We have two objects, m
1
and m
2
. If both objects are moving, we could transform our system to one in which one of the particles is initially stationary. This transformation of coordinates is considered in Chapter 7 of the text. Hence, we will consider here the case of a collision in which one of the objects is initially at rest, and we will not lose any generality by doing so, since after we are finished in this system, we could simply employ the inverse transformation to get back to our initial system where the two objects were both moving initially.
For the system in which one of the particles is initially at rest, the 2-body collision problem reduces to one in two dimensions only. To see this, call the direction of the initial velocity of the one moving particle the x-direction. After the collision, the final total momentum should be the same as the initial total momentum, and hence the total final momentum must be in the x-direction also. If one of the two particles has a component of the final momentum in the y-direction without any in the z-direction, then the second of the two particles must also have a complimentary momentum in the negative y-direction and none in the z-direction.
In the following we assume the initial motion of particle 1 is in the x direction, and the final motion is in the x-y plane. We can obtain equations relating the initial and final quantities by using both conservation of momentum and conservation of energy:
Rectangular form:
Conservation of momentum in the x-direction: p
1xi
+ 0 = p
1xf
+ p
2xf
Conservation of momentum in the y-direction: 0 + 0 = p
1yf
+ p
2yf
Since p =m v , we can write the kinetic energy in terms of p instead of v:
Eq. 1r
Eq. 2r
Conservation of energy, with Q = energy supplied from energy stored in the particles,
E lost
= energy lost due to friction and deformation of the particles, gives: p
1xi
2 /2m
1
+ Q = p
1xf
2 /2m
1 +
p
1yf
2 /2m
1
+ p
2xf
2 /2m
2
+ p
2yf
2 /2m
2
+ E lost
Eq. 3
In these 3 equations , we have 9 quantities : m
1
, m
2
, p
1xi
, p
1xf
, p
1yf
, p
2xf
, p
2yf
, Q and E lost
.
If we know that we have an elastic collision without any internal energies, then Q and
E lost
are zero. We have 3 equations and 7 quantities. If we know both masses and the initial momentum, we still have 3 equations and 4 unknowns. We must know something else - either about the nature of the collision or something about the final motion.

Polar form:
Conservation of momentum in the x-direction: p
1i
+ 0 = p
1f cos(

1
) + p
2f cos(

2
) Eq. 1p
Conservation of momentum in the y-direction:
0 + 0 = p
1f sin(

1
) + p
2f sin(

2
) Eq. 2p
Conservation of energy: p
1i
2 /2m
1
+ 0 + Q = p
1f
2 /2m
1
+ p
2f
2 /2m
2
+ E lost
. Eq. 3
As we had for the rectangular case, if we have Q and E lost
both zero, and if we know both masses and the initial momentum, then we have 3 equations for 4 unknowns: p
1f
, p
2f
,
 and

2
.
1
,
Useful Relations:
We can find out some interesting relations and some limits on the possibilities for different situations by working with our three equations.
One of the advantages of the polar form is that we can sometimes more easily measure an angle such as

1
than we can a rectangular component such as p
1yf
. One of the disadvantages is that

1
and

2
are inside trig functions. However, if we
(1) move all p
1
terms to the left and keep all p
2
terms on the right: p
1i
-p
1f cos(

1
) = p
2f cos(

2
) Eq. 1p
-p
1f sin(

1
) = p
2f sin(

2
) Eq. 2p
(2) square Eq. 1 and square Eq. 2, p
1i
2 - 2 p
1i
p
1f cos(

1
) + p
1f
2 cos 2 (

1
) = p
2f
2 cos 2 (

2
) Eq. 1p p
1f
2 sin 2 (

1
) = p
2f
2 sin 2 (

2
) Eq. 2p and (3) then add these two together, we can use the fact that sin
2 eliminate

2
altogether and also limit

1
to just one term:
(

) + cos
2
(

) = 1 to p
1i
2 - 2p
1i p
1f
cos(

1
) + p
1f
2 = p
2f
2 Eq. 4
We can use Eq. 3 to get an expression for p
2f
so we can eliminate that variable in Eq. 4: p
2f
2 = (m
2
/m
1
)*(p
1i
2 - p
1f
2 ) Eq. 3
Combining Eqs. 4 and 3 gives a quadratic equation for p
1f
in terms of p
1i
and

1
: p
1f
2 [1+(m
2
/m
1
)] + p
1f
[-2p
1i
cos(

1
)] + [p
1i
2 {1-(m
2
/m
1
)}] = 0 Eq. 5

Using the quadratic formula, we find for p
1f
: p
1f =
{ 2p
1i
cos(

1
) +/- [4p
1i
2 cos 2 (

1
) - 4(1+m
2
/m
1
)p
1i
2 (1-m
2
/m
1
)] 1/2 } / 2(1+m
2
/m
1
)
Notice that a p
1i
can be factored out of the right side, so that we now have: p
1f
/p
1i
= { cos(

1
) +/- [cos 2 (

1
) - (1-m
2
2 /m
1
2 )] 1/2 } / (1+m
2
/m
1
) (Eq. 5a)
From Eq. 5a, we can see that to have a real answer we require cos
2
(

1
)

(1 - m
2
2
/m
1
2
).
If m
1
> m
2
, then (1 - m
2
2
/m
1
2
) is always positive, which means that cos
2
(

1
) > 0 always
(i.e., cos
2
(
 ) ≠ 0) , which means that 
1
cannot reach 90 o
, which means that there exists a maximum angle for

1
! Does this agree with what you would have expected for a heavier mass striking a lighter mass? Conversely, if m
1
<m
2
, then (1-m
2
2
/m
1
2
) is negative, and so cos
2
(

1
) can go all the way down to 0, which means that

1
can reach all the way up to 90 o
, which means that there is no maximum angle. Does this agree with what you expect if a lighter mass strikes a heavier mass?