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Momentum and Collision

Momentum and Collision. Chapter 6. Momentum and Impulse. Linear Momentum. Momentum is defined as mass times velocity. Momentum is represented by the symbol p, and is a vector quantity. p = m v momentum = mass  velocity. Momentum.

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Momentum and Collision

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  1. Momentum and Collision Chapter 6

  2. Momentum and Impulse

  3. Linear Momentum • Momentum is defined as mass times velocity. • Momentum is represented by the symbol p, and is a vector quantity. p = mv momentum = mass  velocity

  4. Momentum • Momentum has the dimensions of mass x length/time (kg x m/s) • When do you use the word momentum? • “gaining momentum” or “picking up speed” • Coasting on a bike going down a hill • You accelerate and velocity increases with time.

  5. Bowling ball or basketball? • Picture two lanes at a bowling alley. • One with a bowling ball the other with a basketball going at the same speed. • Which will exert more force on the pins? • Why? • More momentum

  6. Example • A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the momentum of the truck?

  7. Solution • p = mv • (2250 kg)(25 m/s east) • 5.6 x 104 kg x m/s to the east

  8. Your Turn • A deer with a mass of 146 kg is running head-on toward you at a speed of 17 m/s. You are going north find the momentum of the deer. • A 21 kg child on a 5.9 kg bike is riding with a velocity of 4.5 m/s to the northwest. • What is the total momentum of the child and the bike together? • What is the momentum of the child? • What is the momentum of the bike?

  9. Linear Momentum • Impulse • The product of the force and the time over which the force acts on an object is called impulse. • The impulse-momentum theorem states that when a net force is applied to an object over a certain time interval, the force will cause a change in the object’s momentum. F∆t = ∆p = mvf – mvi force  time interval = change in momentum

  10. Linear Momentum • Stopping times and distancesdepend on the impulse-momentum theorem. • Force is reduced when the time interval of an impact is increased.

  11. Example • A 1400 kg car moving westward with a velocity of 15 m/s collides with a utility pole and is brought to rest in 0.30 s. Find the force exerted on the car during the collision.

  12. Solution • F∆t = ∆p = mvf – mvi • F = mvf – mvi ∆t • F = (1400kg)(0 m/s) – (1400 kg)(-15 m/s) 0.30 s • F = 7.0 x 104 N to the east

  13. Your Turn II • A 0.50 kg football is thrown with a velocity of 15 m/s to the right. A stationary receiver catches the ball and brings it to rest in 0.020 s. What is the force exerted on the ball by the receiver? • An 82 kg man drops from rest on a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching the water. What is the net force on the diver as he is brought to rest? • A 0.40 kg soccer ball approaches a player horizontally with a velocity of 18 m/s to the north. The player strikes the ball and cause it to move in the opposite direction with a velocity of 22 m/s. What is the impulse delivered to the ball by the player?

  14. Impulse-Momentum Theorem • Stopping times and distances depend on the impulse-momentum theorem. • Highway safety engineers use the impulse-momentum theorem to determine stopping distances and safe following distances for cars and trucks.

  15. Example • A 2240 kg car traveling to the west slows down uniformly from 20.0 m/s to 5.00 m/s. How long does it take the car to decelerate if the force on the car is 8410 N to the east? How far does the car travel during deceleration?

  16. Solution • Givens: • Mass = 2240 kg • vi = 20.0 m/s to the west = - 20.0 m/s • vf = 5.00 m/s to the west = - 5.00 m/s • F = 8410 N to the east • Equation to use: F∆t = ∆p • ∆t = ∆p / F • ∆t = mvf – mvi / F

  17. Plug and Chug • ∆t = mvf – mvi / F • ∆t = (2240 kg)(-5.00 m/s)-(2240 kg)(-20.0 m/s) 8410 kg x m/s2 • ∆t = 4.00 s • ∆x = ½ (vi+ vf)∆t • ∆x = ½ (-20.0 m/s – 5.00 m/s)(4.00 s) • ∆x = - 50.0 m = 50.0 m to the west

  18. Your Turn III • A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20.0 m/s by a 6250 N braking force acting opposite the car’s motion. Use the impulse-momentum theorem to answer the following questions: • What is the car’s velocity after 2.50 s? • How far does the car move during 2.50 s? • How long does it take the car to come to a complete stop?

