UNIT 2G. Momentum & It s Conservation

Transcription

1 Name: Date:_ UNIT 2G Momentum & It s Conservation

2 Momentum & Newton s 2 nd Law of Motion Newton s 2 nd Law states When an unbalanced force acts upon a body, it accelerates that body in the direction of the force. The acceleration produced is directly proportional to the force. Newton s 3 rd Law states For every action or force there is an equal and opposite reaction or force. Ex. of Newton s Laws and momentum in the real world can be seen with bouncing balls, car crashes, etc. Both Newton s 2 nd & 3 rd Laws are involved in momentum as will be shown in the following notes. Momentum What is harder to stop, a truck or a go-cart? Why? The truck is bigger, heavier, etc. So this plays a factor. What if a go-cart was moving at 100 m/s while the truck was moving at 1 m/s. M truck =100kg while the go-cart s mass (M go-cart ) =5kg. Which would be harder to stop then? This becomes a momentum problem. Momentum Equation is written as p=m v where: p=momentum (kg m/s) m=mass (kg) v=velocity (m/s) Momentum plays a huge factor in the answer. Momentum = mass x velocity Velocity is a vector, therefore Momentum= vector *** Sign convention is important - + Example above using actual information and sign convention truck go-cart p truck = m v = 100 kg x 1 m/s = 100 kg m/s p go-cart = m v = 5 kg x (-100 m/s) = 500 kg m/s (obviously the go-cart has more momentum!) So the go-cart will overcome the momentum of the truck, the net momentum will be 400 kg m/s to the left. 2

3 Example: a) Which would you rather be hit by a 1000 kg car moving at 10.0 km/h or a 0.50 kg ball moving at 10.0 km/h? The easiest way to answer this is to calculate the momentum of the two objects. Obviously the car has more momentum (it also has more inertia). You d probably be able to walk away after being hit by the ball, but you might find it a bit harder to walk away after the car. Car p = m v p = 1000kg (2.78m/s) p = 2.78 x 10 3 kg m/s Ball p = m v p = 0.5kg (2.78m/s) p = 1.4 kg m/s b) How fast would the ball have to be moving to have the same momentum as the car? p = m v v = p/m = 2.78 x 10 3 kg m/s / 0.50 kg = 5.5 x 10 3 m/s = 20,000 km/h!!! 2 nd Law of Motion and Momentum Equation for constant acceleration is a= V= V i -V f t t But we just learned that F Net =ma, & a=f Net /m If we substitute in F Net /m =a for a= V this yields: t Force=F F=m(V i -V f ) or F t= mv i -mv f F Net =f (same meaning in this case) t where f t is the impulse and mv i - mv f is the change in momentum. 2 nd Law of Motion now has t and v in it! We need to revise what it says somewhat as follows: The change in momentum produced by a force acting upon a body is in the same direction as the force and is proportional to the force and to the time it acts. F t= J=Impulse (units are N s OR kg m/s) How does impulse effect sports? Soccer, Tennis, Baseball, Football, Volleyball, ALL SPORTS!!!! 3

4 Other examples besides sports? Car, boat, plane accidents Everyday walking or running you have momentum! Impulse Example Problem A 1.0 kg ball traveling at 4.0 m/s strikes a wall and bounces straight back at 2.0 m/s. What is its change in momentum? What is the impulse applied to the wall? wall wall before after Given m= 1.0 kg v 1 = -4.0 m/s (negative direction) v 2 = 2.0 m/s Soln. Eq. is F t=m v **the change in m v equals the impulse applied to the wall, so find m v and we have the impulse. mv = m v 2 -m v 1 = (1kg) (2 m/s) (1kg) (-4 m/s) = 6.0 kg m/s Example: A 75kg man is involved in a car accident. He was initially traveling at 65km/h when he hit a large truck. a) If he had no airbag in his car and he came to rest against the steering wheel in 0.05s, how much force was exerted on his body? First, change 65 km/h into 18m/s. F t = m v F = (m v)/t = (75kg)(-18m/s) / (0.05s) F = 2.7 x 10 4 N b) If he did have an airbag that inflated and deflated correctly, bringing him to rest over a time of 0.78s, how much force was exerted on his body? F t = m v F = (m v)/t = (75kg)(-18m/s) /(0.78s) F = 1.7 x 10 3 N Which is only about 6% of the force felt without an airbag a definite improvement! 4

