Chapter 13 Two or more particles in space (unconstrained)

There are a plethora of theorems about the momentum and energy of systems of particles. These are discussed in section LABEL:sec:KonigSchmonig. The simplest of these are just the ones that you get from adding up the results for a single particle from section 12.2:

Linear momentum balance.

${\displaystyle {\sum }_{\text{all forces}}}{\stackrel{\rightharpoonup }{𝑭}}_j={\displaystyle \sum m_i{\stackrel{\rightharpoonup }{𝒂}}_i}$.
Either because the forces between particles in a system are usually assumed to come in equal and opposite pairs or because it is an independent postulate of mechanics for general systems, the force sum can be replaced with a sum over all the external forces.

Angular momentum balance.

${\displaystyle {\sum }_{\text{all forces}}}{\stackrel{\rightharpoonup }{𝒓}}_{j/C}\times {\stackrel{\rightharpoonup }{𝑭}}_j={\displaystyle \sum {\stackrel{\rightharpoonup }{𝒓}}_{i/C}\times (m_i{\stackrel{\rightharpoonup }{𝒂}}_i)}$.
As for linear momentum, the force sum can be replaced with only the forces that act externally on the system.

Power balance.

${\displaystyle {\sum }_{\text{all forces}}}{\stackrel{\rightharpoonup }{𝑭}}_j\cdot {\stackrel{\rightharpoonup }{𝒗}}_j={\displaystyle \frac{d}{dt}}{\displaystyle \sum m_i{v}_i^2/2}$.
In this case the sum is over all the forces, internal and external. The simplification to just external forces doesn’t apply to system kinetic energy like it does for momentum and angular momentum.

Let’s review one special problem from the previous section. The ‘one-body’ problem should properly be about the mechanics of a single particle interacting with nothing else. Such a particle moves at constant velocity and is too boring to get a name. Instead, when people refer to the ‘one-body’ problem they are talking about a particle flying around a stationary point mass to which it is attracted. That stationary point mass is held in place by, well, who knows what. It’s just an idealized thing anchored by a massless structure. The ‘one-body’ problem is to find the motion of the particle flying around.

As we discussed in the previous sections, if the gravitational attraction follows an inverse square law then the particle moves on a plane on a curve which is either an ellipse, a circle, a parabola or a hyperbola. These are, quite accurately, the trajectories of the planets and comets around the sun.

If two particles are attracted equally to each other by mutually central forces, and no other forces act, this is called ‘the two-body problem’

.

Assume the two particles are $m_1$ and $m_2$ with positions ${\stackrel{\rightharpoonup }{𝒓}}_1$ and ${\stackrel{\rightharpoonup }{𝒓}}_2$ (relative to the origin of a coordinate system fixed in a Newtonian frame). The force on particle 1 from particle 2 is

where ${\stackrel{\rightharpoonup }{𝒓}}_{12}={\stackrel{\rightharpoonup }{𝒓}}_{2/1}$ is the position of particle 2 relative to particle 1, $r_{12}$ is the distance $|{\stackrel{\rightharpoonup }{𝒓}}_{12}|$ between the particles and $F$ is the magnitude of the attractive force. We assume the force on particle 2 is the opposite of this

The instantaneous velocities are ${\stackrel{\rightharpoonup }{𝒗}}_1$ and ${\stackrel{\rightharpoonup }{𝒗}}_2$. We can find the center of mass G of the pair of particles as

with $m_{tot}=m_1+m_2$.

Either by system linear momentum balance or by adding up $\stackrel{\rightharpoonup }{𝑭}=m\stackrel{\rightharpoonup }{𝒂}$ for each of the particles it is easy to see that

Thus we could put the origin of a good Newtonian reference frame at the center of mass ${\stackrel{\rightharpoonup }{𝒓}}_G$. The positions, velocities and accelerations relative to G, indicated with a prime (’), are

where we can skip use of the prime for the acceleration. Now some facts.

