One-Dimensional Collisions

Consider two particles of mass $m_1$ and $m_2$, respectively, that are free to move in one dimension. Suppose that these two particles collide. Suppose, further, that both particles are subject to zero net force when they are not in contact with one another. Finally, let us assume that we are observing the collision in a convenient inertial reference frame known as the laboratory frame. This situation is illustrated in Figure 1.4.

Both before and after the collision, the two particles move with constant velocity, in accordance with Newton's first law of motion. Let $v_{1i}$ and $v_{2i}$ be the velocities of the first and second particles, respectively, before the collision. Here, velocities to the right in Figure 1.4 are positive. Likewise, let $v_{1f}$ and $v_{2f}$ be the velocities of the first and second particles, respectively, after the collision. During the collision itself, the first particle exerts a large transitory force, $f_{21}$, on the second, whereas the second particle exerts an equal and opposite force, $f_{12}=-f_{21}$, on the first. In fact, we can model the collision as equal and opposite impulses given to the two particles at the instant in time when they come together. (See Section 1.3.1.)

Figure 1.4: A one-dimension collision in the laboratory frame.

We are clearly considering a system in which there is zero net external force (because the forces associated with the collision are internal in nature). Hence, the total (linear) momentum of the system is a conserved quantity. (See Section 1.4.4.) Equating the total momenta before and after the collision, we obtain

$$m_1\,v_{1i} + m_2\,v_{2i} = m_1\,v_{1f} + m_2\,v_{2f}.$$ (1.116)

This equation is valid for any one-dimensional collision, irrespective its nature.

Suppose that the collision is elastic, which means that there is no associated loss of kinetic energy. Equating the net kinetic energies before and after the collision, we obtain

$$\frac{1}{2}\,m_1\,v_{1i}^{\,2} + \frac{1}{2}\,m_2\,v_{2i}^{\,2}= \frac{1}{2}\,m_1\,v_{1f}^{\,2}+\frac{1}{2}\,m_2\,v_{2f}^{\,2}.$$ (1.117)

(See Section 1.3.2.) It follows that

$$m_1\,(v_{1f}^{\,2}-v_{1i}^{\,2}) = -m_2\,(v_{2f}^{\,2}-v_{2i}^{\,2}),$$ (1.118)

or

$$m_1\,(v_{1f}-v_{1i})\,(v_{1f}+v_{1i}) =- m_2\,(v_{2f}-v_{2i})\,(v_{2f}+v_{2i}).$$ (1.119)

However, Equation (1.116) yields

$$m_1\,(v_{1f}-v_{1i})= -m_2\,(v_{2f}-v_{2i}).$$ (1.120)

The previous two equations can be combined to give

$$v_{1f}+v_{1i}= v_{2f}+v_{2i},$$ (1.121)

or

$$(v_{2f}-v_{1f})=-(v_{2i}-v_{1i}).$$ (1.122)

Thus, we conclude that an elastic collision causes the relative velocity of the two particles to reverse direction, while keeping the same magnitude.

Suppose that we transform to a frame of reference that co-moves with the center of mass of the system. The motion of a multi-particle system often looks particularly simple when viewed in such a frame. Because the system is subject to zero net external force, the velocity of the center of mass is invariant [see Equations (1.72)], and is given by

$$V = \frac{m_1\,v_{1i} + m_2\,v_{2i}}{m_1+m_2}= \frac{m_1\,v_{1f} + m_2\,v_{2f}}{m_1+m_2}.$$ (1.123)

[See Equation (1.68).] A particle that possesses a velocity $v$ in the laboratory frame possesses a velocity $v'=v-V$ in the so-called center-of-mass frame. It is easily demonstrated that

$$v_{1i}' = - \frac{m_2}{m_1+m_2}\,(v_{2i}-v_{1i}),$$ (1.124)

$$v_{2i}' = + \frac{m_1}{m_1+m_2}\,(v_{2i}-v_{1i}),$$ (1.125)

$$v_{1f}' = - \frac{m_2}{m_1+m_2}\,(v_{2f}-v_{1f}),$$ (1.126)

$$v_{2f}' = + \frac{m_1}{m_1+m_2}\,(v_{2f}-v_{1f}).$$ (1.127)

Note, incidentally, that the center-of-mass frame is obviously inertial (because it is moving at a constant velocity with respect to the inertial laboratory frame).

Figure 1.5: A one-dimension collision in the center-of-mass frame.

