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: p_1xi + 0 = p_1xf + p_2xf

Conservation of momentum in the y-direction: 0 + 0 = p_1yf + p_2yf

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, E_lost = energy lost due to friction and deformation of particles) :

p_1xi² /2m₁ + Q = p_1xf² /2m₁ + p_1yf² /2m₁ + p_2xf² /2m₂ + p_2yf² /2m₂ + E_lost

In these 3 equations, we have 9 quantities: m₁, m₂, 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 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: p_1i + 0 = p_1f cos(q₁) + p_2f cos(q₂) (Eq. 1)

Conservation of momentum in the y-direction: 0 + 0 = p_1f sin(q₁) + p_2f sin(q₂) (Eq. 2)

Conservation of energy: p_1i²/2m₁ + 0 + Q = p_1f²/2m₁ + p_2f²/2m₂ + 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, 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 p_1yf. 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:

p_1i - p_1f cos(q₁) = p_2f cos(q₂) (Eq. 1a)

-p_1f sin(q₁) = p_2f sin(q₂) (Eq. 2a)

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

p_1i² - 2p_1i p_1f cos(q₁) + p_1f² cos²(q₁) = p_2f² cos²(q₂) (Eq. 1b)

p_1f² sin²(q₁) = p_2f² 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:

p_1i² - 2p_1ip_1f cos(q₁) + p_1f² = p_2f² (Eq. 4)

We can use Eq. 3 [p_1i²/2m₁ + 0 + Q = p_1f²/2m₁ + p_2f²/2m₂ + E_lost with Q and E_lost = 0] to get an expression for p_2f so we can eliminate that variable in Eq. 4:

p_2f² = (m₂/m₁)*(p_1i² - p_1f²) (Eq. 3a)

Combining Eqs. 4 and 3a gives a quadratic equation for p_1f in terms of p_1i and q₁:

p_1f²[1+(m₂/m₁)] + p_1f[-2p_1i cos(q₁)] + [p_1i²{1-(m₂/m₁)}] = 0 (Eq. 5)

Using the quadratic formula, we find for p_1f:

p_1f = { 2p_1i cos(q₁) +/- [4p_1i² cos²(q₁) - 4(1+m₂/m₁)p_1i²(1-m₂/m₁)]^1/2 } / 2(1+m₂/m₁)

Notice that a p_1i can be factored out of the right side, so that we now have:

p_1f/p_1i = { 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 90^o, 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 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?

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