5 Conservation of Momentum (Major Concept) What happens when 2 cars collide? If you look in slow motion, they actually rebound off each other. Examples are the hands on demos in class with rubber balls, erasers, etc. There is no loss of momentum during these collisions instead momentum is transferred. Law of Conservation of Momentum- In a closed system with no outside influence, total momentum of the system remains the same. A closed system in this case consists of two objects that do not have any influence such as friction, wind, and any other outside forces that can effect the system. Remember, any momentum lost by one body is transferred to the other. Mass does not transfer, in other words, mass is not created or destroyed. Only the velocity vector can change, mass does not. Vector Form of Law of Conservation of Momentum The vector sum of the momentum of all bodies in a system remains unchanged in both magnitude and direction. Also can be stated as: - 2 objects that are not subjected to external forces and interact or collide have the same total momentum before the collision as they do after they collide. Simply put, this means the initial momentum of the system must equal the final momentum of the system!! In equation form, this is written m 1 v 1 = m 2 v 2 & or p before = p after!! Elastic & Inelastic Collisions Now, there are two types of collisions, elastic and inelastic. Elastic means that the objects bounce off of each other. m 1 m 2 m 1 m 2 v 1 p before v 2 v 1 p after v 2 Remember, sign convention is extremely important here. Initially, v 1 is positive and v 2 is negative, but after the collision v 1 is negative and v 2 is positive. Also remember that p before = p after!! Note: we place the apostrophe ( ) after the v 1 and v 2 in the p after to signify that this is the final velocity. v 1 and v 2 represent the initial velocity. You can also type v 1initial or v 1final or v 1i and v 1f if you would like. 5

6 Inelastic means the objects hit and stick together. p before p after The key here is that there are two velocities and masses in the before condition, and one mass and velocity in the after condition. The mass after collision would just be m 1 + m 2. p before = p after m 1 v 1 +m 2 v 2 = (m 1 +m 2 )v 3 (we can call this v 3 because both masses have the same velocity after the collision since they stick.) Elastic Collision Example m 1 v 1i m 2 v 2i m 1 v 1f m 2 v 2f Stage 1 Stage 2 Stage 3 A 0.5 kg object traveling at 2m/s (east) collides with a 0.3 kg object traveling at 4 m/s (west). After the collision, the 0.3 kg object is traveling at 2m/s (east). What is the magnitude and direction of the velocity of the first object? 1) Draw schematic and label info 2) Write equation and solve for unknown..5kg.3kg.5kg After.3kg m 1 east west m 2 2 m/s 4 m/s v 1f =? v 2f =2 m/s p before = p after Remember sign convention is important when setting up the equation. - + m 1 v 1i m 2 v 2i = -m 1 v 1f + m 2 v 2f 1kg m/s-1.2kg m/s = -.5kg v 1f + 0.6kg m/s -0.6kg m/s 0.2kg m/s = -0.5kg v 1f -0.8kg m/s= - v 1f 0.5kg OR v 1f = -1.6 m/s Therefore the object moves in the negative direction 6

7 Inelastic Example Same problem above except the two balls stick together in the after condition. m 1 v 1i m 2 m 1 v 1f m 2 v 2f Where v 1f = v 2f OR v 12f Stage 1 Stage 2 Stage 3 A 0.5 kg object traveling at 2m/s (east) collides with a 0.3 kg object traveling at 4 m/s (west). After the collision, both objects are traveling east. What is the magnitude and direction of the velocity of the two objects? 3) Draw schematic and label info 4) Write equation and solve for unknown. p before = p after m 1 v 1i m 2 v 2i = (m 1 +m 2 ) v 12f 1kg m/s-1.2kg m/s =.5kg v 12f 0.2kg m/s = -0.5kg v 12f -0.2kg m/s = v 2f -0.5 kg V 12f = 0.4 kg m/s Note: There are many other similar problems to look at, but if you understand the concept you can apply it to anything you see. More Examples Draw the FBD s on your own!!!!!!!!! Example 1: Objects bounce apart A 0.15kg blue billiard ball moving at 8.0m/s to the right hits a similar red billiard ball at rest. If the blue ball continues to move to the right at 2.5m/s, what is the velocity of the red ball. pbefore = pafter pb + pr = pb` + pr` mb vb + mr vr = mb vb` + mr vr` 0.15kg(8.0m/s) kg(0m/s) = 0.15kg(2.5m/s) kg(vr`) vr` = 5.5m/s [right] Example 2: Objects stick together Two balls of clay, a blue one being 2.3kg and the second red one being 5.6kg, hit each other and stick together. If the blue one was moving to the right at 12m/s, and the red was moving at 8.1m/s to the left, what is their final velocity? pbefore = pafter pb + pr = pb` + pr` mb vb + mr vr = mb vb` + mr vr` mb vb + mr vr = v`(mb + mr) 2.3kg (+12m/s) + 5.6kg (-8.1m/s) = v` (2.3kg + 5.6kg) v` = m/s [left] 7

