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6.1 momentum and Impulse Objectives 1. Compare the momentum of different moving objects. 2. Compare the momentum of the same object moving with different velocities. 3. Identify examples of change in the momentum of an object. 4. Describe changes in momentum in terms of force and time. Questions 1. International regulations specify the mass of official soccer balls. How does the mass of a ball affect the way it behaves when kicked? 2. How does the velocity of the player’s foot affect the final velocity of the ball? 3. Is the pins more likely to move rapidly when the ball travels at a high speed or at a low speed? 4. If two bowling balls of different masses move at the same speed, which is more likely to move the pin? Linear Momentum • The language of physics: – As seen before, words used in our everyday language, such as work and energy, often have precise definitions in physics, momentum is another example of such a word. • Examples of everyday meaning: strength gained by a series of events: – The company has had a successful year and hopes to maintain its momentum by introducing new products. – The movie loses momentum toward the end. • Physics definition of momentum: a property of a moving body, is equal to the product of the body's mass and velocity. Momentum equation Momentum = mass • velocity p=m•v m: mass in kg v: velocity in m/s p: momentum in kg∙m/s • Momentum is a vector quantity • The direction of the momentum vector is the same as the direction of the velocity of the ball. Which is the same as the direction that an object is moving. Mass and velocity have equal importance in momentum • Momentum is directly proportional to mass and momentum is directly proportional to velocity • Consider a 0.5-kg physics cart loaded with one 0.5-kg brick and moving with a speed of 2.0 m/s. Its momentum is 2.0 kg•m/s. If the cart was instead loaded with three 0.5-kg bricks, then the total mass of the loaded cart would be 4.0 kg•m/s. A doubling of the mass results in a doubling of the momentum. • Similarly, if the 2.0-kg cart had a velocity of 4.0 m/s (instead of 2.0 m/s), then the cart would have a momentum of 8.0 kg•m/s (instead of 4.0 kg•m/s). A doubling in velocity results in a doubling of the momentum. Example 1 • Express your understanding of the concept and mathematics of momentum by answering the following questions Determine the momentum of a ... a. 60-kg halfback moving eastward at 9 m/s. b. 1000-kg car moving northward at 20 m/s. c. 40-kg freshman moving southward at 2 m/s. Example 2 2. A car possesses 20 000 units of momentum. What would be the car's new momentum if ... a. its velocity was doubled. b. its velocity was tripled. c. its mass was doubled (by adding more passengers and a greater load) d. both its velocity and mass was doubled. Determine sum of momentum – example 3 • Four billiard balls, each of mass .5 kg, all are traveling in the same direction on a billiard table, with speeds 2 m/s, 4 m/s, 8 m/s and 10 m/s. What is the linear momentum of this system? ptotal = p1 + p2 + p3 + p4 ptotal = (0.5 kg)(2 m/s + 4 m/s + 8 m/s + 10 m/s) ptotal = 12 kg∙m/s • If all four balls collide, what is the total momentum after the collision? 12 kg∙m/s Example 4 - Sample Problem 6A • A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the momentum of the truck? Class work • Page 200 – practice 6A 1. 2500 kg∙m/s 2. a. 120 kg∙m/s NW; b. 94 kg∙m/s NW; c 27 kg∙m/s NW 3. 46 m/s E • Page 232 #12-13, 41,43, 42* 12. a. 8.35 x 10-21 kg∙m/s up; c. 750 kg∙m/s SW; 13. 5.2 toward home plate 41. 42.0 m/s toward 2nd base 42. 3.0 kg; 10. m/s 43. a. 0 kg∙m/s; b. 4.88 kg∙m/s right d. 1.78 x 1029 kg∙m/s forward b. 1.0 kg∙m/s upward A change in Momentum takes force and time • • • • demo p = m∙v ∆p = m∙∆v ∆p = m∙a∙∆t ∆p = F∙∆t a v t v a t A change in momentum takes force and time p F t The equation states that a net external force, F, applied to an object for a certain time interval, ∆t, will cause a change in the object’s momentum. This change in the object’s momentum equals to the product of F and the time interval. Impulse-Momentum Theorem F t p F t mv f mv i In physics, the quantity F•∆t is known as impulse, represented by letter J. F: force in Newton J = F∙∆t t: time in second J: impulse in N∙s •Impulse is a vector quantity, its direction is the same as the net force F. Impulse = Change in Momentum Impulse = Change in momentum F t p • The impulse experienced by the object is the cause of the change in momentum. • The impulse experience by the object equals the change in momentum of the object. • There are two way to determine impulse: – F·t – m∙∆v • There are two way to determine change in momentum: – F·t – m∙∆v Example 1 • A 0.10-kilogram model rocket’s engine is designed to deliver an impulse of 6.0 newton-seconds. If the rocket engine burns for 0.75 second, what average force does it produce? Example 2 • In the diagram, a 60.-kilogram rollerskater exerts a 10.-newton force on a 30.-kilogram rollerskater for 0.20 second. What is the magnitude of the impulse applied to the 30.kilogram rollerskater? Example 3 • If the halfback experienced a force of 800 N for 0.9 seconds, then we could say that the 720 N•s impulse was _______________ • This impulse would cause a momentum 720 kg•m/s change of ____________ • In a collision, the impulse experienced by an object is always equal to the momentum change. Note: the unit N•s = kg•m/s Example 4 • A 0.50-kg cart (#1) is pulled with a 1.0-N force for 1 second; another 0.50 kg cart (#2) is pulled with a 2.0 N-force for 0.50 seconds. a. Which cart (#1 or #2) has the greatest acceleration? Explain. b. Which cart (#1 or #2) has the greatest impulse? Explain. c. Which cart (#1 or #2) has the greatest change in momentum? Explain. Example 5 • Two cars of equal mass are traveling down Lake Avenue with equal velocities. They both come to a stop over different lengths of time. The ticker tape patterns for each car are shown on the diagram