npsm 새물리 New Physics : Sae Mulli

pISSN 0374-4914 eISSN 2289-0041


Research Paper

New Phys.: Sae Mulli 2024; 74: 621-633

Published online June 28, 2024

Copyright © New Physics: Sae Mulli.

Collision of Two Spinning Billiard Balls and the Role of Table

Hyeong-Chan Kim*

School of Liberal Arts and Sciences, Korea National University of Transportation, Chungju 380-702, Korea

Correspondence to:*

Received: February 29, 2024; Accepted: April 30, 2024

This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License( which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.

We study the collision dynamics of a spinning cue ball approaching a static object ball with equal mass on a plane, common in billiards. While typical collisions in billiards are nearly perfectly elastic, with a restitution coefficient close to 1 and low friction, we explore three deviations from ideal elastic collisions: The non-elastic nature, the friction effects between the balls during collision, the friction between the ball and the table. We describe the detailed collision outcomes, emphasizing the importance of considering frictions. We reveal that friction, both between the balls and with the table, significantly influences the post-collision motions, deviating from the expectations of a purely elastic collision. The insights gained contribute to a better understanding of ball dynamics, impacting strategies and gameplay in billiards.

Keywords: Billiards, Collisions, Friction

Billiard players claim that when a cue ball having topspin collides head on with an object ball, the cue ball retreats briefly before moving forward due to friction with the floor. Individuals familiar with the perfectly elastic collision of two objects with equal mass find it difficult to intuitively accept this claim. However, upon examining super-slow-motion videos, such as those on Dr. Dave's YouTube channel[1], it is confirmed that, immediately after the collision, the cue ball briefly lifts off the ground, moves slightly backward, and then advances again due to rotation. In this way, collisions between two billiard balls exhibit phenomena that deviate from intuitive expectations based on perfectly elastic collisions.

To understand these phenomena, a comprehend understanding of collision dynamics is necessary. Theoretical studies on billiards have seen limited development since the early 1830s when Coriolis authored a book on the mathematical theory of spin friction and collision in the game of billiards, translated into English by Nadler[2]. Nearly a century later, Moore attempted the second physics analysis[3]. Subsequent research has covered various aspects, including collisions between billiard balls[4, 5], collisions with cushions[6], high-speed camera analyses[7, 8], and model studies for developing robotic systems for billiards[9, 10]. When using a cue to strike a billiard ball, the ball moves in a direction almost identical to the cue's striking direction. However, due to friction, the ball deviates slightly from the cue's direction, known as the squirt phenomenon[11-16]. Although the angle is small, this deviation is a crucial factor that makes it challenging to hit a distant ball with the desired thickness in actual gameplay.

The motion of a billiard ball is described by the translational motion of the center and the rotational motion around the center. When billiard players use the cue to strike the ball, the ultimate goal is to control the translational and rotational motion appropriately. Additionally, after the cue ball collides with the first object ball, the aim is to guide both balls' trajectories according to the desired outcome. Precisely placing the cue ball at the intended position on the object ball requires extensive training. However, in practical games like 3-cushion or 4-cushion billiards, using the cue to send the cue ball to the object ball (assuming not aiming for a perfectly accurate thickness) is relatively straightforward, excluding the squirt/curve phenomenon. Therefore, during the game, the most critical element is understanding how the moving ball (cue ball) will behave after colliding with the stationary ball (object ball). This study aims to address this aspect.

Firstly, two approximations that closely hold during the collision of two billiard balls are presented to assist in calculations. First, the friction between the balls during collision is considered negligible. The friction coefficient between billiard balls, denoted as μ, varies depending on the state and type of the balls but generally exists in the range of


Assuming no friction, angular momentum is not transferred between the colliding balls. Secondly, billiard balls undergo a perfectly elastic collision. The coefficient of restitution during the collision is approximately


A collision with a restitution coefficient of 1 is considered a perfectly elastic collision. Due to the coefficient of restitution being very close to 1, the collision between the balls closely resembles a perfectly elastic collision. For example, colliding a non-rotating cue ball head-on with a stationary object ball results in the cue ball to stop, and the object ball to move almost at the same speed as the cue ball before the collision. Furthermore, colliding a non-rotating cue ball with a non-zero impact parameter with a stationary object ball shows that the separation angle between them immediately after the collision is very close to 90, as depicted in Fig. 1. Although the actual separation angle is slightly less than 90 due to the collision not being perfectly elastic, it is typically around 85–89 and does not significantly impact real games. It is clear that the collision between balls is never perfectly elastic as sound and heat are generated during the collision. However, in practical gameplay, starting with the approximation of frictionless perfectly elastic collisions and making corrections as needed is often sufficient.

