Momentum. Momentum. Slide 2 / 133. Slide 1 / 133. Slide 3 / 133. Slide 4 / 133. Slide 6 / 133. Slide 5 / 133. Table of Contents.

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1 Slide 1 / 133 Slide 2 / 133 New Jersey Center for Teaching and Learning Progressive Science Initiative This material is made freely available at and is intended for the non-commercial use of students and teachers. These materials may not be used for any commercial purpose without the written permission of the owners. NJCTL maintains its website for the convenience of teachers who wish to make their work available to other teachers, participate in a virtual professional learning community, and/or provide access to course materials to parents, students and others. Momentum Click to go to website: Slide 3 / 133 Slide 4 / 133 Table of Contents Click on the topic to go to that section Momentum Impulse-Momentum Equation The Momentum of a System of Objects Momentum Conservation of Momentum Types of Collisions Collisions in Two Dimensions Return to Table of Contents Slide 5 / 133 Momentum Defined Newton s First Law tells us that objects remain in motion with a constant velocity unless acted upon by an external force (Law of Inertia). In our experience: When objects of different mass travel with the same velocity, the one with more mass is harder to stop. Slide 6 / 133 Momentum Defined Define a new quantity,momentum (p), that takes these observations into account: m o m e nt um = m a s s v e lo c it y p =m v When objects of equal mass travel with different velocities, the faster one is harder to stop. click here for a introductory video on momentum from Bill Nye!

2 Slide 7 / 133 Momentum is a Vector Quantity Since: mass is a scalar quantity velocity is a vector quantity the product of a scalar and a vector is a vector Slide 8 / 133 SI Unit for Momentum There is no specially named unit for momentum - so there is an opportunity for it to be named after a renowned physicist! We use the product of the units of mass and velocity. and: m o m e nt um = m a s s v e lo c it y mass x velocity therefore: Momentum is a vector quantity - it has magnitude and direction Slide 9 / Which has more momentum? kg m/s Slide 9 () / Which has more momentum? A A large truck moving at 30 m/s B A small car moving at 30 m/s B A small car moving at 30 m/s C Both have the same momentum. C Both have the same momentum. A A large truck moving at 30 m/s A Both have the same speed, but the truck has the larger mass 2 What is the momentum of a 20 kg object moving to the right with a velocity of 5 m/s? Slide 10 () / What is the momentum of a 20 kg object moving to the right with a velocity of 5 m/s? Slide 10 / 133 m= 20 kg v = 5 m/s p = mv = (20 kg)(5 m/s) = 100 kg m/s

3 Slide 11 / What is the momentum of a 20 kg object with a velocity of 5.0 m/s to the left? 3 What is the momentum of a 20 kg object with a velocity of 5.0 m/s to the left? Slide 11 () / 133 m= 20 kg v = 5 m/s p = mv = (20 kg)( 5 m/s) = 100 kg-m/s Slide 12 / What is the velocity of a 5 kg object whose momentum is 15 kg-m/s? 4 What is the velocity of a 5 kg object whose momentum is 15 kg-m/s? Slide 12 () / 133 m= 5 kg p = 15 m/s p = mv v = p/m v= (-15 kg-m/s)/5 kg v = -3 m/s 5 What is the mass of an object that has a momentum of 35 kg-m/s with a velocity of 7 m/s? Slide 13 () / What is the mass of an object that has a momentum of 35 kg-m/s with a velocity of 7 m/s? Slide 13 / 133 v= 7 m/s p = 35 m/s p = mv m = p/v m= (35 kg-m/s)/(7 m/s) m = 5 kg

4 Slide 14 / 133 Slide 15 / 133 Change in Momentum Suppose that there is an event that changes an object's momentum. Impulse-Momentum Equation from p0 - the initial momentum (just before the event) by Δp - the change in momentum to pf - the final momentum (just after the event) Return to Table of Contents Slide 16 / 133 The equation for momentum change is: Slide 17 / 133 Momentum Change = Impulse Momentum change equation: Newton's First Law tells us that the velocity (and so the momentum) of an object won't change unless the object is affected by an external force. Look at the above equation. Can you relate Newton's First Law to the Δp term? Δp would represent the external force. Slide 18 / 133 Slide 19 / 133 SI Unit for Impulse There no special unit for impulse. We use the product of the units of force and time. force x time N s Recall that N=kg m/s2, so N s=kg m/s2 x s = kg m/s This is also the unit for momentum, which is a good thing since Impulse is the change in momentum.

