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## Linear Momentum

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**Linear Momentum**Impulse & Collisions**What is momentum?**• Momentum is a measure of how hard it is to stop or turn a moving object. • What characteristics of an object would make it hard to stop or turn? • Mass • Velocity**Calculating Momentum**• For one particle p = mv • Note that momentum is a vector with the same direction as the velocity! • For a system of multiple particles p = Σpi--- add up the vectors • The unit of momentum is… kg m/s or Ns**Practice:**• A 1180-kg car drives along a city street at 13.4 m/s (~30 mi/h). What is the magnitude of the car’s momentum?**Practice**• Calculate the momentum of a system composed of a 65-kg sprinter running east at 10 m/s and a 75-kg sprinter running north at 9.5 m/s.**Change in Momentum**• Like any change, change in momentum is calculated by looking at final and initial momentums. Δp = pf – pi Δp: change in momentum pf: final momentum pi: initial momentum**More Practice**• A rubber ball and a bean bag are each dropped. The 0.10 kg beanbag has a speed of 4 m/s just before it hits the ground. What is its change in momentum when it hits the ground and sticks?**Part 2:**• The rubber ball has a mass of 0.10 kg, and also has a speed of 4 m/s just before it hits the ground. It hits the ground, deforms as the ground pushes it upward, and bounces back, leaving the surface at a speed of 4 m/s upward. What is its change in momentum?**Newton’s 2nd Law (again)**• For momentum to change, a force is required • The net force on the object is equal to the change of its momentum over time: • This equals ma if mass remains constant**Conservation of Momentum**• The falling bean bag took some time to hit the ground. (or the ground took some time to apply a force and stop it) If the total stopping time was 0.2s, what force did the ground apply?**Conservation of Momentum**• If the falling ball’s contact with the ground lasted 0.4s, what was the force exerted by the ground on the ball? • What force did the ball exert on the ground?**Collisions and Impulse**So we said it takes a Force to change momentum… … but in collisions, the force changes considerably over the course of interaction, so we have to use the average force Impulse is the product of force and the time interval, which equals the change in momentum**Collisions and Impulse**• Impulse (J) is the change in momentum • Measured in N•s • Impulse equals the average force multiplied by the time during which it was applied**Calculating Impulse**• An object experiences a force of 19.30 N for a time period of 4.39 s. What is the impulse on the object?**Using Impulse**• An object is traveling along with a velocity of 7.16 m/s. What is its mass if a force of 7.99 N applied for a time period of 6.62 s accelerates it to a velocity of 17.93 m/s?**Conservation of Momentum**• If the resultant external force on a system is zero, then the vector sum of the momentums of the objects will remain constant. Σ Pbefore = Σ Pafter • External forces: forces coming from outside the system of particles whose momentum is being considered. • External forces change the momentum of the system. • Internal forces: forces arising from interaction of particles within a system. • Internal forces cannot change momentum of the system.**External Forces - Golf**• The club head exerts an external impulsive force on the ball and changes its momentum. • The acceleration of the ball is greater because its mass is smaller.**Internal Forces - Pool**• The forces the balls exert on each other are internal and do not change the momentum of the system. • Since the balls have equal masses, the magnitude of their accelerations is equal.**Momentum**Inelastic Collisions**Collisions**• When two moving objects make contact with each other, they undergo a collision. • Conservation of momentum is used to analyze all collisions. • Newton’s Third Law is also useful. It tells us that the force exerted by body A on body B in a collision is equal and opposite to the force exerted on body B by body A.**Collisions**• During a collision, external forces are ignored. • The time frame of the collision is very short. • The forces are impulsive forces (high force, short duration).**Collisions**• Elastic collisions • Also called “hard” collisions • No deformation occurs, no kinetic energy lost • Inelastic collisions • Deformation occurs, kinetic energy is lost • Perfectly Inelastic (stick together) • Objects stick together and become one object • Deformation occurs, kinetic energy is lost**Perfectly Inelastic Collisions**• Simplest type of collisions. • After the collision, there is only one velocity, since there is only one object. • Kinetic energy is lost. • Explosions are the reverse of perfectly inelastic collisions in which kinetic energy is gained!**Explosions**• When an object separates suddenly, as in an explosion, all forces are internal. • Momentum is therefore conserved in an explosion. • There is also an increase in kinetic energy in an explosion. This comes from a potential energy decrease due to chemical combustion.**Practice**• A fish moving at 2 m/s swallows a stationary fish which is 1/3 its mass. What is the velocity of the big fish after dinner?**More Practice**• A car with a mass of 950 kg and a speed of 16 m/s to the east approaches an intersection. A 1300-kg minivan traveling north at 21 m/s approaches the same intersection. The vehicles collide and stick together. What is the resulting velocity of the vehicles after the collision?**Explosion Problem**• An exploding object breaks into three fragments. A 2.0 kg fragment travels north at 200 m/s. A 4.0 kg fragment travels east at 100 m/s. The third fragment has mass 3.0 kg. What is the magnitude and direction of its velocity?**Momentum**Elastic Collisions**Elastic Collisions**• In elastic collisions, there is no deformation of colliding objects, and no change in kinetic energy of the system. Therefore, two basic equations must hold for all elastic collisions • Σ pf = Σ pi (momentum conservation) • Σ Kf = Σ Ki (kinetic energy conservation)**Elastic Collisions: Recoil**• Guns and cannons “recoil” when fired. • This means the gun or cannon must move backward as it propels the projectile forward. • The recoil is the result of action-reaction force pairs, and is entirely due to internal forces. As the gases from the gunpowder explosion expand, they push the projectile forwards and the gun or cannon backwards.**Practice**• Suppose a 5.0-kg projectile launcher shoots a 209 gram projectile at 350 m/s. What is the recoil velocity of the projectile launcher?**Elastic Collisions in 2-D**• Momentum in the x-direction is conserved. Σ Px (before) = Σ Px (after) • Momentum in the y-direction is conserved. Σ Py (before) = Σ Py (after) • Treat x and y coordinates independently. • Ignore x when calculating y • Ignore y when calculating x**Practice**• Calculate velocity of 8-kg ball after the collision.