  19. Impulse-Momentum Theorem • Force is reduced when the time interval of an impact is increased. • Examples • Nets or giant air mattresses used to catch people. • Page 203 in your book shows an image of a girl being tossed in the air and caught in a blanket. • The blanket “gives way” and extends the time of collision to change the momentum over a longer period of time. • Consider the egg on the next slide and explaining what is happening

  20. Impulse-Momentum Theorem

  21. PNBW • Page 204 • Physics 1-3 • Honors 1-5

  22. Conservation of Momentum

  23. Momentum is Conserved • So far we only have considered the momentum of only one object at a time. • Now we will look at two or more objects interacting with each other. • Picture this. . . • You are playing pool. You strike the cue ball it hits the 8 ball. The 8 ball had no momentum before they collided. • During the collision the cue ball loses momentum and the 8 ball gains momentum. • The momentum the cue ball loses is the same amount that the 8 ball gained.

  24. Momentum is Conserved • The Law of Conservation of Momentum: The total momentum of all objects interacting with one another remains constant regardless of the nature of the forces between the objects. m1v1,i+ m2v2,i = m1v1,f + m2v2,f total initial momentum = total final momentum

  25. Momentum is Conserved • The total momentum of all objects interacting with one another remains constant regardless of the nature of the forces between the objects. • Go back to the pool table example. The cue ball and the 8 ball do not have a constant momentum, but the total momentum is constant.

  26. Momentum is Conserved • Consider objects pushing away from each other. • You jump up. You push off the Earth. Take you mv. • Let’s say 60 kg m/s upward. That means that the Earth must have a corresponding momentum of 60 kg m/s downward. However, the Earth has an enormous mass which means its velocity is very small.

  27. Momentum is Conserved • Picture this . . . • Two people on skates facing one another. They push away from one another. Initially, they are both at rest with a momentum of 0. When they push away, they move in opposite directions with equal but opposite momentum, so that the total momentum is 0.

  28. Example • A 76 kg boater, initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock. If the boater moves out of the boat with a velocity of 2.5 m/s to the right, what is the final velocity of the boat?

  29. Solution • Given: m1 = 76 kg m2 = 45 kg v1,i = 0 v2,i = 0 v1,f = 2.5 m/s to the right Unknown: v2,f = ?

  30. Solution • Choose an equation or situation: Because the total momentum of an isolated system remains constant, the total initial momentum of the boater and the boat will be equal to the total final momentum of the boater and the boat. m1v1,i + m2v2,i = m1v1,f + m2v2,f

  31. Solution • Because the boater and the boat are initially at rest, the total initial momentum of the system is equal to zero. Therefore, the final momentum of the system must also be equal to zero. m1v1,f + m2v2,f = 0 Rearrange the equation to solve for the final velocity of the boat.

  32. Solution • Substitute the values into the equation and solve:

  33. Solution • The negative sign for v2,f indicates that the boat is moving to the left, in the direction opposite the motion of the boater. Therefore, v2,f= 4.2 m/s to the left

  34. Your Turn IV • A 63.0 kg astronaut is on a space walk when the tether line to the shuttle breaks. The astronaut is able to throw a spare 10.0 kg oxygen tank in a direction away from the shuttle with a speed of 12.0 m/s, propelling the astronaut back to the shuttle. Assuming that the astronaut starts from rest with respect to the shuttle, find the astronaut’s final speed with respect to the shuttle after the tank is thrown. • An 85.0 kg fisherman jumps from a dock into a 135.0 kg rowboat at rest on the west side of the dock. If the velocity of the fisherman is 4.30 m/s to the west as he leaves the dock, what is the final velocity of the fisherman and the boat?

  35. Your Turn IV • Each croquet ball in a set has a mass of 0.50 kg. The green ball, traveling at 12.0 m/s, strikes the red ball, which is at rest. Assuming that the croquet ball slide on a frictionless surface and all collisions are head-on, find the final speed of the red ball in each of the following situations: • The green ball stops moving after it strikes the red ball. • The green ball continues moving after the collision at 2.4 m/s in the same direction.

