Collisions of Point Masses in Two Dimensions Quiz

PDF page 375: In the previous two sections, we considered only one-dimensional collisions; during such collisions, the incoming and outgoing velocities are all along the same line. But what about collisions, such as those between billiard balls, in which objects scatter to the side? These are two-dimensional collisions, and we shall see that their study is an extension of the one-dimensional analysis already presented. The approach taken (similar to the approach in discussing two-dimensional kinematics and dynamics) is to choose a convenient coordinate system and resolve the motion into components along perpendicular axes. Resolving the motion yields a pair of one-dimensional problems to be solved simultaneously. One complication arising in two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook arms as they pass by one another, they will spin in circles. We will not consider such rotation until later, and so for now we arrange things so that no rotation is possible. To avoid rotation, we consider only the scattering of point masses—that is, structureless particles that cannot rotate or spin. We start by assuming that , so that momentum is conserved. The simplest collision is one in which one of the particles is initially at rest. (See Figure 8.10.) The best choice for a coordinate system is one with an axis parallel to the velocity of the incoming particle, as shown in Figure 8.10. Because momentum is conserved, the components of momentum along the - and -axes will also be conserved, but with the chosen coordinate system, is initially zero and is the momentum of the incoming particle. Both facts simplify the analysis. (Even with the simplifying assumptions of point masses, one particle initially at rest, and a convenient coordinate system, we still gain new insights into nature from the analysis of two-dimensional collisions.)

PDF page 375: In the previous two sections, we considered only one-dimensional collisions; during such collisions, the incoming and outgoing velocities are all along the same line. But what about collisions, such as those between billiard balls, in which objects scatter to the side? These are two-dimensional collisions, and we shall see that their study is an extension of the one-dimensional analysis already presented. The approach taken (similar to the approach in discussing two-dimensional kinematics and dynamics) is to choose a convenient coordinate system and resolve the motion into components along perpendicular axes. Resolving the motion yields a pair of one-dimensional problems to be solved simultaneously. One complication arising in two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook arms as they pass by one another, they will spin in circles. We will not consider such rotation until later, and so for now we arrange things so that no rotation is possible. To avoid rotation, we consider only the scattering of point masses—that is, structureless particles that cannot rotate or spin. We start by assuming that , so that momentum is conserved. The simplest collision is one in which one of the particles is initially at rest. (See Figure 8.10.) The best choice for a coordinate system is one with an axis parallel to the velocity of the incoming particle, as shown in Figure 8.10. Because momentum is conserved, the components of momentum along the - and -axes will also be conserved, but with the chosen coordinate system, is initially zero and is the momentum of the incoming particle. Both facts simplify the analysis. (Even with the simplifying assumptions of point masses, one particle initially at rest, and a convenient coordinate system, we still gain new insights into nature from the analysis of two-dimensional collisions.)

PDF page 375: In the previous two sections, we considered only one-dimensional collisions; during such collisions, the incoming and outgoing velocities are all along the same line. But what about collisions, such as those between billiard balls, in which objects scatter to the side? These are two-dimensional collisions, and we shall see that their study is an extension of the one-dimensional analysis already presented. The approach taken (similar to the approach in discussing two-dimensional kinematics and dynamics) is to choose a convenient coordinate system and resolve the motion into components along perpendicular axes. Resolving the motion yields a pair of one-dimensional problems to be solved simultaneously. One complication arising in two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook arms as they pass by one another, they will spin in circles. We will not consider such rotation until later, and so for now we arrange things so that no rotation is possible. To avoid rotation, we consider only the scattering of point masses—that is, structureless particles that cannot rotate or spin. We start by assuming that , so that momentum is conserved. The simplest collision is one in which one of the particles is initially at rest. (See Figure 8.10.) The best choice for a coordinate system is one with an axis parallel to the velocity of the incoming particle, as shown in Figure 8.10. Because momentum is conserved, the components of momentum along the - and -axes will also be conserved, but with the chosen coordinate system, is initially zero and is the momentum of the incoming particle. Both facts simplify the analysis. (Even with the simplifying assumptions of point masses, one particle initially at rest, and a convenient coordinate system, we still gain new insights into nature from the analysis of two-dimensional collisions.)

PDF page 375: In the previous two sections, we considered only one-dimensional collisions; during such collisions, the incoming and outgoing velocities are all along the same line. But what about collisions, such as those between billiard balls, in which objects scatter to the side? These are two-dimensional collisions, and we shall see that their study is an extension of the one-dimensional analysis already presented. The approach taken (similar to the approach in discussing two-dimensional kinematics and dynamics) is to choose a convenient coordinate system and resolve the motion into components along perpendicular axes. Resolving the motion yields a pair of one-dimensional problems to be solved simultaneously. One complication arising in two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook arms as they pass by one another, they will spin in circles. We will not consider such rotation until later, and so for now we arrange things so that no rotation is possible. To avoid rotation, we consider only the scattering of point masses—that is, structureless particles that cannot rotate or spin. We start by assuming that , so that momentum is conserved. The simplest collision is one in which one of the particles is initially at rest. (See Figure 8.10.) The best choice for a coordinate system is one with an axis parallel to the velocity of the incoming particle, as shown in Figure 8.10. Because momentum is conserved, the components of momentum along the - and -axes will also be conserved, but with the chosen coordinate system, is initially zero and is the momentum of the incoming particle. Both facts simplify the analysis. (Even with the simplifying assumptions of point masses, one particle initially at rest, and a convenient coordinate system, we still gain new insights into nature from the analysis of two-dimensional collisions.)