• •

  $m_1{\stackrel{\rightharpoonup }{𝒓}}_1^{\prime }+m_2{\stackrel{\rightharpoonup }{𝒓}}_2^{\prime }=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$ so ${\stackrel{\rightharpoonup }{𝒓}}_1^{\prime }=-(m_2/m_1){\stackrel{\rightharpoonup }{𝒓}}_2^{\prime }$. For all time the two positions (relative to the center of mass) are in the opposite direction and proportional. Similarly ${\stackrel{\rightharpoonup }{𝒗}}_1^{\prime }=-(m_2/m_1){\stackrel{\rightharpoonup }{𝒗}}_2^{\prime }$ and ${\stackrel{\rightharpoonup }{𝒂}}_1^{\prime }=-(m_2/m_1){\stackrel{\rightharpoonup }{𝒂}}_2^{\prime }$.

• •

  At a given instant there is a single plane defined by ${\stackrel{\rightharpoonup }{𝒓}}_1^{\prime },{\stackrel{\rightharpoonup }{𝒓}}_2^{\prime },{\stackrel{\rightharpoonup }{𝒗}}_1^{\prime },{\stackrel{\rightharpoonup }{𝒗}}_2^{\prime },{\stackrel{\rightharpoonup }{𝒂}}_1^{\prime }$ and ${\stackrel{\rightharpoonup }{𝒂}}_2^{\prime }$ because the positions and accelerations are all parallel (or antiparallel) and the two velocities are (anti) parallel.

• •

  The plane above is constant in time. This is because neither the velocity nor the acceleration has a component orthogonal to the plane, thus there is no tendency to leave the plane.

• •

  Each particle moves as if it was a single particle attracted to a central force at G. Why? Let’s look at the force on mass 1

  $${\stackrel{\rightharpoonup }{𝑭}}_{12}=F(r_{12})\frac{{\stackrel{\rightharpoonup }{𝒓}}_{12}}{r_{12}}=-F(r_{12})\frac{{\stackrel{\rightharpoonup }{𝒓}}_1^{\prime }}{r_1^{\prime }}$$

  because the relative position of the masses passes through the origin G.

• •

  In the special case of inverse-square gravitational attraction

  $$F=\frac{G{m}_1{m}_2}{r_{12}^2}=\frac{G{m}_1{m}_2}{{(r_1^{\prime }+r_2^{\prime })}^2}=\frac{G{m}_{1}M}{r_1^{\prime }{}_{}{}^2}$$

  where $M=m_2/(1+2(m_1/m_2)+{(m_1/m_2)}^2)$ is a fictitious mass at G we find using the substitution $r_2^{\prime }=(m_1/m_2)r_1^{\prime }$.

What we have found here is somewhat remarkable. Two particles are flying around in space attracted to each other by inverse-square gravitational attraction. Instead of doing something wild, they each move, relative to their joint center of mass, as if they were in central force motion with a fixed mass. That is, the 3D two-body problem reduces, exactly, to the 2D one body problem. You just have to use a coordinate system that is on the plane of motion and whose origin is at the center of mass.

Thus, the moon doesn’t really go around the earth. Rather the moon and earth go around their common center of mass (a point about 3/4 of the way out towards the earth’s surface from its center). And Jupiter doesn’t go around the sun, the sun and Jupiter go around their combined center of mass just outside the sun. But both of these examples are, in detail, wrong. Because the earth-moon system is affected by the sun and jupiter. And the Jupiter-sun system is affected by the earth and moon.

With inverse-square attraction, one body goes around a fixed point on one or another conic section. Two bodies go around each other in exactly the same way as one body about a fixed point. The two-body problem reduced to the one-body problem. What about lots of bodies? Let’s start with three. How, in general, do three bodies move that are all mutually attracted with inverse-square gravitation? Great question. Lots of people have asked it. And no-one knows the answer. Given any three masses and their initial conditions we could use a computer program to find out their subsequent positions and velocities. But no-one knows how to categorize all the possible motions of such systems.

Some things are known about ‘the three body problem.’ One is that it is hard, the best minds haven’t been able to solve it in general. Another is that the solutions can be pretty wild. For example, three particles might tumble around each other for a long time and, with no change in the equations, all of a sudden one of the particles will be ejected at high speed and never return (as if on a hyperbolic trajectory relative to the other two particles). A few special solutions of the three-body problem are known. For example, with the right initial conditions, three identical particles can move in either a circle or in Montgomery’s figure 8.

