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## Ch. 8 Momentum and its conservation

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**Ch. 8 Momentum and its conservation**AP Physics C**Citations:**• Wahl, Michael. SparkNote on Conservation of Momentum. 25 Oct. 2008 <http://www.sparknotes.com/physics/linearmomentum/conservationofmomentum>. • Wahl, Michael. SparkNote on Collisions. 25 Oct. 2008 <http://www.sparknotes.com/physics/linearmomentum/collisions>.**Terms:**• Center of Mass - The point at which a given net force acting on the system will produce the same acceleration as if all the mass were concentrated at that point. • Impulse - A force applied over a period of time. • Momentum - The product of an object's mass and velocity. • Conservation of Momentum - The principle stating that for any system with no external forces acting on it, momentum will be conserved.**Center of mass:**• We have been studying the mechanics of single particles. • We will now expand our study to systems of several particles. • The concept of the center of mass allows us to describe the movement of a system of particles by the movement of a single point.**Center of mass equations:**• X-coordinate of the center of mass: • Y-coordinate of the center of mass:**Sample problem 1:**• Calculate the center of mass of the following system: A mass of 5 kg lies at x = 1, a mass of 3 kg lies at x = 4 and a mass of 2 kg lies at x = 0.**Sample problem 2:**• Calculate the center of mass of the following system: A mass of 10 kg lies at the point (1,0), a mass of 2 kg lies at the point (2,1) and a mass of 5 kg lies at the point (0,1).**Center of mass equations:**• Velocity of the center of mass: • Acceleration of the center of mass: • External Net Force**Sample problem 3:**• Consider the system from problem 2, but now with forces acting upon the system. On the 10 kg mass, there is a force of 10 N in the positive x direction. On the 2 kg mass, there is a force of 5 N inclined 45o above horizontal. Finally, on the 5 kg mass, there is a force of 2 N in the negative y direction. Find the resultant acceleration of the system.**Sample problem 4:**• Two masses, m1 and m2, m1 being larger, are connected by a spring. They are placed on a frictionless surface and separated so as to stretch the spring. They are then released from rest. In what direction does the system travel?**Sample problem 5:**• A 50 kg man stands at the edge of a raft of mass 10 kg that is 10 meters long. The edge of the raft is against the shore of the lake. The man walks toward the shore, the entire length of the raft. How far from the shore does the raft move?**Impulse:**• We shall define this concept, force applied over a time period, as impulse. Impulse can be defined mathematically, and is denoted by J: J= FΔt**More on impulse:**• It is a vector quantity. • Can we predict the motion of an object? J = FΔt = (ma)Δt J = m Δt J = mΔv = Δ(mv) = mvf - mvo**Sample problem 6:**• What is the impulse of a force of 10 N acting on a ball for 2 seconds? • The ball has a mass of 2 kg and is initially at rest. What is the velocity of the ball after the force has acted on it?**Momentum:**• From our equation relating impulse and velocity, it is logical to define the momentum of a single particle, denoted by the vector p, as such: p= mv • Again, momentum is a vector quantity, pointing in the direction of the velocity of the object. From this definition we can generate two every important equations, the first relating force and acceleration, the second relating impulse and momentum.**Sample problem 7:**• A particle has linear momentum of 10 kg-m/s, and a kinetic energy of 25 J. What is the mass of the particle?**Relation between force and acceleration:**• If we take a time derivative of our momentum expression we get the following equation: • Newton’s Second Law of Motion**Sample problem 8:**• A 2 kg bouncy ball is dropped from a height of 10 meters, hits the floor and returns to its original height. What was the change in momentum of the ball upon impact with the floor? What was the impulse provided by the floor?**Sample problem 9:**• A ball of 2 kg is thrown straight up into the air with an initial velocity of 10 m/s. Using the impulse-momentum theorem, calculate the time of flight of the ball.**Momentum and kinetic energy:**• Recall: • Impulse is a change in momentum. • Work done is a change in kinetic energy.**Total momentum of a system:**• Suppose we have a system of N particles, with masses m1, m2,…, mn. Assuming no mass enters or leaves the system, we define the total momentum of the system as the vector sum of the individual momentum of the particles: • P = p1 + p2 + ... + pn= m1v1 + m2v2 + ... + mnvn**Total momentum of a system:**• Recall: vcm= (m1v1 + m2v2 + ... + mnvn)**Conservation of momentum:**• If the net external force is zero, then the total momentum of the system is constant.