The previous four equations yield

$$-p_{1i}' = p_{2i}' = \mu\,(v_{2i}-v_{1i}),$$ (1.128)

$$-p_{1f}' = p_{2f}' = \mu\,(v_{2f}-v_{1f}),$$ (1.129)

where $\mu = m_1\,m_2/(m_1+m_2)$ is the so-called reduced mass (see Section 1.10.7), and $p_{1i}' = m_1\,v_{1i}'$ is the initial momentum of the first particle in the center-of-mass frame, et cetera. In other words, when viewed in the center-of-mass frame, the two particles approach one another with equal and opposite momenta before the collision, and diverge from one another with equal and opposite momenta after the collision. See Figure 1.5. Thus, the center-of-mass momentum conservation equation,

$$p_{1i}' + p_{2i}' = p_{1f}' + p_{2f}',$$ (1.130)

is trivially satisfied, because both the left- and right-hand sides are zero. Incidentally, this result is valid for both elastic and inelastic collisions.

Equations (1.122), (1.128), and (1.129) can be combined to give

$$p_{1f}' = -p_{1i}',$$ (1.131)

$$p_{2f}' =-p_{2i}'.$$ (1.132)

In other words, in the center-of-mass frame, an elastic collision causes the equal and opposite momenta of the two particles to both reverse direction, but keep the same magnitude. The previous two expressions imply that

$$v_{1f}' = -v_{1i}',$$ (1.133)

$$v_{2f}' =-v_{2i}'.$$ (1.134)

In other words, in the center-of-mass frame, an elastic collision also causes the velocity of each particle to reverse direction, but keep the same magnitude. Thus, the total kinetic energy of the system is obviously a conserved quantity in the center-of-mass frame.

Equations (1.124) and (1.125) can be combined with the previous two equations to give

$$v_{1f}' = \frac{m_2}{m_1+m_2}\,(v_{2i}-v_{1i}),$$ (1.135)

$$v_{2f}' = -\frac{m_1}{m_1+m_2}\,(v_{2i}-v_{1i}).$$ (1.136)

However, $v_{1f}= v_{1f}'+V$ and $v_{2f}= v_{2f}'+V$, which allows us to express the velocities of the two particles after the collision in the laboratory frame in terms of the corresponding velocities before the collision:

$$v_{1f} = \left(\frac{m_1-m_2}{m_1+m_2}\right) v_{1i} + \left(\frac{2\,m_2}{m_1+m_2}\right)v_{2i},$$ (1.137)

$$v_{2f} = \left(\frac{2\,m_1}{m_1+m_2}\right)v_{1i} - \left( \frac{m_1-m_2}{m_1+m_2}\right)v_{2i}.$$ (1.138)

Let us, now, consider some special cases. Suppose that two equal-mass particles collide elastically. If $m_1=m_2$ then Equations (1.137) and (1.138) yield

$$v_{1f} = v_{2i},$$ (1.139)

$$v_{2f} = v_{1i}.$$ (1.140)

In other words, the two particles simply exchange velocities when they collide. For instance, if the second particle is stationary and the first particle strikes it head-on with velocity $v$ then the first particle is brought to a halt whereas the second particle moves off with velocity $v$. It is possible to reproduce this effect in snooker or pool by striking the cue ball with great force in such a manner that it slides, rather that rolls, over the table; in this case, when the cue ball strikes another ball head-on it comes to a complete halt, and the other ball is propelled forward very rapidly. Incidentally, it is necessary to prevent the cue ball from rolling, because rolling motion is not taken into account in our analysis, and actually changes the answer.

Suppose that the second particle is much more massive than the first (i.e., $m_2\gg m_1$), and is initially at rest (i.e., $v_{2i} =0$). In this case, Equations (1.137) and (1.138) yield

$$v_{1f} \simeq - v_{1i},$$ (1.141)

$$v_{2f} \simeq 0.$$ (1.142)

In other words, the velocity of the light particle is effectively reversed during the collision, whereas the massive particle remains approximately at rest. Indeed, this is the sort of behavior we expect when an object collides elastically with an immovable obstacle; for instance, when an elastic ball bounces off a brick wall.

Suppose, finally, that the second particle is much lighter than the first (i.e., $m_2\ll m_1$), and is initially at rest (i.e., $v_{2i} =0$). In this case, Equations (1.137) and (1.138) yield

$$v_{1f} \simeq v_{1i},$$ (1.143)

$$v_{2f} \simeq 2\,v_{1i}.$$ (1.144)

In other words, the motion of the massive particle is essentially unaffected by the collision, whereas the light particle ends up moving twice as fast as the massive one.