5 Slide 20 / There is a battery powered wheeled cart moving towards you at a constant velocity. You want to apply a force to the cart to move it in the opposite direction. Which combination of the following variables will result in the greatest change of momentum for the cart? Select two answers. Slide 20 () / There is a battery powered wheeled cart moving towards you at a constant velocity. You want to apply a force to the cart to move it in the opposite direction. Which combination of the following variables will result in the greatest change of momentum for the cart? Select two answers. A Increase the applied force. B Increase the time that the force is applied. A, B B Increase the time that the force is applied. C Maintain the same applied force. C Maintain the same applied force. D Decrease the time that the force is applied. D Decrease the time that the force is applied. Slide 21 / From which law or principle is the Impulse-Momentum equation derived from? A Increase the applied force. Slide 21 () / From which law or principle is the Impulse-Momentum equation derived from? A Conservation of Energy. B Newton's First Law. B Newton's First Law. C Newton's Second Law. C Newton's Second Law. D Conservation of Momentum. D Conservation of Momentum. A Conservation of Energy. C 8 Can the impulse applied to an object be negative? Why or why not? Give an example to explain your answer. Students type their answers here Slide 22 () / Can the impulse applied to an object be negative? Why or why not? Give an example to explain your answer. Students type their answers here Slide 22 / 133 Yes, since Impulse is a vector quantity, it has magnitude and direction. Direction can be described as negative. An example would be if a baseball is thrown in the positive direction, and you are catching it with a mitt. The mitt changes the baseball's momentum from positive to zero - hence applying a negative impulse to it.

6 Slide 23 / 133 Slide 24 / 133 Effect of Collision Time on Force Impulse = F(# t ) = F(# t ) Since force is inversely proportional to Δt, changing the t of a given impulse by a small amount can greatly change the force exerted on an object! F (newtons) Every Day Applications Impulse = F(# t ) = F(# t ) The inverse relationship of Force and time interval leads to many interesting applications of the Impulse-Momentum equation to everyday experiences such as: car structural safety design car air bags landing after parachuting martial arts hitting a baseball catching a basebal t (seconds) Slide 25 / 133 Slide 26 / 133 Every Day Applications Car Air Bags I= F# t = # p I= F# t = # p Let's analyze two specific cases from the previous list: car air bags hitting a baseball Whenever you have an equation like the one above, you have to decide which values will be fixed and which will be varied to determine the impact on the third value. For the car air bags, we'll fix Δp, vary Δt and see its impact on F. For the bat hitting a ball, we'll fix F, varyδt and see the impact on Δp. Slide 27 / 133 In the Dynamics unit of this course, it was shown how during an accident, seat belts protect passengers from the effect of Newton's First Law by stopping the passenger with the car, and preventing them from striking the dashboard and window. They also provide another benefit explained by the ImpulseMomentum equation. But, this benefit is greatly enhanced by the presence of air bags. Can you see what this benefit is? Slide 28 / 133 Car Air Bags Car Air Bags I= F# t = # p I= F# t = # p The seat belt also increases the time interval that it takes the passenger to slow down - there is some play in the seat belt, that allows you to move forward a bit before it stops you. The Air bag will increase that time interval much more than the seat belt by rapidly expanding, letting the passenger strike it, then deflating. Earlier it was stated that for the Air bag example, Δp would be fixed and Δt would be varied. So, we've just increased Δt. Why is Δp fixed? Δp is fixed, because as long as the passenger remains in the car, the car (and the passengers) started with a certain velocity and finished with a final velocity of zero, independent of seat belts or air bags. Rearranging the equation, we have: F represents the Force delivered to the passenger due to the accident.