Despite the difficulty of analytic description, there is no special impediment to finding solutions to any 3-body problem with computer simulation.

With many particles all manner of complicated motion is possible. And there are few solutions which are known analytically. One solution has the $n$ particles chasing each-other around in a circle, with the particles forming a regular polygon. Another amazing approximate solution, the Buck solution, is that a string of thousands of particles will all chase each-other around an arbitrary curve in 3-dimensional space. At least approximately, for a while.

By applying $\stackrel{\rightharpoonup }{𝑭}=m\stackrel{\rightharpoonup }{𝒂}$ to 3 or 1000 interacting particles you can see all manner of $n$-body solutions on your computer.

Location of the center-of-mass. A structure is made up of three point masses, $m_1=1\mathrm{kg},m_2=2\mathrm{kg}$ and $m_3=3\mathrm{kg}$. At the instant of interest, the coordinates of the three masses are (1.25$\mathrm{m}$, 3$\mathrm{m}$), (2$\mathrm{m}$, 2$\mathrm{m}$), and (0.75$\mathrm{m}$, 0.5$\mathrm{m}$), respectively. At the same instant, the velocities of the three masses are $2\mathrm{m}/\mathrm{s}\hat{\mathit{ı}},\mathrm{\hspace{0.33em}2}\mathrm{m}/\mathrm{s}(\hat{\mathit{ı}}-1.5\hat{\mathit{ȷ}})$ and $1\mathrm{m}/\mathrm{s}\hat{\mathit{ȷ}}$, respectively.

1. 1.

   Find the coordinates of the center-of-mass of the structure.

2. 2.

   Find the velocity of the center-of-mass.

A spring-mass system in space. A spring-mass system consists of two masses, $m_1=10\mathrm{kg}$ and $m_2=1\mathrm{kg}$, and a weak spring with stiffness $k=1\mathrm{N}/\mathrm{m}$. The spring has zero relaxed length. The system is in 3-D space where there is no gravity. At the instant of observation, i.e., at $t=0$, ${\stackrel{\rightharpoonup }{𝒓}}_1=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$, ${\stackrel{\rightharpoonup }{𝒓}}_2=1\mathrm{m}(\hat{\mathit{ı}}+\hat{\mathit{ȷ}}+\widehat{𝒌}),{\dot{\stackrel{\rightharpoonup }{𝒓}}}_1=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$, and ${\dot{\stackrel{\rightharpoonup }{𝒓}}}_2=\sqrt{6}\mathrm{m}/\mathrm{s}(-\hat{\mathit{ı}}+\hat{\mathit{ȷ}})$. Track the motion of the system for the next 20 seconds. In particular,

1. 1.

   Plot the trajectory of the two masses in space.

2. 2.

   Plot the trajectory of the center-of-mass of the system.

3. 3.

   Plot the trajectory of the two masses as seen by an observer sitting at the center-of-mass.

4. 4.

   Compute and plot the total energy of the system and show that it remains constant during the entire motion.

From the linear momentum balance of the two masses, we can write the equations of motion as follows.

Let ${\stackrel{\rightharpoonup }{𝒓}}_1=x_1\hat{\mathit{ı}}+y_1\hat{\mathit{ȷ}}+z_1\widehat{𝒌}$ and ${\stackrel{\rightharpoonup }{𝒓}}_2=x_2\hat{\mathit{ı}}+y_2\hat{\mathit{ȷ}}+z_2\widehat{𝒌}$. Substituting above and dotting the two equations with $\hat{\mathit{ı}}$, $\hat{\mathit{ȷ}}$, and $\widehat{𝒌}$, we get

Thus we get six second order coupled linear ODEs as equations of motion.

1. 1.

   To plot the trajectory of the two masses, we need to solve for ${\stackrel{\rightharpoonup }{𝒓}}_1(t)$ and ${\stackrel{\rightharpoonup }{𝒓}}_2(t)$, i.e., for $x_1(t),y_1(t),z_1(t)$, and $x_2(t),y_2(t),z_2(t)$. We can do this by first writing the six second order equations as a set of 12 first order equations and then solving them using a numerical ODE solver. Here is a pseudocode to accomplish this task.