Figure 1. (Color online) The perfectly elastic collision between two balls. The velocities of the two balls perpendicular to the collision plane are exchanged. In this illustration, the collision plane is represented by the vertical line. After the collision, the two balls move perpendicular to each other.

If gravity is disregarded, three factors are involved in the adjustment: friction between the colliding balls (proportional to the friction coefficient μ), deviation from a perfectly elastic collision (e*-1), and friction between the ball and the table. Recent studies by Peskin[17] considered the collision of rotating billiard balls, taking into account the friction between the balls. However, no solutions considering the effects of inelastic collisions and friction with the floor currently exist.

The billiard table serves two roles during the collision of balls. Firstly, it provides normal forces to the balls during the collision that prevent them from falling. Secondly, it imparts frictional force to the balls, proportional to the vertical normal force. Frequently, it was assumed that significant changes in the vertical normal force do not occur during the collision to ignore the effect of this frictional force. If this assumption holds, the collision between two balls can be much simplified. On the contrary, if this assumption is incorrect, friction with the floor will affect the post-collision motion of the balls. In reality, billiard balls roll on a smooth cloth laid over a hard surface. Therefore, the object providing vertical support force to the balls is the hard surface, and the cloth plays the role of providing friction with the balls.

The collision time between two colliding billiard balls is approximately 250μs<Δt<300μs. Let the cue ball's velocity before the collision be Ui and its mass be M. Then, the impact force of approximately MUi/Δt acts between the balls during the collision. This force, delivering an impact roughly equivalent to the magnitude of momentum during the collision time, is approximately 400 times greater than the force of gravity acting on the balls for Ui=1m/sec. Using the friction coefficient between the balls (1) (μ0.05), the magnitude of the frictional force (fμF) between the balls due to rotation is approximately 20 times greater than the gravity. Only when the cue ball collides with the object ball at a very low speed (5cm/s), the frictional force's magnitude becomes comparable to gravity. When friction acts in the direction of lifting the cue ball upward, the force pressing the cue ball against the table disappears. Consequently, the vertical normal force exerted by the table on the cue ball would become zero. The magnitude of the impulse imparted to the cue ball by this frictional force is approximately fΔt~μFΔt=μMUi, and the resulting vertical height gained by the ball, regardless of collision time, is approximately given by:


For a collision with an initial speed of Ui=10m/s and a friction coefficient of μ0.05, this yields h1.3cm, which is an observable value. However, for slower cue ball speeds, such as Ui1m/s, the height h becomes ~102cm, making it practically imperceptible. In a high-speed situation, if the assumption mentioned in the previous paragraph is correct, the object ball would move in the downward direction with the same speed. Consequently, the object ball would collide with and rebound off the surface of the table instantly. In typical situations, when the cue is used to strike the cue ball horizontally, causing the two balls to collide with equal heights, one will observe that one of the balls (either the cue ball or the object ball) tends to rise, while the other moves horizontally1, depending on the direction of rotation, which is contrary to the previous expectation.

These observations suggest that the assumption may not be appropriate that significant changes in the vertical normal force do not occur during the collision. On the other hand, if the frictional force lifts the cue ball, the reactional force tends to push the object ball downward. The magnitude of this frictional force is significantly greater than gravity, making it impossible to ignore. Then, the normal force exerted by the floor on the object ball cannot be neglected either. Additionally, the frictional force between the object ball and the table would also large, particularly due to the static friction present when the object ball is at rest before the collision. The static friction between the table and the ball, which brings the ball to a stop, is determined by the material of the table surface. Since the collision occurs in a very short time, the fiber material cannot respond during the collision. Therefore, the table and the ball experience the maximum static friction during the collision time. This static friction is significantly larger than other frictions making it a crucial factor that causes significant deviation from the results expected in a perfectly elastic collision.

The equations of motion for the collision between two balls with equal masses will be formulated in Section II. Section III will demonstrate how these equations are utilized during the collision process. Section IV will analyze the collision results, and finally, Section V will summarize the research findings in this paper.

Let the mass of the cue ball and the object ball be denoted as M, and their radii as R. Assume both balls move on a flat table, which is parallel to the (x,y) plane. Consider colliding the cue ball, which is in motion with a center velocity Ui=(Uix,Uiy,0) and angular velocity ωi before the collision, with the static object ball. Assume that both balls are of the same size, the table is horizontal, and the balls do not rise before the collision; hence, the collision plane is perpendicular to the table, as shown in Fig. 2. Define the direction from the center of the cue ball through the collision point to the center of the object ball as the x-direction and the intersection of the collision plane (the (y,z)-plane in Fig. 2) and the table as the y-direction. Here, let the center velocity Ui of the cue ball before the collision be inclined by an angle φ with respect to the x-axis, as shown in Fig. 1. Finally, choose the z-axis as the vertical upward direction perpendicular to the table. This arrangement is illustrated in Fig. 2.

