To play billiards spanish12/19/2023 ![]() ![]() Since the billiard ball only moves in the x-direction a Gy = 0, so the above equation becomes Where a Gy is the acceleration of the center of mass in the y-direction. This is a frictional force.į Py is the y-component of the force exerted on the ball by the billiard table, at point P.īy Newton's Second Law, the general force equation in the x-direction is:Ī Gx is the acceleration of the center of mass in the x-directionīy Newton's Second Law, the general force equation in the y-direction is: G is the acceleration due to gravity, which is 9.8 m/s 2 P is the point of contact of the ball with the billiard tableį Px is the x-component of the force exerted on the ball by the billiard table, at point P. In this analysis, we can represent the ball + cue system with a free-body diagram as shown below.į is the force the cue exerts on the ball when it strikes We wish to find the height h so that no (horizontal) frictional force develops at point P when the ball is struck by the cue. ![]() Knowing the location of this sweet spot can give you an idea of where to hit the ball so that it develops backspin or forward spin, which can be useful when making a shot.Ĭonsider the figure below showing the position of the cue at height h. The physics of billiards is similar to the Physics Of Hitting A Baseball, in that there is also a sweet spot on a billiard ball where you can strike with the cue stick so that no friction force develops between the ball and the billiard table. This essentially means that the velocity of ball A is completely transferred to ball B.įor a more detailed and complete analysis, in which the trajectory of ball A is calculated (after impact), under the influence of friction between the ball and billiard table, see the problem, Cue ball trajectory with table friction. This is because only a very small fraction of the momentum of ball A (and therefore velocity) is transferred to ball B, due to the obliqueness of the impact.įor the case where the impact is head on (θ = 90°) the above solution does not apply. There are two additional special cases to consider, involving ball collision.įor the case where the target ball B must be hit at an angle θ very close to zero (such as to sink it in the side pocket), ball A needs to be moving at a high speed V 1A (meaning you would have to hit ball A quite hard with the cue). Thus, after impact ball A moves in a direction perpendicular to the direction of ball B. Therefore, the vector equation for conservation of momentum can be drawn as shown below. Since the masses m A and m B are equal, this equation simplifies to:īy the Pythagorean theorem, this last equation tells us that the vectors V 1A, V 2A, V 2B form a right-angled triangle. Since the masses m A and m B are assumed equal, this equation simplifies to:įor an elastic collision kinetic energy is conserved, and the equation is: This interesting result can be proven as follows.įor the two colliding balls, the general vector equation for conservation of linear momentum is: Notice that, after impact, ball A moves in a direction perpendicular to the direction of ball B. Thus, ball B moves in the direction of this impulse. This is because the force (impulse) delivered by ball A to ball B acts normal to the surface of ball B, assuming there is no friction between the balls (a good assumption since billiard balls are smooth). Due to geometry, L 1 also makes an angle θ with the vertical, and the line passing through the center of the balls makes an angle θ with the horizontal.Īfter impact at CP, ball B moves in the direction of the line joining the center of the two balls, as shown. Due to geometry, L 1 is perpendicular to the line passing through the center of the two balls and the contact point CP. The line L 1 is drawn at a tangent to both balls at the point of contact. After impact, ball A moves at velocity V 2A in the direction shown, and ball B moves at velocity V 2B in the direction shown. It is assumed that balls A and B have the same mass and that ball B is initially at rest (zero velocity). For the general case, the collision is not head on, which is what the figure shows. The figure below shows a collision between two billiard balls. For a simplified case assuming no friction (discussed below), we can combine this fact with the elastic-collision assumption to find the trajectory of two colliding billiard balls after impact. Therefore, for simplicity one can assume that for collisions involving billiard balls, the collision is perfectly elastic.įor collisions between balls, momentum is always conserved (just like in any other collision). An elastic collision is one in which the kinetic energy of the system is conserved before and after impact. When two billiard balls collide the collision is nearly elastic. The physics behind billiards (or the physics behind pool), in large part, involves collisions between billiard balls. ![]()
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