**Conservation of momentum:**• That is, • Remember that momentum is a vector quantity.**Sample problem 10:**• A 60 kg man standing on a stationary 40 kg boat throws a .2 kg baseball with a velocity of 50 m/s. With what speed does the boat move after the man throws the ball? Assume no friction between the man and the boat.**Sample problem 11:**• A .05 kg bullet is fired at a velocity of 500 m/s, and embeds itself in a block of mass 4 kg, initially at rest and on a frictionless surface. What is the final velocity of the block?**Sample problem 12:**• An object at rest explodes into three pieces. Two, each of the same mass, fly off in different directions with velocity 50 m/s and 100 m/s, respectively. A third piece is also formed in the explosion, and has twice the mass of the first two pieces. What is the magnitude and direction of its velocity?**Sample problem 13:**• A spaceship moving at 1000 m/s fires a missile of mass 1000 kg at a speed of 10000 m/s. What is the mass of the spaceship it slows down to a velocity of 910 m/s?**Collisions:**• Collision - The brief direct contact between two bodies that results in a net impulse on each body. • Elastic Collision - Any collision in which kinetic energy is conserved. • Inelastic Collision - Any collision in which kinetic energy is not conserved. • Completely Inelastic Collision - Any collision in which the two bodies stick together.**Elastic collisions:**• Why are these collisions special? We know with all collisions that momentum is conserved. If two particles collide we can use the following equation: m1v1o+ m2v2o = m1v1f + m2v2f • However, we also know that, because the collision is elastic, kinetic energy is conserved. For the same situation we can use the following equation: m1v1o2 + m2v2o2 = m1v1f2 + m2v2f2**Sample problem 14:**• Two balls, each with mass 2 kg, and velocities of 2 m/s and 3 m/s collide head on. Their final velocities are 2 m/s and 1 m/s, respectively. Is this collision elastic or inelastic?**Sample problem 15:**• Two balls of mass m1 and m2, with velocities v1 and v2 collide head on. Is there any way for both balls to have zero velocity after the collision? If so, find the conditions under which this can occur.**Sample problem 16:**• Two balls with equal masses, m, and equal speed, v, engage in a head on elastic collision. What is the final velocity of each ball, in terms of m and v?**Sample problem 17:**• One pool ball traveling with a velocity of 5 m/s hits another ball of the same mass, which is stationary. The collision is head on and elastic. Find the final velocities of both balls.**Inelastic collisions:**• So what if energy is not conserved? Our knowledge of such situations is more limited, since we no longer know what the kinetic energy is after the collision. However, even though kinetic energy is not conserved, momentum will always be conserved.**Completely inelastic collision:**• Consider the case in which two particles collide, and actually physically stick together. In this case, called a completely inelastic collision we only need to solve for one final velocity, and the conservation of momentum equation is enough to predict the outcome of the collision. The two particles in a completely inelastic collision must move at the same final velocity, so our linear momentum equation becomes: m1v1o+ m2v2o = m1vf + m2vf Thus m1v1o+ m2v2o = Mvf**Sample problem 18:**• A car of 500 kg, traveling at 30 m/s rear ends another car of 600 kg, traveling at 20 m/s. in the same direction The collision is great enough that the two cars stick together after they collide. How fast will both cars be going after the collision?**2-d collisions:**• Two balls of equal masses move toward each other on the x-axis. When they collide, each ball ricochets 90 degrees, such that both balls are moving away from each other on the y-axis. What can be said about the final velocity of each ball?**Sample problem 19:**• Two pool balls traveling in opposite directions collide. One ball travels off at an angle θ to its original velocity, as shown below. Is there any possible way for the second ball to be completely stopped by this collision? If so state the conditions under which this could occur.**Sample problem 20:**• Two objects are traveling perpendicular to each other, one moving at 2 m/s with a mass of 5 kg, and one moving at 3 m/s with a mass of 10 kg, as shown below. They collide and stick together. What is the magnitude and direction of the velocity of both objects?**Sample problem 21:**• A common pool shot involves hitting a ball into a pocket from an angle. Shown below, the cue ball hits a stationary ball at an angle of 45o, such that it goes into the corner pocket with a speed of 2 m/s. Both balls have a mass of .5 kg, and the cue ball is traveling at 4 m/s before the collision. Recalling that this collision is elastic, calculate the angle with which the cue is deflected by the collision.