Two-Dimensional Collisions

Suppose that an object of mass $m_1$, moving with initial velocity ${\bf v}_{1i}$, strikes a second object, of mass $m_2$, that is initially at rest. Suppose, further, that the collision is not head-on, so that after the collision the first object moves off at an angle $\theta_1$ to its initial direction of motion, whereas the second object recoils at an angle $\theta_2$ to this direction. Let the final velocities of the two objects be ${\bf v}_{1f}$ and ${\bf v}_{2f}$, respectively. See Figure 1.6.

Figure 1.6: A two-dimensional collision in the laboratory frame.

We are again considering a system in which there is zero net external force (because the forces associated with the collision are internal in nature). It follows that the total momentum of the system is a conserved quantity. However, unlike before, we must now treat momentum as a vector quantity, because we are no longer dealing with one-dimensional motion. Momentum conservation implies that

$$m_2\,{\bf v}_{1i} = m_1\,{\bf v}_{1f} + m_2\,{\bf v}_{2f}.$$ (1.148)

As before, it is convenient to transform to a frame of reference that co-moves with the center of mass of the system. The invariant velocity of the center of mass is given by

$${\bf V} = \frac{m_1\,{\bf v}_{1i}}{m_1+m_2} = \frac{m_1\,{\bf v}_{1f}+m_2\,{\bf v}_{2f}}{m_1+m_2}.$$ (1.149)

An object that possesses a velocity ${\bf v}$ in the laboratory frame possesses a velocity ${\bf v}' = {\bf v}-{\bf V}$ in the center-of-mass frame. Hence, it follows that

$${\bf v}_{1i}' =\left( \frac{m_2}{m_1+m_2}\right){\bf v}_{1i},$$ (1.150)

$${\bf v}_{2i}' = - \left(\frac{m_1}{m_1+m_2}\right){\bf v}_{1i},$$ (1.151)

$${\bf v}_{1f}' =-\left(\frac{m_2}{m_1+m_2}\right)({\bf v}_{2f} - {\bf v}_{1f}),$$ (1.152)

$${\bf v}_{2f}' = \left(\frac{m_1}{m_1+m_2}\right)({\bf v}_{2f}-{\bf v}_{1f}).$$ (1.153)

Furthermore, the momenta in the center-of-mass frame take the form

$$-{\bf p}_{1i}' = {\bf p}_{2i}' = -\mu\,{\bf v}_{1i},$$ (1.154)

$$-{\bf p}_{1f}' = {\bf p}_{2f}' = \mu\,({\bf v}_{2f}-{\bf v}_{1f}),$$ (1.155)

where $\mu = m_1\,m_2/(m_1+m_2)$. (Of course, ${\bf p}_{1i}' = m_1\,{\bf v}_{1i}'$, et cetera.) As before, in the center-of-mass frame, the two objects approach one another with equal and opposite momenta before the collision, and diverge from one another with equal and opposite momenta after the collision. Let $\theta$ be the direction subtended between the final and initial momenta of each object in the center-of-mass frame. See Figure 1.7. It follows that in the $x$-$y$ coordinate system shown in the figure,

$${\bf p}_{1i}' = p_{1i}'\,(1,\,0),$$ (1.156)

$${\bf p}_{2i}' = -p_{1i}'\,(1,\,0),$$ (1.157)

$${\bf p}_{1f}' = p_{1f}'\,(\cos\theta,\,\sin\theta),$$ (1.158)

$${\bf p}_{2f}' = -p_{1f}'\,(\cos\theta,\,\sin\theta),$$ (1.159)

where $p_{1i}'=\vert{\bf p}_{1i}'\vert$, et cetera. Finally, if the collision is elastic then the kinetic energy before the collision must equal that after the collision (in both the laboratory and the center-of-mass frames). It follows that

$$p_{1f}' = p_{1i}'.$$ (1.160)

Hence,

$${\bf v}_{1f}' \equiv \frac{{\bf p}_{1f}'}{m_1}= \left(\frac{m_2\,v_{1i}}{m_1+m_2}\right)(\cos\theta,~\sin\theta),$$ (1.161)

$${\bf v}_{2f}' \equiv \frac{{\bf p}_{2f}'}{m_2}= -\left(\frac{m_1\,v_{1i}}{m_1+m_2}\right)(\cos\theta,~\sin\theta),$$ (1.162)

It can be seen from the previous two equations that an elastic two-dimensional collision is fully characterized once the initial velocity, $v_{1i}$, and the scattering angle, $\theta$, are specified. In general, we would expect $\theta$ to be able to take all values in the range 0 to $\pi$. In fact, a head-on collision corresponds to $\theta=\pi$, whereas a glancing collision corresponds to $\theta\ll 1$.

