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Slide 1 / 140 Slide 2 / 140 Momentum www.njctl.org Slide 3 / 140 Table of Contents Click on the topic to go to that section Conservation of Linear Momentum · Impulse - Momentum Equation · · Collisions in One Dimension · Collisions in Two Dimensions · Center of Mass · It is Rocket Science Ballistic Pendulum ·

Slide 4 / 140 Conservation of Linear Momentum Return to Table of Contents Slide 5 / 140 Conservation of Momentum The most powerful concepts in science are called "conservation principles". Without worrying about the details of a process, conservation principles can be used to solve problems. If we were to take a snapshot of the initial and final system, comparing the two would provide a lot of information. The last unit presented the Conservation of Energy, which proved helpful in solving problems where the situation was too complex for Newton's Laws to be effectively used. Conservation of Energy is not enough. Slide 6 / 140 Conservation of Momentum If two ice skaters are holding hands, but then push away from each other, Conservation of Energy will not be able to determine each skater's velocity after the push. Assume a closed system with no external forces. We need something new. Let's start with Newton's Third Law. Assume the ice is frictionless. Then the force exerted by one skater on the second is equal and opposite to the force exerted by the second skater on the first and there are no external forces. Stating Newton's Third Law and doing a little bit of substitution

Slide 7 / 140 Conservation of Momentum Make the assumption that we're not dealing with relativistic conditions (the skaters are moving way slower than the speed of light), and the skaters don't lose any mass in the process. Then, the constant mass terms can be put within the derivative. This is good, as it looks like we can now say something about the skaters's velocities. The time rate of change of the sum of each skater's mass times velocity is zero - it doesn't change. The quantity is defined as linear momentum and will be represented as - it is a vector. Slide 8 / 140 Conservation of Momentum The initial linear momentum is then equal to the final linear momentum of the system. Thus, linear momentum is conserved in a closed system with no external forces. For two particles, , and since the initial and final momentum remain constant, we can write: Slide 9 / 140 Conservation of Momentum The Energy chapter of this course discussed how potential energy could be calculated for a system that had only conservative forces. The Conservation of Momentum does not put this restriction on forces - as long as the forces are internal, momentum is conserved. We will leave the Conservation of Momentum for now, and apply it to problems involving collisions between objects in an upcoming chapter. Many of the great discoveries in nuclear and particle physics involve smashing atoms and nuclei into each other and seeing what comes out. The Conservation of Momentum is key in these experiments.

Slide 10 / 140 Newton's Second Law restated Let's take a brief detour and examine Newton's Second Law as it is taught today, and use the definition of momentum. The sum of forces on a particle is equal to ma. Since we assume the mass is constant, we can bring it inside the derivative. This is how Newton presented his Second Law in Principia - the sum of forces on an object changes its momentum over time. The more general statement of the Second Law has the benefit that it can be extended to cases where the mass of the objects change - not just the velocity. More on this later. Slide 11 / 140 Momentum is a Vector Quantity A key difference between momentum and energy is that energy is a scalar, while momentum is a vector. When there is more than one object in a system, the total momentum of the system is found by the vector addition of the each object's momentum. Another key difference is that momentum comes in only one flavor (there is no kinetic, potential, elastic, etc.). The unit for momentum is kg-m/s. There is no special unit for momentum, which presents an excellent opportunity to honor another physicist. Slide 12 / 140 1 What is the momentum of a 20 kg object with a velocity of +5.0 m/s? A -100 kg-m/s B -50 kg-m/s C 0 kg-m/s D 50 kg-m/s E 100 kg-m/s

Slide 12 (Answer) / 140 1 What is the momentum of a 20 kg object with a velocity of +5.0 m/s? A -100 kg-m/s B -50 kg-m/s Answer C 0 kg-m/s E D 50 kg-m/s E 100 kg-m/s [This object is a pull tab] Slide 13 / 140 2 What is the momentum of a 20 kg object with a velocity of -5.0 m/s? A -100 kg-m/s B -50 kg-m/s C 0 kg-m/s D 50 kg-m/s E 100 kg-m/s Slide 13 (Answer) / 140 2 What is the momentum of a 20 kg object with a velocity of -5.0 m/s? A -100 kg-m/s B -50 kg-m/s Answer A C 0 kg-m/s D 50 kg-m/s E 100 kg-m/s [This object is a pull tab]