Figure 2. (Color online) Various forces acting during the collision of the cue ball and the object ball. The cue/object ball is placed on the left/right.

In the following, we denote the center velocities of the cue ball and the object ball as U and V, respectively, and their angular velocities as ω and Ω.

As illustrated in Fig. 2, the force F exerted by the cue ball on the object ball during the collision, consists of the sum of the force Fx=(F,0,0) in the x-direction perpendicular and the friction force f acting parallel to the collision plane. The friction force acts in the opposite direction to the relative velocity of the collision points and its magnitude is generally given by the product of the friction coefficient μ and the normal force F, with its direction determined by the angle θ with the horizontal direction (choose the positive y-direction so that -π/2θπ/2):

f=(0,fcosθ,fsinθ),  f=μF.

Thus, if there is topspin or draw on the cue ball (ωy0), the angle θ satisfies θ0.

Listing the forces acting on the cue ball, -F=(-F,0,0)-f is the reaction to the action of the cue ball on the object ball (due to the friction and the normal force). ft is the friction on the cue ball exerted by the table, and FN is the normal force supporting the cue ball on the table, respectively. Finally, Fg represents the gravitational force acting on the cue ball. As discussed earlier, the normal force FN supporting the cue ball on the table must counterbalance the sum of the vertical forces due to gravity and friction with the object ball. Therefore,

FNFNz^   FN=(Mg+fsinθ)Θ(θ+θc),

where θcarcsin(Mg/μF) and Θ(x) is the Heaviside step function. If the frictional force in the vertical direction is greater than or equal to the gravitational force (Mg+μFsinθ<0), the normal force becomes zero, and the cue ball moves upward due to the collision. The friction force ft on the table is given by the product of the normal force magnitude FN and the coefficient of friction between the ball and the table. If the cue ball was rolling before the collision, the friction coefficient is very small (μrolling10-2), and the friction force can be neglected during the rolling motion. Assuming a smooth table surface around the cue ball, the friction force does not significantly affect the ball's motion before the collision ends. If the collision occurs in a sliding state before the collision, the sliding friction coefficient μbt0.1 comes into play, and in this case, the frictional force is relatively large compared to the rolling case. The magnitude of the friction force with the table is given by


Depending on the situation, if the vertical force becomes very large, the friction with the table cannot be ignored. However, it still acts as a second-order correction since its magnitude is proportional to the product of the two friction coefficients. Therefore, in this paper, we assume that the friction between the cue ball and the table can be ignored compared to F.

Similarly, for the object ball, the forces include the force exerted by the cue ball F, the friction force from the table f't, and the normal force supporting the object ball F'N. Finally, the gravitational force on the object ball is represented by Fg. During the collision, the normal force on the object ball is given by

F'NF'Nz^   F'N=(Mg-fsinθ)Θ(θc-θ).

The object ball stands still before the collision and then starts moving. As discussed in the introduction, the object ball, being supported by the soft fibers of the table, experiences motion only influenced by these fibers during the short moment of collision. In this brief moment, the displacement of the object ball is extremely small, to the extent that it is difficult to consider it moving through other fiber materials. Therefore, during the collision, the frictional force between the object ball and the table is given by static friction. The maximum static friction coefficient2 is approximately


This value, being relatively larger than the coefficients of kinetic friction or rolling friction, cannot be ignored, and its influence is notable in the context of the interaction with the table. The relative velocity, V==-z^×z^×(V+Rz^×Ω), of the contact point of the object ball with the table will be given by the sum of the translational velocity and the rotational velocity of the object ball. The rotational motion of the object ball is developed by the friction between the two balls, which can be considered negligible compared to the translational motion. In addition, consulting the result of the perfectly elastic collision, the motion of the object ball mainly directed to the x-direction. The friction force acts in the opposite direction of relative motion. Therefore, for the friction force f't, it can be approximated as


where V^= is defined as V^=V=|V=|x^. The magnitude of the friction force between the objective ball and the table is given by



where F'N is the normal force supporting the object ball. If we neglect gravity, this frictional effect is only significant when there is topspin of the cueball.

As discussed in the introduction, if the cue ball is moving rapidly (Ui1m/sec)and collides with the objective ball at an angle θ0, the vertical component of the frictional force becomes significantly larger than the gravitational force. Therefore, in this case, it is permissible to approximate the normal force as

FNfsinθΘ(θ),   F'N-fsinθΘ(-θ).

The subsequent calculations will utilize this approximation.