Figure 1.7: A two-dimensional collision in the center-of-mass frame.

Now, ${\bf v}={\bf v}' + {\bf V}$, where

$${\bf V}=\left( \frac{m_1\,v_{1\,i}}{m_1+m_2}\right)(1,\,0).$$ (1.163)

It follows that, in the $x$-$y$ coordinate system shown in Figure 1.6, the laboratory-frame velocities of the two objects after the collision are

$${\bf v}_{1f} =\left(\frac{v_{1i}}{m_1+m_2}\right)(m_1+m_2\,\cos\theta,~m_2\,\sin\theta),$$ (1.164)

$${\bf v}_{2f} = \left(\frac{m_1\,v_{1i}}{m_1+m_2}\right)(1-\cos\theta,~-\sin\theta).$$ (1.165)

Hence, according to Figure 1.6,

$$\tan\theta_1 = \frac{\sin\theta}{\cos\theta+m_1/m_2},$$ (1.166)

$$\tan\theta_2 = \frac{\sin\theta}{1-\cos\theta}=\tan\left(\frac{\pi}{2}-\frac{\theta}{2}\right).$$ (1.167)

The last equation implies that

$$\theta_2 = \frac{\pi}{2}-\frac{\theta}{2}.$$ (1.168)

Differentiating Equation (1.166) with respect to $\theta$, we obtain

$$\frac{d\tan\theta_1}{d\theta} = \frac{1+(m_1/m_2)\,\cos\theta}{(\cos\theta+m_1/m_2)^{\,2}}.$$ (1.169)

Thus, $\tan\theta_1$ attains an extreme value, which can be shown to correspond to a maximum possible value of $\theta_1$, when the numerator of the previous expression is zero; that is, when

$$\cos\theta = -\frac{m_2}{m_1}.$$ (1.170)

Note that it is only possible to solve the previous equation when $m_1> m_2$. If this is the case then Equation (1.166) yields

$$\tan\theta_{1\,{\rm max}} = \frac{m_2/m_1}{\sqrt{1-(m_2/m_1)^{\,2}}},$$ (1.171)

which reduces to

$$\theta_{1\,{\rm max}} = \sin^{-1}\left(\frac{m_2}{m_1}\right).$$ (1.172)

Hence, we conclude that when $m_1> m_2$ there is a maximum possible value of the scattering angle, $\theta_1$, in the laboratory frame. This maximum value is always less than $\pi/2$, which implies that there is no backward scattering (i.e., $\theta_1>\pi/2$) at all when $m_1> m_2$. For the special case when $m_1=m_2$, the maximum scattering angle is $\pi/2$. However, for $m_1<m_2$ there is no maximum value, and the scattering angle in the laboratory frame can thus range all the way to $\pi$.

Suppose that the two particles have equal masses, so that $m_1=m_2$. In this case, Equation (1.166) yields

$$\tan\theta_1 = \frac{\sin\theta}{\cos\theta+1} =\tan\left(\frac{\theta}{2}\right).$$ (1.173)

Hence,

$$\theta_1= \frac{\theta}{2}.$$ (1.174)

In other words, the scattering angle of the first particle in the laboratory frame is half of the scattering angle in the center-of-mass frame. The previous equation can be combined with Equation (1.168) to give

$$\theta_1 + \theta_2 =\frac{\pi}{2}.$$ (1.175)

Thus, in the laboratory frame, the two particles move off at right-angles to one another after the collision. 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 obliquely then the two balls move off at right-angles to one another. 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. Finally, it is easily demonstrated that the fractions of the initial kinetic energy carried off by the two particles after the collision are

$$\frac{E_{1f}}{E_{1i}} = \cos^2\theta_1,$$ (1.176)

$$\frac{E_{2f}}{E_{1i}} = \sin^2\theta_1.$$ (1.177)