Video Transcript
A smooth sphere of mass six 𝑚
kilograms was sliding on a smooth horizontal plane when it collided with another,
initially stationary, smooth sphere of the same size and of mass nine 𝑚
kilograms. The first sphere’s direction of
motion after the collision was at right angles to its direction of motion before the
collision. Let 𝜃 be the angle the first
sphere’s direction of motion made with the line of centers of the spheres just
before impact. Find an expression for tan squared
𝜃 in terms of the coefficient of restitution between the two spheres, 𝑒. Is it (a) two 𝑒 minus three over
five, (b) three 𝑒 plus two over five, (c) three 𝑒 minus two over five, (d) three
𝑒 minus two over three, or (e) two 𝑒 plus three over five?
In an oblique impact, there are two
physical laws which govern what will happen. The first is the conservation of
linear momentum, which states that the sum of the products of masses and velocities
of both objects will be the same both before and after the collision. So 𝑚 one 𝑢 one plus 𝑚 two 𝑢 two
equals 𝑚 one 𝑣 one plus 𝑚 two 𝑣 two, where 𝑚 one and 𝑚 two are the masses of
the object one and two, respectively, 𝐮 one and 𝐮 two are the velocities before
the collision of object one and two, respectively, and 𝐯 one and 𝐯 two are the
velocities after the collision of object one and two, respectively.
Since this is a vector equation, it
will also apply component-wise along any axis. In particular, we can separate the
vector equation into orthogonal component equations equal to the number of
dimensions in the problem. Since the two spheres are moving in
a horizontal plane, this is a two-dimensional problem, so we can separate the vector
equation into two different component equations.
This first equation is the
component equation along the line of action of the forces during the collision. Let’s call this horizontal for
simplicity. So, for example, 𝑢 one 𝐻 is the
horizontal component of the vector 𝑢 one, and likewise for the rest of the
vectors. And this second equation is the
component equation along the line of axis perpendicular to the line of action. But we’ll call this vertical for
simplicity. So 𝑢 one 𝑃 is just the vertical
component of 𝑢 one, and likewise for the rest of the vectors.
The second physical law that
governs what will happen during a collision is the conservation of energy. Energy will be generally conserved,
but the kinetic energy may not necessarily be conserved. The amount of kinetic energy lost
to during the collision is determined by the coefficient of restitution, 𝑒. In a more useful form, 𝑒 is the
ratio of the magnitudes of the relative velocities of the two colliding objects
after the collision to the magnitudes of the relative velocities of the two
colliding objects before the collision along the line of action of forces during the
collision.
So we have 𝑒 equals the magnitude
of 𝑣 two 𝐻 minus 𝑣 one 𝐻 over the magnitude of 𝑢 two 𝐻 minus 𝑢 one 𝐻, where
𝑢 one 𝐻 is the same as we had in the conservation of momentum equations, and
likewise for the other vectors. So to summarize, we have three
equations. The conservation of linear momentum
in the horizontal axis or along the line of action, 𝑚 one 𝑢 one 𝐻 plus 𝑚 two 𝑢
two 𝐻 equals 𝑚 one 𝑣 one 𝐻 plus 𝑚 two 𝑣 two 𝐻. The second, the conservation of
linear momentum in the vertical axis or perpendicular to the line of action, 𝑚 one
𝑢 one 𝑃 plus 𝑚 two 𝑢 two 𝑃 equals 𝑚 one 𝑣 one 𝑃 plus 𝑚 two 𝑣 two 𝑃. And the third, the coefficient of
restitution, 𝑒, equals the magnitude of 𝑣 two 𝐻 minus 𝑣 one 𝐻 over the
magnitude of 𝑢 two 𝐻 minus 𝑢 one 𝐻.
Now it’s time to consider our
scenario. We have two spheres of the same
size, one of mass six 𝑚 and the other of mass nine 𝑚. The line of action of the collision
— that is, the axis along which all forces between the spheres will act — is this
line connecting the two centers and the point of impact.
The first sphere approached the
collision with initial velocity 𝐮 one at an angle of 𝜃 to the line of action. The second sphere was stationary
before the collision. The first sphere then rebounded
with some velocity 𝐯 one at an angle right angles to the velocity 𝐮 one. The second sphere rebounded with
velocity 𝐯 two. This velocity was generated by a
force acting along the line of action. So the velocity must also be along
the line of action so it’s perfectly horizontal.
Our goal is to express 𝜃 or, more
specifically, tan squared 𝜃 as some function of the coefficient of restitution,
𝑒. And we can find this function by
eliminating all of the other variables, the velocities 𝐮 one, 𝐯 one, and 𝐯 two,
from these three equations on the left. The fact that the question asks us
to find tan squared 𝜃 suggests that this quantity will arise naturally in the
derivation. And indeed it does.
So let’s start substituting in the
variables to these three equations, starting with the first equation, the
conservation of linear momentum in the horizontal direction. So 𝑚 one is just the mass of our
first object, which is equal to six 𝑚. 𝑢 one H is the horizontal
component of the vector 𝐮 one. So it’s the magnitude of 𝑢 one
times cos of this angle here, 𝜃. To make things more readable, let’s
write the magnitude of the vector 𝐮 one as simply the scalar 𝑢 one. And we’ll do this for all of the
other vectors as well.