Following the discussion above, the equations of motion for the changes in the center velocities of the cue ball and the objective ball can be written as:

MdUdt =-F+ft+Fg+FN F-x^-μcosθy^-μsinθΘ(-θ)z^,MdVdt =F+f't+Fg+F'N F[1+μsμsinθΘ(-θ)x^    +μcosθy^+μsinθΘ(θ)z^],

where F is the perpendicular component of the collision force. The rotational motion equations for the changes in the angular velocities ω and Ω of the cue ball and the objective ball are given by:

Idωdt =τ-f+τft=Rx^×(-f)-Rz^×ft RμF(sinθy^-cosθz^),IdΩdt =τf+τf't =(-Rx^)×f-Rz^×f't μRFsinθ1+μsΘ(-θ)y^-cosθz^,

where, I=νMR2 represents the rotational inertia of the ball with ν=2/5 for a uniformly dense sphere. In these equations, the first/second term represent the torques due to the collision between the balls and due to the friction between the balls and the table, respectively. As observed, neglecting the friction with the table (μs0), would imply equal changes in the angular velocities of both balls. However, as discussed earlier, the value of the static friction coefficient cannot be easily dismissed.

The frictional forces between the cue ball and the object ball act in the opposite direction of the relative velocity at the point of collision. To calculate the relative velocity at the point of collision for each ball, let's denote the displacement from the center of the cue ball to the collision point as Rc=Rx^. The velocity at the collision point of the cue ball is given by U+ω×Rc, and for the objective ball, it is V+Ω×Ro, where Ro=-Rc. Therefore, the relative velocity of the objective ball with respect to the cue ball at the two contact points is given by


Conversely, the relative velocity of the cue ball with respect to the objective ball at the contact point is -v. The initial values of the relative velocities are given by

vx(0) =v(0)=v0=-Uicosφ,vy(0) =-Uisinφ-Rωz(0),vz(0) =Rωy(0).

Here, the initial directional angle θ0 for the frictional force satisfies tanθ0=vz0/vy0, and

v(0)=v0 =[Uisinφ+Rωz(0)]2+ R2ωy2(0).

Because the friction acts along the opposite direction to the relative velocity, the angle satisfies

cosθ0 =-Uisinφ/R+ωz0(Uisinφ/R+ωz0)2+ωy02,sinθ0 =-ωy0(Uisinφ/R-ωz0)2+ωy02.

The time-dependent change in the relative velocity is then given by


where we use Eqs. (9) and (10).

Let's denote the normal (perpendicular to the collision plane) impulse exerted on the balls during the collision as


This value monotonically increases during the collision. We will use this impulse in place of time. Define the impulses in the directions perpendicular and parallel to the collision plane during the short time dt as

dp dp=Fdt=MdV,dp =Fdt=fdt=μFdt=MdV=μdp,

respectively. Now, the equations of motion (9) can be expressed by multiplying both sides by dt and using the defined expressions as follows:

MdU =dp-x^-μcosθy^-μsinθΘ(-θ)z^,MdV =dp[1+μsμsinθΘ(-θ)x^    +μcosθy^+μsinθΘ(θ)z^].

Regarding the change in angular velocity in (10), when expressed in terms of impulse, it becomes:

Idωy μRsinθdpIdΩy1+μsΘ(-θ),Idωz -μRdpcosθIdΩz.

Here, ≈ indicates the approximation neglecting gravity and friction between the cue-ball and the table.

The change in relative velocity with respective to the impulse can be obtained as follows:


Here, recalling that

vy-vcosθ,   vz=-vsinθ,

and utilizing vdθ=sinθdvy-cosθdvz, we find


As observed from this equation, the angle θ, which indicates the direction in which frictional force is acting, changes over time. Because ν(=2/5)>μs, dθ becomes negative when θ<0, causing the angle θ to decrease. The change in the angle becomes zero at θ=0,±π/2 and reaches its maximum at θ=±π/4, happening in the direction of increasing |θ|.

In both cases, there is no event during the collision that reverses the sign of the angle θ. Moreover, as evident from the equations of motion, the friction coefficient μs always comes with a step function Θ(-θ), and the sign of θ does not change during the motion. Therefore, for simplicity, we will denote μsΘ(-θ) as μs later in this work.

In this section, we aim to provide a detailed chronological description of the process that occurs at the moment when billiard balls collide. To illustrate the sequence of physical phenomena during the collision process, we focus on the impulse pp acting in the perpendicular direction to the collision plane. If there is no adhesive force between the balls on the collision plane, the impulse p monotonically increases over time during the collision process.

1. Sliding state

The moment the cue ball reaches the object ball, both balls enter a state of sliding against each other. At the moment of collision, the cue ball simultaneously pushes the object ball vertically with a speed of Uicosφ while sliding on the surface of the ball with a speed of Uisinφ. This sliding is gradually slowed down by friction, and if the friction coefficient is denoted as μ, the impulse due to frictional force is given by, from Eq. (16),


Here, we assume that the friction coefficient is independent of the sliding speed or vertical pressure, which is valid as long as the sliding speed on most surfaces is not too fast. In collisions between balls, the friction coefficient μ is not large, making it difficult to enter a stick state where sliding stops. During sliding, changes in the horizontal and vertical components of relative velocity are given by, from Eq. (17),

dvydp =-2μMν+1νvyv,dvzdp =-μMν+2+μsνvzv,dvdp =2+μμssinθM.