   Filename:tfigure8-ang-accel-ex

       ODEs = {x1dot = u1,
               u1dot = k/m1*(x2-x1),
               y1dot = v1,
               v1dot = k/m1*(y2-y1),
               z1dot = w1,
               w1dot = k/m1*(z2-z1),
               x2dot = u2,
               u2dot = -k/m2*(x2-x1),
               y2dot = v2,
               v2dot = -k/m2*(y2-y1),
               z2dot = w2,
               w2dot = -k/m2*(z2-z1) }
       IC   = {x1(0)=0, y1(0)=0, z1(0)=0,
               u1(0)=0, v1(0)=0, w1(0)=0,
               x2(0)=1, y2(0)=1, z2(0)=1,
               u2(0)=-sqrt(6), v2(0)=sqrt(6), w2(0)=0}
       Set  k=1, m1=10, m2=1
       Solve ODEs with IC for t=0 to t=20
       Plot {x1,y1,z1} and {x2,y2,z2}
   

   The 3-D plot showing the trajectory of the two masses obtained from the numerical solution is shown in fig. 13.5. From the plot, it seems like the smaller mass goes around the bigger mass as the bigger mass moves on its trajectory.

   Filename:tfigure8-ang-accel1

2. 2.

   We can find the trajectory of the center-of-mass using the following relationships.

   $$x_{\mathrm{cm}}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2},y_{\mathrm{cm}}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2},z_{\mathrm{cm}}=\frac{m_1{z}_1+m_2{z}_2}{m_1+m_2}.$$

   Since there is no external force on the system if we consider the two masses and the spring together, the center-of-mass of the system has zero acceleration. Therefore, we expect the center-of-mass to move on a straight path with constant velocity. The center-of-mass coordinates $x_{\mathrm{cm}}$, $y_{\mathrm{cm}}$, and $z_{\mathrm{cm}}$ are plotted against time in fig. 13.6 which show that the center-of-mass moves on a straight line in a plane parallel to the $xy$-plane ($z$ is constant). This is expected since the initial velocity of the center of mass has no $z$-component:

   Filename:sfig8-2-3

   	${\stackrel{\rightharpoonup }{𝒗}}_{\mathrm{cm}}$	$=$	${\displaystyle \frac{m_1{\stackrel{\rightharpoonup }{𝒗}}_1+m_2{\stackrel{\rightharpoonup }{𝒗}}_2}{m_1+m_2}}$
   		$=$	${\displaystyle \frac{m_1\cdot \stackrel{\rightharpoonup }{\mathrm{𝟎}}+1\mathrm{kg}\cdot \sqrt{(}6)\mathrm{m}/\mathrm{s}(-\hat{\mathit{ı}}+\hat{\mathit{ȷ}})}{10\mathrm{kg}+1\mathrm{kg}}}$
   		$=$	$0.22\mathrm{m}/\mathrm{s}(-\hat{\mathit{ı}}+\hat{\mathit{ȷ}}).$

3. 3.

   The trajectory of the two masses with respect to the center-of-mass can be easily obtained by the following relationships.

   	$x_{1/cm}=x_1-x_{\mathrm{cm}},$	$y_{1/cm}=y_1-y_{\mathrm{cm}},$	$z_{1/cm}=z_1-z_{\mathrm{cm}}$
   	$x_{2/cm}=x_2-x_{\mathrm{cm}},$	$y_{2/cm}=y_2-y_{\mathrm{cm}},$	$z_{2/cm}=z_2-z_{\mathrm{cm}}$

   The trajectories thus obtained are shown in fig. 13.7. It is clear that the two masses have closed orbits with respect to the center-of-mass. These closed orbits are actually conic sections as we would expect in a central force problem.

   Filename:sfig8-2-3a

4. 4.

   We can calculate the kinetic energy of the two masses and the potential energy of the spring at each instant during the motion and add them up to find the total energy.

   	${(E_k)}_{m_1}$	$=$	${\displaystyle \frac{1}{2}}{m}_1(u_1^2+v_1^2+w_1^2)$
   	${(E_k)}_{m_2}$	$=$	${\displaystyle \frac{1}{2}}{m}_2(u_2^2+v_2^2+w_2^2)$
   	$E_p$	$=$	${\displaystyle \frac{1}{2}}k[{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2]$
   	$E_{\mathrm{total}}$	$=$	${(E_k)}_{m_1}+{(E_k)}_{m_2}+E_p$

   The energies so calculated are plotted in fig. 13.8. It is clear from the plot that the total energy remains constant during the entire motion.

