2 minus v1 cosine 40 degrees
divided by v2 and again i'm going to move to decimal approximations
keep things a little bit simpler so this gives me 2 minus cosine 40 is 0.7660
v1 over v2 now i apply my trig identity and i go the square of this equation
plus the square of this equation must be equal to one so i have squaring 0.6428 v1 i get 0.413
v1 squared divided by v2 squared then i add to that the square of cosine theta
so i'm going to have a 2 minus 0.7660 v1 squared over v2 squared
and that's got to be equal to 1. so now my goal is to take this equation
and combine it with equation number 3 in our list that's two equations and two unknowns and
at least now it's algebraic instead of having trigonometric functions mixed in
so i'm going to multiply both sides by v2 squared plus the quantity 2 minus 0.7660v1 squared equals
v2 squared i'm going to simplify a little bit on the left hand side so i end up with a 0.413
v1 squared when i expand that squared binomial i'm going to get a 4 and then i'll get minus 4 times
0.7660 and that gives me 3.064 v1 then i'll have plus 0.7660 squared v1 squared so that's 0.5868 v1
squared and that's all equal to v2 squared finally combining the v1 squared terms i get 0.9998 v1
squared which to our level of precision is just one so i'm going to have v1 squared minus 3.064
v1 plus 4 equals v2 squared
now that that's cleaned up i'm going to sub into equation 3
and replace v2 squared with all this stuff down here
so i'll go ahead and move up to the top to get that done so i'm going to replace v2 squared
with v1 squared minus 3.064
v1 plus 4. and immediately i notice something nice happens the 4s cancel
and i get 0 on the left hand side equals 2v1 squared
minus 3.064 v1 so one possible solution here
is that v one is zero but it's not the solution we're looking for
what would happen if v one is zero it would mean this ball is completely stationary
it also would imply if i go back to this equation sine theta is v1 sine 40 over v2 if i plug in v1
equals zero it would turn out the angle theta was zero so one possibility for this collision
would actually be a head-on elastic collision with two equal masses and that would mean the original
ball would stop the other one would proceed with the original velocity at theta equals zero and
momentum and energy would be conserved it's not the solution that we're interested in however
really what i want is the solution i get when i set 2v1 minus 3.064 equal to 0 which implies
that v1 is equal to 1.532 meters per second and we've got our first answer
so then i head back into my old equations and find the rest of the variables
i'm going to go back to equation 3 to find v2 and it's going to be the square root
of 4 minus v1 squared which is the square root of 4 minus 1.532 squared which comes out to 1.286
meters per second
all right we've got our second answer the last thing we need to do
is find out the angle for that ball that was struck so i'm going to go back and get that out of
this equation right here i have sine theta equals v1 sine 4d over v2
and i have all those velocities
1.532 sine 40 over 1.286
and so theta is going to be the inverse sine if i just get a decimal on this real quick
0.766 and that means theta is equal to
practically 50 degrees so 50.0
now if you would like to check your answer on this and i actually did that because i
came up with this problem myself so there's no back of the book
you could go back and check energy x momentum and y momentum and make sure that they all check
out so i get the energy in the initial state now that i have these two velocities v1 being 1.532
meters per second and v2 being 1.286 meters per second i can get the kinetic energy
of each of these balls after the collision make sure it adds up to the same total value
i could check the x momentum by getting the x component of the momentum for the first and
second ball adding them together make sure it's the same as what it started with
and then check the y momentum and make sure that the upward momentum of this
ball equals the downward momentum of this ball and everything checks out
you