For our second term, 𝑚 two is just
the mass of our second sphere, nine 𝑚, but 𝑢 two 𝐻 is the horizontal component of
the velocity of the second sphere before the collision, which was zero. So this whole term will just be
zero. For our next term, we have 𝑚 one
again, which is just six 𝑚, times the horizontal component of the velocity after
the collision of the first sphere, which is the magnitude 𝑣 one times cos of this
angle in here, which is 90 minus 𝜃. But cos of 90 minus 𝜃 is just the
same as sin of 𝜃. So this is in fact just 𝑣 one sin
𝜃.
However, in resolving in the
horizontal axis, we have already assumed by convention that rightward velocity is
positive. Notice that this first sphere has a
velocity going towards the left. So this is a negative velocity. So this term is in fact negative 𝑣
one sin 𝜃. And for our final term, we just
have 𝑚 two, which again is nine 𝑚, times the horizontal component of the velocity
after the collision of the second object, which is just the magnitude, 𝑣 two, since
it’s perfectly horizontal.
Plugging all these in, we get six
𝑚 𝑢 one cos 𝜃 equals negative six 𝑚𝑣 one sin 𝜃 plus nine 𝑚𝑣 two. We have a common factor here of
three 𝑚 in every term, so we can divide through by three 𝑚 to simplify the
equation. This gives two 𝑢 one cos 𝜃 equals
negative two 𝑣 one sin 𝜃 plus three 𝑣 two.
Now for our second equation, the
conservation of linear momentum in the axis perpendicular to the line of action, if
we proceed similarly, for this first term we get six 𝑚𝑢 one sin 𝜃, for the second
term, we get zero once again, for the third term, we get six 𝑚𝑣 one cos 𝜃, and
for the fourth term, we get zero. Plugging all these in and dividing
through by the common factor of six 𝑚, we get 𝑢 one sin 𝜃 equals 𝑣 one cos
𝜃.
Now for our final equation the
coefficient of restitution. This first term, 𝑣 two 𝐻, is the
horizontal component of the second sphere’s velocity after the collision, which is
just its magnitude 𝑣 two, since it’s perfectly horizontal. 𝑣 one 𝐻 is the horizontal
component of the first object’s velocity after the collision, which is 𝑣 one cos 90
minus 𝜃. And again, since this is leftward,
this is negative. And again, the cos of 90 minus 𝜃
is just the same as sin of 𝜃. So this is equal to negative 𝑣 one
sin 𝜃.
𝑢 two 𝐻 is the horizontal
component of the second object’s velocity before the collision, which is just
zero. And 𝑢 one 𝐻 is the horizontal
component of the first object’s velocity before the collision, which is just 𝑢 one
cos 𝜃. Plugging all these in, we get the
magnitude of 𝑣 two minus negative 𝑣 one sin 𝜃 all over the magnitude of zero
minus 𝑢 one cos 𝜃. This simplifies to 𝑣 two plus 𝑣
one sin 𝜃 all over 𝑢 one cos 𝜃.
We now have our three
equations. And the goal is to find an
expression for tan squared 𝜃 in terms of only the coefficient of restitution, 𝑒,
meaning that we have to eliminate 𝑢 one, 𝑣 one, and 𝑣 two. Let’s rename these equations one,
two, and three, respectively, for simplicity. Starting with equation two, we can
easily rearrange this to give sin 𝜃 over cos 𝜃 equals 𝑣 one over 𝑢 one. But sin 𝜃 over cos 𝜃 is
identically equal to tan 𝜃. So this is, in fact, tan 𝜃 equals
𝑣 one over 𝑢 one. Let’s call this equation equation
four and take a last look at the diagram of the scenario because we’re going to need
a little space.
Now we could simply square equation
four and then use the other equations to eliminate 𝑣 one and 𝑢 one. But it’s actually easier to leave
this for now and eliminate 𝑣 two in the other two equations. From equation one, we can rearrange
to get 𝑣 two equals two-thirds times 𝑢 one cos 𝜃 plus 𝑣 one sin 𝜃. Substituting this expression for 𝑣
two into equation three, we get 𝑒 equals two-thirds times 𝑢 one cos 𝜃 plus 𝑣 one
sin 𝜃 plus 𝑣 one sin 𝜃 all over 𝑢 one cos 𝜃.
Expanding the parentheses and
simplifying gives us 𝑒 equals two-thirds plus five 𝑣 one sin 𝜃 over three 𝑢 one
cos 𝜃. But sin 𝜃 over cos 𝜃 is again
identically equal to tan 𝜃. And from equation four, 𝑣 one over
𝑢 one is also equal to tan 𝜃. Clearing a little space and
continuing up here, we then have 𝑒 equals two plus five tan squared 𝜃 all over
three. Rearranging for tan squared 𝜃
gives three 𝑒 minus two over five. Comparing this with our possible
answers, we can see this matches with (c), three 𝑒 minus two over five.