As mentioned earlier, μs represents μsΘ(-θ). The last equation indicates that for θ>0, v linearly varies with the vertical impulse. On the other hand, for θ<0, v no longer linearly varies with p.

Due to the step function, the equation (23) needs to be solved separately for each case where θ is negative or positive. Fortunately, the results for the case of θ>0 can be obtained by solving the case of θ<0 first and then taking the limit, μs0. Therefore, it is sufficient to solve the case of θ<0. In this case, vy<0, vz>0. Integrating the first two equations yields a conserved quantity during the collision process:

v¯ vz-2(1+ν)μs-ν|vy|2+ν+μsμs-ν=|vy|vyvz2(1+ν)μs-ν=vzvyvz2+ν+μsμs-ν=constant.

Thus, the magnitude of the component parallel to the collision plane of the relative velocity is given by v=|vy|1+xy1/ay =|vz|1+xz1/az. Here,

ay-1+νν-μs, az2+ν+μs2(ν-μs),


xyvyv¯=(tan2θ)ay,    xzvzv¯=(tan2θ)-az.

From the initial values (11), it follows that


Using this, the first two equations in Eq. (23) become

1+xk1/akdxk =-μβkmv¯dp,  k=y,z,


βy=ν+1ν,   βz=ν+2+μs2ν.

The differential equations (27) can be integrated using hypergeometric functions:

Vk=Vk0 -μβkpmv¯.


Vk Fak(xk) txk1+x1/akdx =xk2F1(-1/2,ak,1+ak,-xk1/ak),

and Vk0 (k=y,z) is determined by the relative velocity components vy(0), vz(0) at p=0. This function monotonically increases in the region, x>0. Finally, considering the motion perpendicular to the collision plane from the last equation of (23), we get

v-v(0) =pm+μμsMtpsinθdp =pm+μs[vz-vz(0)]2βz.

Here, mM/2 and Eq. (A1) in appendix is used.

The results for positive θ can be obtained by taking the limit as μs0. For example, in the case of relative velocity perpendicular to the collision plane, equation (31) simplifies to v=p/m. If v0, even with a small friction coefficient, a stick state can happen in a short time, but this paper does not consider such possibility.

2. Maximally compressed state

As the relative vertical velocity decreases, the two balls are compressed, and energy is accumulated. Let's denote the vertical impulse at the moment when this vertical compression stops, i.e., when v=0, as p=pc. At that moment, the relative vertical velocity is 0, and the following equations hold:

v(pc) =v0+pcm+μs[vz(pc)-vz(0)]2βz=0,Vk(pc) =Vk(0)-μβkpcmv¯.

By solving the first equation and the second equation for k=z, we can determine the value of pc. In the case of θ>0, taking the μs0 limit yields the result that the vertical impulse up to the moment of maximum compression is pc=-mv0=mUicosφ.

The total energy absorbed during the compression process can be obtained by integrating the force acting in the vertical direction over the displacement.

If the vertical component of the relative velocity varies linearly with the impulse during compression, the magnitude of the work done as the vertical impulse increases from 0 to pc can be expressed using equations (23), (31), and (30):

W(pc)=0pc v dp=v0pc+pc2 2m+μs 2βz [vz0pc+0pcdpvz]=v0pc+pc2 2mμs 2βz [vz0pc+Mv¯2 4μF2az (xz2)F2az (xz2(0))βz ].

In the case of θ>0, taking the limit as μs0 simplifies the result to the first two terms of equation (33).

3. Restoration process

At the moment when the vertical velocities of the cue ball and the object ball become equal, both balls stop compressing and enter the restoration process. During this period, as the shapes of the two balls return to their original state, the elastic energy accumulated during the compression process is utilized to push each other away. Let's denote the final value of the vertical impulse as pf, and explore the kinetic energy restored during this process. Since both balls are in a sliding state during the restoration process, the vertical and the horizontal impulses satisfy a similar form of equation as in (32). Integrating this equation from pc to pf determines the final state.

The final horizontal relative velocity is given by (29):

Vk(pf) =Vk(pc)-μβkmv¯(pf-pc) =Vk0-μβkmv¯pf.

Therefore, the current assumption that the friction is small enough not to reach a stick state still holds if μβkmpf<Vk0 for k=y,z. As it is still in a sliding state, the vertical relative velocity also satisfies equation (31). Thus,

vf =v(0) +pfm+μs2βz[vzf-vz(0)] =pf-pcm+μs2βz[vzf-vz(pc)] ,

where, in the second equality, the first equation of (32) is used. If θ is positive, v(pf)=pf-pcm can be expressed more succinctly.