1. (a)

   Find an initial value for $v_0$ so that all the masses spiral in and then bounce out again. Plot the trajectories of all the masses on one plot for a long-enough time so the plot is pleasing to the eye.

2. (b)

   Find a value for $v_0$ so all the particles travel on circular trajectories.

3. (c)

   Can you find a formula for $v_0$ above in terms of the other parameters in the problem? Answer: $v_0=\sqrt{\frac{3m}{r}{\displaystyle \sum _{i=2}^{n-1}}\cot ((i-1)\frac{\pi }{n})}$

1. (a)

   Assume inverse square attraction and find a set of values for $m,G,R$ and $\omega$ so the assumed circular path is a solution of the equations of motion. Answer: $\omega =\sqrt{\frac{3M}{2R}}$

2. (b)

   A third mass $m$ is introduced which is gravitationally attracted to the other two. Pick initial conditions for the two big masses that are consistent with their circular motion solution. For the third mass use initial conditions ${[{\stackrel{\rightharpoonup }{𝒓}}_0]}_{xyz}=[00z_0]$. Run a simulation for some time.

   • •

     Does the third mass stay exactly on the $z$ axis for all time in the simulation? Would it if the simulation was exact? If so, why? If not, why not? Answer: The mass would stay on the $z$ axis if the solution was exact.

   • •

     Is the motion of the third mass exactly periodic in the computer simulation? Would it be if the solution was exact? If so, why? If not, why not? Answer: The solution would be exactly periodic if the ratio of the masses was infinite rather than just 1000. There are special initial conditions for which the motion is periodic for any mass ratio, the oscillations of the light mass need to be synchronous with the in-and-out oscillations of the heavier nearly-circular-motion masses.

and initial velocities

For each of the problems below show accurate computer plots and explain any curiosities. (Note, this solution was discovered independently, and nearly simultaneously, by Richard Montgomery and Chris Moore.)

1. (a)

   Use computer integration to find and plot the motions of the particles. Plot each with a different color. Run the program for 2.1 time units. Answer: These three trajectories are all parts of the same figure 8.

2. (b)

   Same as above, but run for 10 time units. Answer: The trajectories trace and retrace the same figure 8. If your integration is not accurate, the curves will not exactly retrace.

3. (c)

   Same as above, but change the initial conditions slightly. Answer: The trajectories make a beautiful swirl resembling a figure 8.

4. (d)

   Same as above, but change the initial conditions more and run for a much longer time. Answer: The trajectories get wild, possibly ejecting one or more masses off to infinity.

The conceptually simplest collision is a sticking collision also called a perfectly plastic no-slip collision (see fig. 13.12). Here the word ‘plastic’ is used in its old latin meaning malleable or ‘clay like’. Imagine two lumps of wet clay colliding in space and just sticking together. The plastic collision model applies, for example,

• •

  when a projectile gets embedded in its target,

• •

  when two cars crash and get entangled so they move together after the collision, or

• •

  when two machine parts engage at contact because of a mechanism like a door catch.

In short, the constitutive law for plastic collisions is

And the impulse is what it is, as determined by momentum balance for the two particles. Here’s the simplest collision problem.

The impulse on the particle is

And here is the general two-particle sticking collision problem.

There are three velocities to consider, the before-collision velocities ${\stackrel{\rightharpoonup }{𝒗}}_1^{-}$ and ${\stackrel{\rightharpoonup }{𝒗}}_2^{-}$ and the common after-collision velocity ${\stackrel{\rightharpoonup }{𝒗}}^{+}$. Also relevant is the interaction impulse $\stackrel{\rightharpoonup }{𝑷}$. That’s 4 vector quantities (8 scalars in 2D, 12 in 3D). The governing equations are momentum balance for the two particles

making up 2 vector equations (4 scalar equations in 2D, 6 in 3D). Thus to solve a problem in 2D, 4 scalar quantities need to be given so that the other quantities can be found from the momentum balance equations. In 3D, 6 scalar quantities have to be given.