7 Slide 29 / 133 Slide 30 / If a car is in a front end collision, which of the below factors will help reduce the injury to the driver and passengers? Select two answers. Car Air Bags I= F# t = # p A An absolutely rigid car body that doesn't deform. Since Δp is fixed, by extending the time interval (Δt increases) that it takes a passenger to come to rest (seat belt and air bag), the force, F delivered to the passenger is smaller. Less force on the passenger means less physical harm. Also, another benefit needs a quick discussion of Pressure. B Deployment of an air bag for each adult in the car. C Deployment of an air bag only for the driver. D The proper wearing of a seatbelt or child seat for each person in the car. Pressure is Force per unit area. By increasing the area of the body that feels the force (the air bag is big), less pressure is delivered to parts of the body - reducing the chance of puncturing the body. Also good. Slide 30 () / If a car is in a front end collision, which of the below factors will help reduce the injury to the driver and passengers? Select two answers. Slide 31 / Which of the following variables are fixed when a moving car strikes a barrier and comes to rest? A Force delivered to the car. B Deployment of an air bag for each adult in the car. B Interval of time for the collision. C Deployment of an air bag only for the driver. C Momentum change. B, D D The proper wearing of a seatbelt or child seat for each person in the car. D Acceleration of the car. A An absolutely rigid car body that doesn't deform. Slide 31 () / Which of the following variables are fixed when a moving car strikes a barrier and comes to rest? B Interval of time for the collision. Hitting a Baseball I= F# t = # p A Force delivered to the car. C Momentum change. Slide 32 / 133 C D Acceleration of the car. Now, we're going to take a case where there is a given force, and the time interval will be varied to find out what happens to the change of momentum. This is different from the Air Bag example just worked (Δp was constant, Δt was varied, and its impact on F was found). Assume a baseball batter swings with a given force that is applied over the interval of the ball in contact with the bat. The ball is approaching the batter with a positive momentum. What is the goal of the batter?

8 Slide 33 / 133 Slide 34 / 133 Hitting a Baseball Hitting a Baseball I= F# t = # p I= F# t = # p The batter wants to hit the ball and get the largest Δp possible for his force which depends on his strength and timing. The greater the Δp, the faster the ball will fly off his bat, which will result in it going further, hopefully to the seats. In this case, the batter wants to maximize the time that his bat (which is providing the force) is in contact with the ball. This means he should follow through with his swing. Hitting a baseball is way more complex than the analysis that will follow. If you're interested in more information, please check out the book, The Physics of Baseball, written by Robert Adair, a Yale University physicist. The batter needs to apply a large impulse to reverse the ball's large momentum from the positive direction (towards the batter) to the negative direction (heading towards the center field bleachers). Slide 35 / 133 Every Day Applications Slide 36 / An external force of 25 N acts on a system for 10.0 s. What is the magnitude of the impulse delivered to the system? I= F# t = # p Now, discuss the other examples. Make sure you decide which object in the collision is more "affected" by the force or the change in momentum, and which variables are capable of being varied. Consider the Air Bag example - the car experiences the same impulse as the passenger during an accident, but a car is less valuable than a human being - so it is more important for the passenger that less force is delivered to his body - and more force is absorbed by the car. car structural safety design landing after parachuting martial arts catching a basebal Slide 36 () / 133 An external force of 25 N acts on a system for 10.0 s. What is the magnitude of the impulse delivered to the system? 11 F = 25 N Δt = 10 s I = F Δt = (25N) (10 s) = 250 N s Slide 37 / In the previous problem, an external force of 25 N acted on a system for 10.0 s. The impulse delivered was 250 N-s. What was the change in momentum of the system?