8 Name Momentum Notesheet Definitions: Date 1. Momentum 2. Impulse 3. Closed system 4. Law of Conservation of Momentum 5. Newton s second law of motion Equation Variables/ constants Units Can be used to find F Net = m a p = m v p before = p after J = F t =Δp Algebra review Show how Newton s second law and impulse are equivalent Graph review Label the vertical axis Force and the horizontal axis time. Calculate the Impulse in each case. 8 (OVER)

9 Problem-solving methods for momentum: 1) Coordinate system required -x +x 2) Scaling (ex. 1 cm = 5 kg.m/s) 3) Use two columns to represent moment before and momentum after Example: A 5- kg cart moving at 2.4 m/s (+x) collides and sticks together to a 3-kg cart initially at rest. Sketch before after 5 kg 3 kg 5kg + 3kg 2.4 m/s +x +x A a) Calculate the initial momentum of the 5 kg cart b) Calculate the initial momentum of the 3 kg cart c) Calculate the final (after the collision)momentum of both carts d) Calculate the final velocity of both carts e) Using 1 cm = 2 kg.m/s, Draw the momentum vector for both carts at point A on 3 kg cart. Example: A 7- kg cart and a 3- kg cart are held together at rest by a string. When the string is cut the 7- kg cart moves away from the 3-kg cart with a velocity of m/s. Sketch a) What is the initial momentum of the system? b) What is the final momentum of the system? c) What is the final velocity of the 3-kg cart? 9

10 Name: Momentum Worksheet #1 Date: 1. A 20-Kilogram car traveling east at 6 meters per second collides with a 30- Kilogram car that is traveling west. If both the cars come to rest immediately after the collision, what was the speed of the west bound car just before the collision? (1 ) 6 meters per second (2 ) 2 meters per second (3 ) 3 meters per second (4 ) 4 meters per second 2. An unbalanced 6.0-newton force acts eastward on an abject for 3.0 seconds. The impulse produced by the force is (1 ) 18 N s east (2 ) 2.0 N s east (3 ) 18 N s west (4 ) 2.0 N s west 3. What is the magnitude of the velocity of a 25-kilogram mass that is moving with a momentum of 100. kilogram-meters per second? (1 ) 0.25 m/s (2 ) 2500 m/s (3 ) 40. m/s (4 ) 4.0 m/s 4. In the diagram below, a 0.4-kilogram steel sphere and a 0.1-kilogram wooden sphere are located 2.0 meters above the ground. Both spheres are allowed to fall from rest. Which statement best describes the spheres after they have fallen 1.0 meter? [neglect air resistance.] (1 ) Both spheres have the same speed and momentum. (2 ) Both spheres have the same speed and the steel sphere has more momentum than the wooden sphere. (3 ) The steel sphere has greater speed and has less momentum that the wooden sphere. (4 ) The steel sphere has greater speed than the wooden sphere and both spheres have the same momentum. (OVER) 10

11 5. In a baseball game, a batter hits a ball for a home run. Compared to the magnitude of the impulse imparted to the ball, the magnitude of the impulse to the bat is... (1 ) less (2 ) greater (3 ) the same 6. A bullet traveling at 5.0 x 10 2 meters per second is brought to rest by an impulse of 50. newton-seconds. What is the mass of the bullet? (1 ) 2.5 x 10 4 kg (2 ) 1.0 x 10 1 kg (3 ) 1.0 x 10-1 kg (4 ) 1.0 x 10-2 kg 7. What is the magnitude of net force acting on a 2.0 x 10 3 kilogram car as it accelerates from rest to a speed of 15 meters per second in 5 seconds? (1 ) 6.0 x 10 3 N (2 ) 2.0 x 10 4 N (3 ) 3.0 x 10 4 N (4 ) 6.0 x 10 4 N 8. The direction of an object s momentum is always the same as the direction of the object s (1) inertia (2) potential energy (3) velocity (4) weight 9. As a freely falling object approaches the earth s surface, the impulse required to stop the object (1) decreases (2) increases (3) stays the same 10. A 1-kg object falls freely from rest. The magnitude of its momentum after 1.0 second of fall is (1) 1.0 kg-m/s (2) 4.9 kg-m/s (3) 9.8 kg-m/s (4) 20 kg-m/s 11