There are all different ways to involute such problems, say by taking one of the masses as unknown. Here is the most straightforward example.

Given $m_1,m_2,{\stackrel{\rightharpoonup }{𝒗}}_1^{-}$ and ${\stackrel{\rightharpoonup }{𝒗}}_2^{-}$ we find by solving the momentum balance equations that

The answer can be interpreted like this. The final velocity is the same as the pre-collision average velocity. This is also the system’s initial (and final) center of mass velocity. The impulsive interaction is associated with the change of velocity of ${\stackrel{\rightharpoonup }{𝒗}}_1^{-}-{\stackrel{\rightharpoonup }{𝒗}}_2^{-}$ of an effective ‘reduced mass’ with a value of $m_{red}=m_1{m}_2/(m_1+m_2)$ (see box 13.2 on page 13.2).

This is the most common model used in elementary mechanics courses. It is originally due to Newton, at least in the 1D case we discussed in section 10.5. Two particles collide and then separate. There is no interaction force in their common contact tangent plane (hence ‘frictionless’). See fig. 13.13. The impulse is such that the particles separate at a speed that is a fixed ratio $e$ of the speed at which they approached. The speed of approach and separation are measured in the $\widehat{𝒏}$ direction.

The speed of approach is the rate at which the distance between the particles decreases just before the collision. Really, this only makes precise sense if

• •

  the masses are round, or

• •

  the masses are not rotating.

The approach speed is the relative velocity dotted with the $\widehat{𝒏}$ direction

The separation speed, measured just after the collision, has the same definition but with a sign change

Newton’s law of collisional restitution

is

We use the coefficient of restitution for approximate collisional modeling but,

Two particles $m_1$ and $m_2$ have pre-collision velocities of ${\stackrel{\rightharpoonup }{𝒗}}_1^{-}$ and ${\stackrel{\rightharpoonup }{𝒗}}_2^{-}$ and collide frictionlessly with coefficient of restitution $e$ on the tangent plane with normal $\widehat{𝒏}$. The post collision velocities ${\stackrel{\rightharpoonup }{𝒗}}_1^{+}$ and ${\stackrel{\rightharpoonup }{𝒗}}_2^{+}$, as well as the impulse $\stackrel{\rightharpoonup }{𝑷}$ are found by simultaneously solving these equations.

In 2D this makes up 5 scalar equations for 5 scalar unknowns. In 3D it’s 7 equations for 7 unknowns. The most direct solution is to set up and solve these equations using a computer.

Rather it is an approximate empirical observation. Or, to put it another way, the value of the coefficient of restitution $e$ depends on the material, the shape, the orientation and the speed of the colliding particles. It is not a true constant. Nonetheless, eqn. (13.4) is a reasonable approximation for some engineering purposes. Just don’t assume that predictions it makes will generally be highly accurate.

The ‘frictionless’ part of this collision law is expressed by the assumption that the net impulse of interaction is in the $\widehat{𝒏}$ direction. So $\stackrel{\rightharpoonup }{𝑷}=P\widehat{𝒏}$ with no component in the $\widehat{𝝀}$ direction.

Generally one assumes that the coefficient of restitution is between zero and one:

For the masses have to pass through each other. For $e>1$ the collision would involve a gain in energy. This might happen if there was explosive gunpowder in the contact region of the collision.

In the case $e=0$ the collision is perfectly plastic but still frictionless. This is generally not a sticking collision because the masses can enter and hence leave the collision with some relative velocity in the common contact plane (the $\widehat{𝝀}$ direction).

If one particle explodes into pieces it’s as if the pieces had a collision. It’s just that the initial velocities of the pieces were all the same and the total kinetic energy of the system increases during the ‘collision’. See fig. 13.14.

The overall treatment is extremely similar to that for sticking collisions, but in some sense backwards. Instead of the particles entering the collision with different velocities and leaving with the same velocity, they enter with the same velocity and leave with different velocities. But the same momentum principles apply. There is no collision law or coefficient of restitution to apply, all of the post-collision relative velocity is restituted from nothing. Rather one just has to know (or find) the action-reaction impulse between the masses.