The elastic energy restored during the restoration process can be calculated using the above equation:

W(pf)W(pc)0pc v dp=pcpf[ppc m+μs 2βz [vzvz(pc)]]dp=(pfpc)2 2mμs 2βz [vz(pc)(pfpc)+Mv¯2 4μβz (F2az (xzf2)F2aa (xzc2))].

From the above results, the ratio of absorbed elastic energy during the compression process to the restored energy can be related to the square of the (energetic) coefficient of restitution, given by:


Here, e* represents the energetic coefficient of restitution between the cue ball and the object ball[15].

From this equation, one can write the value of pf in terms of the initial value.

Generally, although this equation presents a complex relationship, in the case of θ>0, the right side of this equation is simplified, and e*2=pfpc-12 is given. Therefore, in this case,


In general, pf has a correction term proportional to μs in addition to the above equation:


Using the above results to find the post-collision velocity of the cue ball, with the help of (17) and (A1), we have:

Uf=Ui1M0pfdp=1e* 2Uicosϕμsδpf M,Uyf=UyiμM0pf cosθdp=Uisinϕ12βy [vy(pf)vy(0)],UzfUziμΘ(θ)M0pf sinθdp=12βz [vz(pf)vz(0)]Θ(θ),

where vk(pf) is obtained from equation (34), and vk(0) is obtained from equation (11). Also, Uzi=0. The post-collision velocity of the object ball is given by:

Vf =1M(1+μμssinθ)dp =1+e*2Uicosφ +μsδpfM +μs2βzvz(p)-vz(0),Vyf =μMcosθdp =12βy[vy(pf)-vy(0)] ,Vzf =μΘ(θ)Msinθdp =12βz[vz(pf)-vz(0)]Θ(θ).

In the y-direction, the law of conservation of momentum holds, but it does not hold in the x and z-directions.

This phenomenon occurs due to the counteracting effect of vertical forces, neutralizing the impact of frictional forces.

The angular velocities of the two balls after the collision become

ωf-ωi =0=Ωf,ωyf-ωyi =μRIdpsinθ=MRIvzf-vzi2βz =Ωyf1+μsΘ(-θ),ωzf-ωzi =-μRIdpcosθ =-MRIvyf-vyi2βy=Ωzf,

as derived from (18). Utilizing the above results to compare the energy before and after the collision, one may notice that the conservation of kinetic energy does not hold when there is friction and the coefficient of restitution is not 1.

In the previous section, we discussed the collision between two balls in a general form.

However, in the case of a collision between billiard balls, the magnitude of the coefficient of friction μ between the balls is not large, resulting in a small change in the relative velocity in the direction parallel to the collision plane.

In such cases, it is sufficient to use linear approximation by expanding the function Vk with k=y,z around the initial values.

Firstly, from the definition of the function Fa in Eq. (30), we obtain the following expansion:


We have described up to quadratic approximation, as it is necessary for expanding the coefficient of restitution formula (37) to the linear order in the friction coefficient μ. Conversely, we can also use


First, we calculate vzf-vzi, which is necessary to obtain the value needed for expanding the coefficient of restitution formula, up to quadratic order in μ.

Using Eqs. (29) and (43), we get

|vz||vz0|βzμp|sinθ0|m1μs ν2νcos2θ0μpmv0 +.

The calculation process uses Eq. (20) and Eq. (26) also.

If θ<0, vz>0, so removing the absolute value on the left-hand side is not a problem. On the right-hand side, the - sign can be absorbed into sinθ0. If θ>0, the entire left-hand side gets a - sign, allowing us to cancel the absolute value and the - sign on the right. Therefore, the following generally holds:

vz-vz0 =βzμpsinθ0m 1+ν-μs2νcos2θ0 μpmv0 +.

In future calculations, vyf-vyi only needs to be approximated up to the first order in μ. Using Eq. (29) and vy<0, we get:

vy-vy0 =-(|vy|-|vy0|) -v¯Vy-Vy01+xy01/ay+ =βyμpcosθ0m+.

Then, let's solve for pc using Eq. (32) by determining it around the initial value. From the first equation, we get:

v(pc)=v0+pcm+μμspsinθ02m×1μs ν2νcos2θ0μpmv0 +=0.

Therefore, the following holds:

v0=-pcm1 +μμssinθ02+.

Now, using the above results and the coefficient of restitution formula (37), let's determine pf. First, during the contraction, the elastic energy accumulated in the ball can be obtained from (33). Using (44) and (46), we get:

W(pc)=v0pc+pc22m+μsΘ(θ)2βz×[vz0pcM v¯24μβz[F2az(xz2)F2az(xz2(0))]]1+μμs sinθ0 2pc22m.