12 Name: Momentum and It s Conservation Date: Note: Remember to use sign convention!!!!!!!! (-) (+) 0 1. Calculate the momentum of a 70 kg person traveling at 30 m/s in a car. 2. Using a diagram (FBD), show the direction and speed of the following: A 7 kg bowling ball moving at 27 m/s collides with an 8 kg bowling ball at rest. The 8 kg ball moves away from the collision at 15 m/s. What is the velocity (speed and direction) of the 7 kg ball? 3. A car of mass 1.6 x 10 3 kg is traveling at 30 m/s. The car collides with a hot dog stand of mass 200 kg. The hot dog stand and the car become locked together and continue moving. What is the velocity (speed and direction) of the hot dog stand/car system? 4. What force is needed to reduce the speed of a 90 kg body from 27 m/s to rest (0 m/s) in 0.05 seconds? 12

13 Name: Date: Momentum Investigations 1. Find the momentum of a heavy, American automobile which has a mass of 2360 kg when it is traveling at a velocity of 21.1 m/s, west. 2. Find the momentum of a lighter, foreign vehicle of mass 1170 kg traveling at the same velocity. 3. What is the value of the speed of a bullet, whose mass is 1.00 x 10-3 kg, if it is to have the same size momentum as a 1.8 x 10-3 kg bullet traveling at 3.0 x 10 2 m/s. 4. How long does it take a net force of N to increase the speed of a 50.0 kg rocket from 100 m/s to 150 m/s? 5. What braking force is needed to bring a 1000 kg car, traveling at 30 m/s, to a stop in a time of 50.0 seconds? 6. A.050 kg bullet leaves the muzzle of a rifle with a speed of 40.0 m/s when the 4.0 kg gun is fired. What is the recoil velocity of the gun if it is free to move? 13

14 Name:_ Momentum Warm Up Date: 1. A rocket ship emits hot gases with a speed of 400 m/s. If it shoots out 200 kg of gas every second, what is the resultant thrust of the engine? 2. A 20-kg boy running at 2.0 m/s jumps into a 15-kg wagon from the rear. If there is no friction, how fast will the wagon go with the boy in it? 3. A 3.0 kg ball hits a wall with a speed of 5.4 m/s & rebounds with a v=3.0 m/s. a. What is the change in momentum of the ball? b. What is the magnitude of the impulse given to the wall? c. If the ball is in contact with the wall for.010 seconds, what is the magnitude of the average force acting on it? 4. A 10 kg wagon is pushed with a force of 7.0 N for 3 seconds, then with a force of 2.6 N for 2.0 seconds, and finally with a force of 6.5 N for 1.2 seconds. a. What was the total impulse applied to the wagon? b. What was the change in the velocity of the wagon? 5. A 40-kg boy is riding in the back of a wagon traveling over a frictionless surface. The wagon has a mass of 10 kg and is traveling at 4.0 m/s, east. The boy jumps from the back of the wagon with a horizontal velocity with respect to the ground of 1.0 m/s, west. What is the new velocity of the wagon? 14

15 Name Date Momentum Practice Problems 1. A 150-kg cart rolls in the +x direction with a speed of 0.5 m/s. The 150-kg cart collides and sticks to a 200-kg cart initially at rest. Both carts then move together at a certain speed. Sketch and coordinate system a) Calculate the momentum of the system before the collision: b) Calculate the momentum of the system after the collision: a) c) Calculate the velocity of the carts after the collision: b) c) 2. A 550-kg cannon fires a 10-kg cannonball in the +x direction at a speed of 150 m/s. Sketch and coordinate system a) Calculate the momentum of the system before the firing: b) Calculate the momentum of the system after the firing: a) c) Calculate the velocity of the cannon after the firing: b) c) 15