below. a. Which car (A or B) experiences the greatest acceleration? Explain. b. Which car (A or B) experiences the greatest change in momentum? Explain. c. Which car (A or B) experiences the greatest impulse? Explain. Example 6 • Calculate the change of momentum for a 1.0-kilogram cart for the following situations. Which situation has the greatest change of momentum? a. b. c. d. accelerating it from rest to 3.0 m/s accelerating it from 2.0 m/s to 4.0 m/s applying a net force of 5.0 N for 2.0 s applying a net force of 10.0 N for 0.5 s Example 7 • A 0.15-kilogram baseball moving at 20. meters per second is stopped by a catcher in 0.010 second. What is the average force stopping the ball? Example 8 • A 2.0-kilogram laboratory cart is sliding across a horizontal frictionless surface at a constant velocity of 4.0 meters per second east. What will be the cart’s velocity after a 6.0-newton westward force acts on it for 2.0 seconds? Example 9 • A 1,000-kilogram car traveling due east at 15 meters per second is hit from behind and receives a forward impulse of 8,000 newton-seconds. Determine the magnitude of the car’s change in momentum due to this impulse. Example 10 - Sample problem 6B • 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 magnitude of the force exerted on the car during the collision. Class work • Page 211 – practice 6B 1. 2. 3. 4. 380 N right 1100 N up 16.0 kg∙m/s S a. 9 m/s right; b. 15 m/s left • Page 232 #14-15, 47, 55, 56* 14. 15. 47. 55. 56. 160 N right 18 N 400 N a. 0.83 m/s right; b. 1.2 m/s left a. 9.9 m/s down; b. 1800 N up Real-World Applications • The effect of collision time upon the amount of force an object experiences • The effect of rebounding upon the velocity change and hence the amount of force an object experiences. The relationship between Collision Time and Force of Impact m∙∆v = F∙∆t •The greater the time over which the collision occurs, the smaller the force acting upon the object. •To minimize the effect of the force on an object involved in a collision, the time must be increased. •To maximize the effect of the force on an object involved in a collision, the time must be decreased. Reduce force by increase time • Air bags are used in automobiles because they are able to minimize the affect of the force on an object involved in a collision. Air bags accomplish this by extending the time required to stop the momentum of the driver and passenger. • Padded dashboards also reduces force by increase time. • A boxer rides the punch in order to extend the time of impact of the glove with their head. • Nylon ropes are used in the sport of rock-climbing because of its ability to stretch. The rock climber can appreciate minimizing the effect of the force through the use of a longer time of impact . The Effect of Rebounding • Bouncing off each other is known as rebounding. Rebounding involves a change in the direction of an object; rebounding situations are characterized by – a large velocity change F∙∆t = m∙∆v – a large momentum change. – a large impulse • The importance of rebounding is critical to the outcome of automobile accidents. • Automobiles are made with crumple zones. Crumple zones minimize the affect of the force in an automobile collision in two ways. – By crumpling, the car is less likely to rebound upon impact, thus minimizing the momentum change and the impulse. – The crumpling of the car lengthens the time over which the car's momentum is changed; by increasing the time of the collision, the force of the collision is greatly reduced. Increase velocity by increasing time: F∙∆t = m∙∆v • In racket and bat sports, hitters are often encouraged to follow-through when striking a ball. • In this situation, both the force applied (as hard as you can) and the mass (the mass of the ball) are constant. By following through, the hitter increases the time, the result is increasing the ball’s velocity. examples • A constant force can act on an object for different lengths of time. As the length of time increases, 1. the impulse imparted to the object a) decreases b) increases c) remains the same 2. The momentum of the object a) decreases b) increases c) remains the same Example 3 • A cannonball shot from a long-barrel cannon travels faster than one shot from a short-barrel cannon because the cannonball receives a greater a. force. b. impulse. A cannonball shot from a cannon receive the same force regardless of the length of c. both A and B its barrel. A long-barrel will take longer d. neither A nor B time (t) for the cannonball to travel. Since J = F∆t = m∆v, the longer the time, the bigger the impulse, the faster it will travel. Example 4 • A student drops two eggs of equal mass simultaneously from the same height. Egg A lands on the tile floor and breaks. Egg B lands intact, without bouncing, on a foam pad lying on the floor. Compared to the magnitude of the impulse on egg A as it lands, the magnitude of the impulse on egg B as it lands is _________________ (less, greater, the same) as egg A. Stopping times and distances depend on the impulse-momentum theorem F∙∆t = m∙∆v • Two identical trucks travels with the same velocity. One is empty with mass m, and the other is loaded with mass 2m. 1. Why is loaded truck’s stopping time twice as much as the empty truck’s when acted on by the same force? Loaded is 2x unloaded 2. How do the stopping distances compare? Loaded is 2x unloaded Example 5 • If the maximum coefficient of kinetic friction between a car and a road is 0.50, what is the minimum stopping distance for car moving at 29 m/s? (about 65 mi/hr) Example 6 - Sample problem 6C • A 2250 car traveling to the west slows down uniformly from 20.0 m/s to 5.00 m/s. 1. How long does it take the car to decelerate if the force on the car is 8450 N to the east? 2. How far does the car travel during the deceleration? Class work • Page 213 practice 6C 1. 5.33 s; 53.3 m E 2. a. 14 m/s N; b. 42 m/s N; 3. a. 1.22 x 104N E; b. 53.3 m W • Page 232 #16, 47 16. 0.010 s; 0.13 m 47. 400 N c. 8.0 s