Slide 14 / 140 Momentum of a System of Objects If a system contains more than one object, its total momentum is the vector sum of each object's individual momentum: Slide 15 / 140 Momentum of a System of Objects In order to determine the total momentum of a system : Select a direction to be positive for each dimension that's · being considered. Assign positive values to each momentum in that direction. · Assign negative values to each momentum in the opposite · direction. Add the momenta together independently for each · dimension. Vectorially add each dimension's momentum together to get · the total momentum - the Pythagoras theorem is used to find the magnitude, and trigonometry is used to find its direction. Slide 16 / 140 Momentum of a System of Objects Let's work an example in one dimension. Determine the momentum of a system of two objects: m 1, has a mass of 15 kg and a velocity of 16 m/s towards the east and m 2 ,has a mass of 42 kg and a velocity of 6.0 m/s towards the west. Choose East as positive. or to the west

Slide 17 / 140 3 Determine the magnitude of the momentum of a system of two objects: m 1 , which has a mass of 6.0 kg and a velocity of 13 m/s north and m 2 , which has a mass of 14 kg and a velocity of 7.0 m/s south. Assume north is positive. A 10 kg m/s B 15 kg m/s C 20 kg m/s D -15 kg m/s E -20 kg m/s Slide 17 (Answer) / 140 3 Determine the magnitude of the momentum of a system of two objects: m 1 , which has a mass of 6.0 kg and a velocity of 13 m/s north and m 2 , which has a mass of 14 kg and a velocity of 7.0 m/s south. Assume north is positive. A 10 kg m/s Answer E B 15 kg m/s C 20 kg m/s D -15 kg m/s [This object is a pull tab] E -20 kg m/s Slide 18 / 140 4 Determine the momentum of a system of 3 objects: m 1 , which has a mass of 7.0 kg and a velocity of 23 m/s north, m 2 , which has a mass of 9.0 kg and a velocity of 7.0 m/s north and m 3 , which has a mass of 5.0 kg and a velocity of 42 m/s south. Assume north is positive. A -12 kg m/s B 12 kg m/s C -14 kg m/s D 14 kg m/s E 15 m/s

Slide 18 (Answer) / 140 4 Determine the momentum of a system of 3 objects: m 1 , which has a mass of 7.0 kg and a velocity of 23 m/s north, m 2 , which has a mass of 9.0 kg and a velocity of 7.0 m/s north and m 3 , which has a mass of 5.0 kg and a velocity of 42 m/s south. Assume north is positive. A -12 kg m/s Answer D B 12 kg m/s C -14 kg m/s D 14 kg m/s [This object is a pull tab] E 15 m/s Slide 19 / 140 Impulse - Momentum Equation Return to Table of Contents Slide 20 / 140 Impulse-Momentum Theorem The Conservation of Linear Momentum applies to an isolated system of particles. The overall momentum is conserved, but what about the momentum of each particle? Start with Newton's Second Law, as expressed in Principia , where we look at all the forces on one of the particles. Assume the force acts over a time interval t 0 to t f , and integrate this expression. The particle's momentum will change.

Slide 21 / 140 Impulse-Momentum Theorem We will examine the specific case where a single, constant, very large force acts on the particle for a very short time, so all other forces need not be considered. The equation simplifies to: Define Impulse as: The Impulse-Momentum equation is then: Impulse is a vector, and it is in the same direction as the change of momentum or velocity of the particle acted on by the force. Slide 22 / 140 Impulse-Momentum Theorem The force is not always constant - for example when a tennis racquet strikes a tennis ball, the force starts out small, and increases as the ball increases its contact time with the racquet, then decreases as it leaves. The large force at the peak results in a deformation of the ball. F(N) t (s) Slide 23 / 140 Impulse-Momentum Theorem F(N) F(N) The shaded areas are equal in magnitude. F avg t (s) t (s) The force - time graph can be used to find the Impulse delivered by the racquet in two ways: Find the area underneath the curve, either by integration if the · force is specified as a function of time, or by numerical methods. Find the average force delivered by the racquet and multiply if by · the time interval.