16 Name Momentum Practice Problems #2 Sketch Date 1. A 7-kg ball moving at 7 m/s (+x) strikes a 5-kg ball at rest. After the collision the 5-kg ball has a velocity of 5 m/s. (The balls separate after the collision) a) Calculate the momentum of the system before the collision: b) Calculate the momentum of the system after the collision: a) c) Calculate the velocity of the 7-kg ball after the collision: b) c) 2. What impulse is required to bring your mass to rest from a 12.3 m/s initial velocity in 1.5 seconds? a) Impulse b)force required?_ c) acceleration your body undergoes? Sketch 16

17 Name Date Momentum, Friction & Impulse Problem Show all work. No work, no credit 1. A box of mass 5.0 kg initially traveling at +3.0 m/s slides on a rough surface of μ k = The box comes to rest. +y v i = v f = 0 m/s +x a) Calculate the p before of the box: Draw the p before vector on the box a) direction b) Calculate the Normal force on the box: Draw the F normal vector on the box direction b) c) Calculate the Friction force on the box: Draw the F friction vector on the box direction c) d) Calculate the acceleration on the box: Draw the acceleration vector between v i, v f direction d) e) How long does it take for the box to come to rest? e) f) What was the impulse given to the box? Draw the impulse vector on the box direction f) g) What displacement did the box undergo during the deceleration? g) direction 17

18 Name Momentum Pre-Test Show all work Sketch p before p after Date_ N E 1. An kilogram ball is fired horizontally from a 1.00 x kg cannon initially at rest. After having been fired, the momentum of the ball is 2.40 x 10 3 kg. m/s East. a) Calculate the velocity of the ball after it has been fired: a) b) Calculate the momentum of the cannon after the ball has been fired: b) c) Calculate the velocity of the cannon after the ball has been fired: c) d) Calculate impulse given to the ball during the firing d) e) Calculate the force applied to the ball if the acceleration lasted 0.15 seconds e) (OVER) 18

19 Sketch p before p after 2. A 60.0 kg stick figure running at 5.0 m/s (+x) jumps on a 20.0 kg cart at rest. The stick figure and cart then move as one unit. a) Calculate the initial momentum of the figure a) b) Calculate the initial momentum of the cart b) c) Calculate the momentum of the system after the stick figure lands on the cart c) d) Calculate the velocity of the stick figure and cart unit after the stick figure landed on the cart. d) (OVER) 19

20 Sketch p before p after 3. A 200-kg cart traveling at 2.5 m/s (+x) collides with a Crumple Zone Crash Control Device. The cart is brought to rest in 0.75 seconds. a) Calculate the initial momentum of the cart: a) b) Calculate the force exerted on the cart by the CZCCD. b) c)calculate the displacement of the cart as it was decelerated by the CZCCD c) d) Calculate the impulse on the cart exerted by the CZCCD: d) e) Scale: 1 cm = 100 kg m/s Draw the initial momentum vector on the cart below +x 20

21 NAME DATE Impulse Practice Problem Sketch velocity initial = _ velocity final = _ a) A 20-kg cart is traveling 15 m/s in the +x direction. Calculate the initial momentum of the cart: Sketch the initial momentum vector on the above diagram +x a) b) What impulse is needed to bring the cart to rest in 3 seconds? Sketch the impulse vector on the above diagram b) c) What acceleration does the cart undergo in the 3 second interval? Sketch the acceleration vector on the above diagram c) d) Calculate the force needed to bring the cart to rest in the 3 second interval: Sketch the force vector in the above diagram d) e) What was the displacement of the cart in the 3 second interval? e) f) Compare the momentum before and the momentum after in this exercise. Does this exercise violate the law of conservation of momentum? Yes No Explain your answer : 21

22 Name: Date: Momentum Worksheet SHOW ALL WORK!!!!!!!!! 1. Draw a FBD that shows a 4.0-kilogram cart moving to the right and a 6.0-kilogram cart moving to the left on a horizontal frictionless surface. The 4 kg cart has a vi=4 m/s & the 6 kg cart has a vi=10 m/s B. When the two carts collide they lock together. Determine the magnitude of the total momentum of the two-cart system after the collision. 3. During a collision, an 84-kilogram driver of a car moving at 24 meters per second is brought to rest by an inflating air bag in 1.2 seconds. Determine the magnitude of the force exerted on the driver by the air bag. Show all work including an FBD. 4. Show calculations to explain why the magnitude of the momentum is the same for both carts? Z:\Physics\\Class Material\Unit 2G Momentum doc 22