  36. Momentum is Conserved • Newton’s third law leads to conservation of momentum • During the collision, the force exerted on each bumper car causes a change in momentum for each car. • The total momentum is the same before and after the collision.

  37. PNBW • Page 211 • Physics 1-3 • Honors 1-4

  38. Elastic and Inelastic Collisions

  39. Collisions • Perfectly inelastic collision • A collision in which two objects stick together after colliding and move together as one mass is called a perfectly inelastic collision. • Example: The collision between two football players during a tackle. • Conservation of momentum for a perfectly inelastic collision: m1v1,i + m2v2,i = (m1 + m2)vf total initial momentum = total final momentum

  40. Example • A 1850 kg luxury sedan stopped at a traffic light is struck from behind by a compact car with a mass of 975 kg. The two cars become entangled as a result of the collision. If the compact car was moving with a velocity of 22.0 m/s to the north before the collision, what is the velocity of the entangled mass after the collision? Given: m1 = 1850 kg m2 = 975 kg v1,i = 0 m/s v2,i = 22.0 m/s north Unknown: vf

  41. Solution • Choose your equation: m1v1,i + m2v2,i = (m1 + m2)vf vf =m1v1,i + m2v2,i (m1 + m2) vf =(1850 kg)(0 m/s) + (975 kg)(22.0 m/s) (1850 kg + 975 kg) vf =7.59 m/s north

  42. Your Turn V • A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is initially at rest at a stoplight. The car and truck stick together and move together after the collision. What is the final velocity of the two-vehicle mass? • You are shopping at Publix and toss a 9.0 kg bag of rice into a stationary 18.0 kg shopping cart. The bag hits the cart with a horizontal speed of 5.5 m/s toward the front of the cart. What is the final speed of the cart and the bag? • A 47.4 kg student runs down the sidewalk and jumps with a horizontal velocity of of 4.20 m/s onto a stationary skateboard. The student and the skateboard move down the sidewalk with a speed of 3.95 m/s. Find the following: • The mass of the skateboard • How fast the student would have to jump to have a final speed of 5.00 m/s

  43. Kinetic Energy in Inelastic Collisions • In an inelastic collision the total kinetic energy does not remain constant when the objects collide and stick together. • Some energy is converted into sound energy and internal energy as the objects deform during the collision. • Elastic in physics refers to a material that when work is done to deform the material during a collision the same amount of work is done to return the material to its original shape. • Inelastic material does not return to its original shape and therefore some energy is converted to sound or heat.

  44. Example • Two clay balls collide head-on in a perfectly inelastic collision. The first ball has a mass of 0.500 kg and an initial velocity of 4.00 m/s to the right. The second ball has a mass of 0.250 kg and an initial velocity of 3.00 m/s to the left. What is the decrease in kinetic energy during the collision?

  45. Solution • Given: m1= 0.500 kg m2 = 0.250 kg v1,i = 4.00 m/s to the right v1,i = +4.00 m/s v2,i = 3.00 m/s to the left v2,i = –3.00 m/s Unknown: ∆KE = ?

  46. Solution • The change in kinetic energy is simply the initial kinetic energy subtracted from the final kinetic energy. ∆KE = KEi– KEf Determine both the initial and final kinetic energy.

  47. Solution • Use the equation for a perfectly inelastic collision to calculate the final velocity.

  48. Solution • Next calculate the initial and final kinetic energy.

  49. Solution • Finally, calculate the change in kinetic energy.

  50. Your Turn VI • A 0.25 kg arrow with a velocity of 12 m/s to the west strikes and pierces the center of a 6.8 kg target. What is the final velocity of the combined mass? What is the decrease in kinetic energy during the collision? • A clay ball with a mass of 0.35 kg hits another 0.35 kg ball at rest, and the two stick together. The first ball has an initial speed of 4.2 m/s. What is the final speed of both balls? Calculate the decrease in kinetic energy that occurs during the collision. • A 56 kg ice skater traveling at 4.0 m/s to the north meets and joins hands with a 65 kg skater traveling at 12.0 m/s in the opposite direction. Without rotating the two skaters continue skating together. What is the final velocity of the two skaters and what is the decrease in kinetic energy durning the collision?

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