PDF page 377: The equations of conservation of momentum along the -axis and -axis are very useful in analyzing twodimensional collisions of particles, where one is originally stationary (a common laboratory situation). But two equations can only be used to find two unknowns, and so other data may be necessary when collision experiments are used to explore nature at the subatomic level.

PDF page 377: find two unknowns because we have two independent equations: the equations describing the conservation of momentum in the - and -directions. Solution Solving for and for and , taking the ratio yields an equation (in which θ2 is the only unknown quantity. Applying the identity we obtain:

PDF page 377: Suppose the following experiment is performed. A 0.250-kg object is slid on a frictionless surface into a dark room, where it strikes an initially stationary object with mass of 0.400 kg . The 0.250-kg object emerges from the room at an angle of with its incoming direction. The speed of the 0.250-kg object is originally 2.00 m/s and is 1.50 m/s after the collision. Calculate the magnitude and direction of the velocity and of the 0.400-kg object after the collision. Strategy Momentum is conserved because the surface is frictionless. The coordinate system shown in Figure 8.11 is one in which is originally at rest and the initial velocity is parallel to the -axis, so that conservation of momentum along the - and -axes is applicable.

PDF page 377: Suppose the following experiment is performed. A 0.250-kg object is slid on a frictionless surface into a dark room, where it strikes an initially stationary object with mass of 0.400 kg . The 0.250-kg object emerges from the room at an angle of with its incoming direction. The speed of the 0.250-kg object is originally 2.00 m/s and is 1.50 m/s after the collision. Calculate the magnitude and direction of the velocity and of the 0.400-kg object after the collision. Strategy Momentum is conserved because the surface is frictionless. The coordinate system shown in Figure 8.11 is one in which is originally at rest and the initial velocity is parallel to the -axis, so that conservation of momentum along the - and -axes is applicable.

PDF page 377: Suppose the following experiment is performed. A 0.250-kg object is slid on a frictionless surface into a dark room, where it strikes an initially stationary object with mass of 0.400 kg . The 0.250-kg object emerges from the room at an angle of with its incoming direction. The speed of the 0.250-kg object is originally 2.00 m/s and is 1.50 m/s after the collision. Calculate the magnitude and direction of the velocity and of the 0.400-kg object after the collision. Strategy Momentum is conserved because the surface is frictionless. The coordinate system shown in Figure 8.11 is one in which is originally at rest and the initial velocity is parallel to the -axis, so that conservation of momentum along the - and -axes is applicable.

PDF page 377: Everything is known in these equations except and , which are precisely the quantities we wish to find. We can

PDF page 378: Thus, 8.73 Discussion It is instructive to calculate the internal kinetic energy of this two-object system before and after the collision. (This calculation is left as an end-of-chapter problem.) If you do this calculation, you will find that the internal kinetic energy is less after the collision, and so the collision is inelastic. This type of result makes a physicist want to explore the system further.

PDF page 378: Some interesting situations arise when the two colliding objects have equal mass and the collision is elastic. This situation is nearly the case with colliding billiard balls, and precisely the case with some subatomic particle collisions. We can thus get a mental image of a collision of subatomic particles by thinking about billiards (or pool). (Refer to Figure 8.10 for masses and angles.) First, an elastic collision conserves internal kinetic energy. Again, let us assume object 2 is initially at rest. Then, the internal kinetic energy before and after the collision of two objects that have equal masses is

PDF page 378: Some interesting situations arise when the two colliding objects have equal mass and the collision is elastic. This situation is nearly the case with colliding billiard balls, and precisely the case with some subatomic particle collisions. We can thus get a mental image of a collision of subatomic particles by thinking about billiards (or pool). (Refer to Figure 8.10 for masses and angles.) First, an elastic collision conserves internal kinetic energy. Again, let us assume object 2 is initially at rest. Then, the internal kinetic energy before and after the collision of two objects that have equal masses is

PDF page 379: There are three ways that this term can be zero. They are • : head-on collision; incoming ball stops • : no collision; incoming ball continues unaffected • : angle of separation is after the collision All three of these ways are familiar occurrences in billiards and pool, although most of us try to avoid the second. If you play enough pool, you will notice that the angle between the balls is very close to after the collision, although it will vary from this value if a great deal of spin is placed on the ball. (Large spin carries in extra energy and a quantity called angular momentum, which must also be conserved.) The assumption that the scattering of billiard balls is elastic is reasonable based on the correctness of the three results it produces. This assumption also implies that, to a good approximation, momentum is conserved for the two-ball system in billiards and pool. The problems below explore these and other characteristics of two-dimensional collisions.