2-Body Collisions: The Basics

We have two objects, m1 and m2 . 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 Symon text. Hence, we will consider here the case of a collision in which one of the objects, m2, 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, m1, 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: p1xi + 0  =  p1xf + p2xf

Conservation of momentum in the y-direction: 0 + 0  =  p1yf + p2yf

Since p=mv, we can write the kinetic energy in terms of p instead of v:

Conservation of energy (with Q = energy supplied from energy stored in particles, Elost = energy lost due to friction and deformation of particles) :

p1xi2 /2m1  +  Q    =    p1xf2 /2m1  +  p1yf2 /2m1  +  p2xf2 /2m2  +  p2yf2 /2m2  +  Elost

In these 3 equations, we have 9 quantities: m1, m2, p1xi, p1xf, p1yf, p2xf, p2yf, Q and Elost.

If we know that we have an elastic collision without any internal energies, then Q and Elost are both 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:    p1i + 0   =   p1f cos(q1) + p2f cos(q2)                 (Eq. 1)

Conservation of momentum in the y-direction:      0 + 0   =   p1f sin(q1) + p2f sin(q2)                   (Eq. 2)

Conservation of energy:       p1i2/2m1  +  0  +  Q    =    p1f2/2m1  +  p2f2/2m2  +  Elost .                (Eq. 3)

As we had for the rectangular case, if we have Q and Elost both zero, and if we know both masses and the initial momentum, then we have 3 equations for 4 unknowns: p1f, p2f, q1, and q2 .

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 q1 than we can a rectangular component such as p1yf. One of the disadvantages is that mathematically q1 and q2 are inside trig functions. However, if we

(1) move all the p1 terms to the left and keep all the p2 terms on the right:

p1i - p1f cos(q1)   =   p2f cos(q2)                         (Eq. 1a)

-p1f sin(q1)  =  p2f sin(q2)                                  (Eq. 2a)

(2) square Eq. 1 and square Eq. 2:

p1i2 - 2p1i p1f cos(q1) + p1f2 cos2(q1)   =   p2f2 cos2(q2)                     (Eq. 1b)

p1f2 sin2(q1)   =   p2f2 sin2(q2)                                            (Eq. 2b)

and then (3) add these two together, we can use the fact that sin2(q) + cos2(q) = 1 to eliminate q2 altogether and also limit q1 to just one term:

p1i2 - 2p1ip1f cos(q1) + p1f2   =   p2f2                            (Eq. 4)

We can use Eq. 3 [p1i2/2m1  +  0  +  Q    =    p1f2/2m1  +  p2f2/2m2  +  Elost  with Q and Elost = 0] to get an expression for p2f so we can eliminate that variable in Eq. 4:

p2f2  =  (m2/m1)*(p1i2 - p1f2)                                    (Eq. 3a)

Combining Eqs. 4 and 3a gives a quadratic equation for p1f in terms of p1i and q1:

p1f2[1+(m2/m1)] + p1f[-2p1i cos(q1)] + [p1i2{1-(m2/m1)}]   =   0                        (Eq. 5)

Using the quadratic formula, we find for p1f:

p1f    =    { 2p1i cos(q1) +/- [4p1i2 cos2(q1) - 4(1+m2/m1)p1i2(1-m2/m1)]1/2 } / 2(1+m2/m1)

Notice that a p1i can be factored out of the right side, so that we now have:

p1f/p1i   =   { cos(q1) +/- [cos2(q1) - (1-m22/m12)]1/2 } / (1+m2/m1)                         (Eq. 5a)

From Eq. 5a, we can see that to have a real answer we require cos2(q1) ³ (1-m22/m12).

If m1 > m2, then (1-m22/m12) is always positive, which means that cos2(q1) > 0 always, which means that q1 cannot reach 90o, which means that there exists a maximum angle for q1 ! Does this agree with what you would have expected for a heavier mass striking a lighter mass?

Conversely, if m1<m2, then (1-m22/m12) is negative, and so cos2(q1) can go all the way down to zero, which means that q1 can reach all the way up to 90o, which means that there is no maximum angle. Does this agree with what you expect if a lighter mass strikes a heavier mass?