Two particles $m_1$ and $m_2$ are stuck together and moving at ${\stackrel{\rightharpoonup }{𝒗}}^{-}$ when they explode and an impulse $\stackrel{\rightharpoonup }{𝑷}$ separates them. After the collision

The full range of behavior for sticking collisions to explosions can be captured with a single restitution coefficient $e_g$ (see box 13.2).

Our avoiding of frictional collisions is not because there generally is no friction during collisions. Friction is a fact of the mechanical world. We avoid friction here because a host of special assumptions are needed to make frictional problems deterministic. And no given set of assumptions is known to yield accurate predictions. Frictional collision models have too dis-satisfyingly low a ratio of accuracy to complexity for inclusion in a book at this level.

If one particle is involved in two collisions at one time then we have not explained how to calculate the resulting motion. In an attempt to make the situation clear one is tempted to say “Let’s make it ideal and assume the collisions are exactly instantaneous and at exactly the same time.” Then, unfortunately, one is making the situation exactly ambiguous.

Unfortunately for our hope of making reliable predictions, simultaneous collisions are not rare events. Why? Imagine B is touching C and both are stationary. Then A comes and bangs into B. Because B and C are already touching one must assume that there are impulsive forces not just between A and B, but also between B and C. And we have no reliable rules for sorting out the result. Nor will we find such rules if we make it a life’s work.

Imagine 15 accurately-machined nominally-identical spheres laid out in a tight triangle (5 in one row, 4 in the next, then 3,2, and 1) on a very flat smooth surface. Then imagine a 16th ball rolls in and hits the apex of the triangle. How do the 15 balls move?

This experiment is performed in smoky rooms full of intoxicated people night after night. It’s the ‘break’ in a pool game. And the game depends on the result being unpredictable. Each time, due to tiny differences, the results are different.

And, according to theory, the more rigid and perfect the balls are, the more sensitive are the results to the smallest of differences in the initial conditions.

What is the source of the problem?

Consider the one-dimensional collision of three identical particles. B and C are in line, stationary and touching and then A comes along with $v_{A0}=v^{-}$. Let’s assume that the collision(s) whatever they are, are completely elastic and conserve energy ($e=1,e_g=0$). Here are two ways to predict the outcome:

• •

  A hits B and C, being all the way at the other side of B, is oblivious to the interaction between A and B until it is complete. Thus A comes to rest and B is moving to the right with $v^{-}$. Then B collides elastically with C and B comes to rest and C shoots off with the $v^{-}$.

• •

  B and C are touching and act as a single rigid object throughout the collision with A. Thus the result is like that between a particle with mass $m$ and another with mass $2m$. Such an elastic collision would leave B and C going forwards at $2v^{-}/3$ and A going the other way with $v_A^{+}=-v^{-}/3$. A different result.

• •

  actually there is a one-parameter family of results that are consistent with energy conservation and momentum balance. We have three outcomes (the velocities of the three particles) and only 2 scalar equations restricting them (momentum and energy balance).

But what would really happen? That would depend on details that are not stated. Of course if the exact shape and configuration of the balls was known, and the exact rules for elastic and inelastic deformation, then one could calculate the resulting motion solving partial differential equations or with atomic simulations. In principle. But we generally do not know such details nor have such calculation abilities. And crowding that which we don’t know into concepts like ‘rigid-object’ and ‘exactly simultaneous’ crowds the prediction of the outcome to dependence on infinitesimal things.

So, as an engineer, what are you supposed to do when calculating in situations involving simultaneous collisions?

• •

  first relax and remember that no collision calculation is likely to be very accurate (unless the result only depends on balance of momentum). So simultaneous collisions, while philosophically worse in that even the equations are indeterminate, are not that much worse than the usual deterministic, but not accurate, collisional relations.

• •

  do experiments, and

• •

  take account of the range of outcomes depending on assumptions about the collision details.

Samples 13.18 and 13.22 starting on page 13.18 illustrate the ambiguity of simultaneous collisions in 2D.

This is the second of three sections about collisions. Section 10.5 was about collisions in 1D, then this section about particles in 2D and 3D, and finally the ideas in this section will be extended from particles to rigid objects in section 17.5.