Here, vz0=-v0sinθ0 and if μs0, θ<0 is used.

The elastic energy recovered in the restoration process can be obtained from (36). Using a similar process as above, we get:

W(pf)W(pc)= (pf pc )22mμs2βz[vz(pc)(pfpc)M v¯24μβz(F2az(xzf2)F2aa(xzc2))] (pf pc )22m1+μμs 2sinθc.

Now, using (37), we have:

e*2=pfpc-12 2+μμssinθc 2+μμssinθ0+O(μ2).

Here, θc-θ0O(μ), so we obtain:


In other words, up to the linear-order approximation in μ, the vertical impulse is determined by the coefficient of restitution. Therefore, using (46) and (11), the final vertical impulse is given by:

pf =(1+e*)mUicosφ1-μμssinθ02+.

Comparing this equation with (38), the impulse correction is:


Now, we write the velocities of the cue ball and the object ball after the collision given by Eqs. (39) and (40):

UfUicosϕ[1e* 2+μμs (1+e* )sinθ0 4x^+tanϕμ(1+e* )cosθ0 2y^μ(1+e*)2sinθ0Θ(θ)z^], Vf1+e*2Uicosϕ[1+μμs sinθ0 2x^+μcosθ0y^+μsinθ0Θ(θ0)z^].

In the case where there is a draw, i.e., θ0>0, one can take the limit μs0 in the above results. As seen from these equations, for θ0<0, the friction between the object ball and the ground reduces the speed in the direction perpendicular to the collision plane by the same amount for both the cue ball and the object ball.

Now, let's examine the changes in angular velocity for the cue ball and the object ball. Firstly, there is no change in angular velocity component in the direction perpendicular to the collision plane. On the other hand, the angular velocity components along y and z-directions change according to Eq. (41):

ωf-ωi = μ(1+e*)2νUicosφRsinθ0y^-cosθ0z^,Ωf = μ(1+e*)2νUicosφR (1+μs)sinθ0y^-cosθ0z^

This means that the object ball obtains a larger angular velocity with a higher initial horizontal speed and a larger friction coefficient. Additionally, we observe that the friction between the object ball and the ground only contributes to the change in the angular velocity of the object ball when there is a draw.

In this work, we discuss the phenomenon that occurs when a cue ball, rotating on a plane, approaches and collides with a static object ball of equal mass. In the case of billiard balls, the coefficient of restitution is close to 1, and the friction coefficient takes a value close to 0, demonstrating a collision phenomenon close to a perfectly elastic collision. In this scenario, immediately after the collision, the trajectories of the cue ball and the object ball form a right angle with each other. The cue ball moves in the direction parallel to the collision plane, while the object ball moves in the direction perpendicular to the collision plane. This situation is illustrated in Fig. 3 using red arrows, where Ui represents the velocity of the cue ball before the collision, and U and U represent the velocities of the object ball and the cue ball, respectively, after a perfectly elastic collision on the table.

Figure 3. (Color online) The velocity changes of the cue ball and the object ball before and after the collision (ignoring the component of velocity in the direction perpendicular to the table).

In this paper, three effects which develop deviations from the results of a perfectly elastic collision in the collision of balls are discussed. These three effects are: 1) the collision between balls is not perfectly elastic, 2) the effect of friction between balls during collision, and 3) the effect of friction between the ball and the table at the moment of collision.

Before discussing the detailed results of all these effects, let's first consider the scenario where a cue ball with topspin collides head-on with an object ball, taking into account the presence of frictions, the topic we mentioned at the beginning of the introduction. Firstly, assume that the cue ball has only rolling motion before the collision. Therefore, θ0=-π/2. In a head-on collision, φ=0. To simplify the discussion, we assume a perfectly elastic collision, implying e*=1. As discussed earlier, the effects of friction between the object ball and the table are valid only when there is topspin in the cue ball. Therefore, we interpret the friction coefficient μs in Eqs. (52) and (53) as μsΘ(-θ0), where Θ is the step function. Then, the velocities of the cue ball and the object ball after the collision are: Uf μUi-μs2x^ +z^, Vf Ui1-μμs2x^. If there is no friction between the balls (μ=0), as expected, the cue ball comes to a stop, and the object ball moves with the same speed as the cue ball. However, when there is friction between the balls and between the object ball and the table, after the collision, the cue ball moves backward at a rate slower by the factor μs/2 compared to the speed at which it would jump into the air. Of course, this backward motion will soon turn into forward motion due to the friction when the cue ball collides with the table again under the influence of gravity. Similarly, the object ball also moves slower by the same amount compared to its speed without friction with the table. This result clearly demonstrates that the friction between the object ball and the table cannot be ignored in the collision of two balls.

Now, let's describe the detailed results of the collision. Firstly, the phenomena arising from the difference with a perfectly elastic collision, due to the absence of friction, are proportional to (1-e*). This manifests as the appearance of the vertical velocity Uf of the cue ball in Eq. (52) and the decrease of Vf of the object ball in Eq. (53). The phenomena solely caused by the friction between the cue ball and the object ball are proportional to μ. Both balls exhibit changes in the velocity components parallel to the collision plane. The changes in angular velocities, as seen in Eq. (54), confirm that the presence of friction between the balls is necessary for such changes to occur. In Fig. 3, the blue arrows illustrate the trajectories of the cue ball and the object ball immediately after the collision.

Ignoring the motion perpendicular to the billiard table and comparing the results with a perfectly elastic collision, the differences in the speed and the direction of the object ball are proportional to (1-e*) and μ, respectively. Let's examine the post-collision motion of the object ball (53) using the linear order approximation. The object ball's speed is given by:

Vf1+e*2 Uicosφ1+μμssinθ02.

If we interpret (1+e*)/2 as 1-(1-e*)/2, we can understand how it differs from a perfectly elastic collision. The velocity correction due to the friction coefficient is proportional to μμs and exists only when θ is negative (in the presence of the cue ball's topspin).

As seen in Eq. (39), without friction, the object ball moves along the x-direction, perpendicular to the collision plane. If there is friction, the object ball's velocity has components in both the y-direction and the vertical upward z-direction parallel to the collision plane. The magnitude of the upward velocity component is given by:

Vfz=μ(1+e*)2Uicosφsinθ0 Θ(θ0).

This vertical motion is linearly proportional to the friction coefficient and occurs only when the cue ball has an angular velocity about the y-axis, i.e., there is topspin. The directional difference is denoted by ψ in Fig. 3 is approximately given by:


This angle is maximized when there is no initial rotational angular velocity about the y-axis, and for a given friction coefficient according to Eq. (1), it varies by approximately 1.7 to 4.8. As seen, the primary cause for the object ball deviating from the direction perpendicular to the collision plane is the frictional force between the two balls during the collision.

When two balls of equal mass undergo a frictionless perfectly elastic collision, the cue ball moves in the y-direction after the collision. Neglecting the vertical upward motion, the motion of the cue ball, as seen from the first equation in (52), shows two changes. First, due to the effect of friction, the speed in the y-direction parallel to the collision plane decreases. Second, as the collision is not perfectly elastic, the cue ball gains a velocity component in the x-direction perpendicular to the collision plane. As a result, the cue ball deviates from the y-direction by an angle ψc as depicted in Fig. 3, given by:


If the cue ball engages in a perfectly elastic collision (e*=1) with the object ball and has topspin (i.e., θ<0), the cue ball's direction deviates from the perpendicular direction to the collision plane because of the frictions. On the contrary, if there is no topspin, the direction of the cue ball is towards the y-axis. If a perfectly elastic collision does not occur, the cue ball's direction is determined by the combined effects of (1-e*) and the topspin effect through the friction between the object ball and the table. Calculating the angular deviation using μ=0 and the range of restitution coefficient given by Eq. (2), at φ=45, for e*=0.92, there is an angular deviation of approximately 2.3, and for e*=0.98, an angular deviation of about 0.57. At φ=30, the deviations are approximately 4.0 and 1.0, respectively. For φ=10, an angular deviation of around 13 is observed for e*=0.92. These deviations are substantial and cannot be ignored. As the angle φ approaches 0 degrees (meaning closer to a head-on collision), the speed of the cue ball significantly decreases, and its direction is strongly influenced by the friction.

As time progresses after the collision, the angle formed by the velocities of the two balls changes due to the rotational motions. The post-collision trajectories of the balls are determined by the velocities/angular velocities resulting from the collision and the friction with the table. If there is angular velocity around the axis of the ball’s heading direction, the ball follows a curved path. In the case of the object ball, the rotation induced by friction is not significant, and bending is not visibly apparent. However, the cue ball, depending on the situation, can exhibit significant curvilinear motion. Further research is needed to investigate these trajectories in detail.

The results of this study contribute to the understanding of ball movements and collisions in billiards, providing insights into strategies and gameplay. Additionally, this research can offer valuable information in the field of physics education and experiments.

We attach an integration formula which is used frequently in the text:

t0pdpcosθ = -vyvdpdvydvy =M2μβyvy(p)-vy(0), 0pdpsinθ = -vzvdpdvzdvz =M2μβzvz(p)-vz(0).

In deriving this formula, we us Eq. (23).

This work was supported by the National Research Foundation of Korea grants funded by the Korea government RS-2023-00208047 and Korea National University of Transportation Industry-Academy Cooperation Foundation in 2023.

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