9 Slide 37 () / Slide 38 / 133 In the previous problem, an external force of 25 N acted on a system for 10.0 s. The impulse delivered was 250 N-s. What was the change in momentum of the system? 13 The momentum change of an object is equal to the. A force acting on it B impulse acting on it C velocity change of the object D object's mass times the force acting on it I = 250 N s I = Δp Δp = 250 N s Slide 38 () / Slide 39 / 133 The momentum change of an object is equal to the. 14 A increase the force with which a passenger hits the dashboard A force acting on it B increase the duration (time) of the passenger's impact B impulse acting on it C decrease the momentum of a collision C velocity change of the object D object's mass times the force acting on it Air bags are used in cars because they: D decrease the impulse in a collision B B Slide 39 () / Air bags are used in cars because they: A increase the force with which a passenger hits the dashboard One car crashes into a concrete barrier. Another car crashes into a collapsible barrier at the same speed. What is the difference between the two crashes? Select two answers. B increase the duration (time) of the passenger's impact C decrease the momentum of a collision D decrease the impulse in a collision 14 Slide 40 / 133 A change in momentum B force on the car B By increasing the amount of time during the collision, the force required to reduce the passenger's momentum is reduced. This in turn reduces or prevents injury. B C impact time D final momentum

10 Slide 40 () / One car crashes into a concrete barrier. Another car crashes into a collapsible barrier at the same speed. What is the difference between the two crashes? Select two answers. Slide 41 / In order to increase the final momentum of a golf ball, the golfer should: Select two answers: A maintain the speed of the golf club after the impact (follow through). A change in momentum B force on the car B Hit the ball with a greater force. B, C C Decrease the time of contact between the club and the ball. C impact time D Decrease the initial momentum of the golf club. D final momentum Slide 41 () / In order to increase the final momentum of a golf ball, the golfer should: Select two answers: Slide 42 / An external force acts on an object for s. During that time the object's momentum increases by 400 kg-m/s. What was the magnitude of the force? A maintain the speed of the golf club after the impact (follow through). B Hit the ball with a greater force. C Decrease the time of contact between the club and the ball. A, B D Decrease the initial momentum of the golf club. Slide 42 () / 133 An external force acts on an object for s. During that time the object's momentum increases by 400 kg-m/s. What was the magnitude of the force? Δt = s 17 Δp = 400 kg m/s I = FΔt = Δp F = Δp/Δt = (400 kg m/s)/(0.002 s) = 200,000 N Slide 43 / A 50,000 N force acts for s on a 2.5 kg object that was initially at rest. What is its final velocity?

11 Slide 43 () / A 50,000 N force acts for s on a 2.5 kg object that was initially at rest. What is its final velocity? Slide 44 / A 1200 kg car slows from 40.0 m/s to 5.0 m/s in 2.0 s. What force was applied by the brakes to slow the car? F = 50,000 N Δt = 0.03 s m = 2.5 kg v0 = 0 FΔt = Δp ; Δp = mδv FΔt = mδv ; Δv = (vf-v0) FΔt = m(vf-v0) = mvf (since v0 = 0) vf = (F/m) Δt [This object pull tab]s) = 600 m/s = (50,000 N)/(2.5 kg) isxa(0.03 Slide 44 () / A 1200 kg car slows from 40.0 m/s to 5.0 m/s in 2.0 s. What force was applied by the brakes to slow the car? Slide 45 / 133 Graphical Interpretation of Impulse Graphs are a very valuable method to solve physics problems. So far, we've dealt with a constant force exerted over a given time interval. But forces are not always constant - most of the time, they are changing over time. Δt = 2 s m = 1200 kg v0 = 40 m/s vf = 5 m/s In the baseball example, the force of the bat starts out very small as it makes initial contact with the ball. It then rapidly rises to its maximum value, then decreases as the ball leaves the bat. This would be a very difficult problem to handle without a graph. FΔt = Δp ; Δp = mδv FΔt = m(vf-v0) F = m(vf-v0) /Δt = (1200 kg)(5 m/s -40 m/s)/(2 s) = 21,000 N Slide 46 / 133 Start with representing a constant force over a time interval, Δt, and plot Force vs. time on a graph. Slide 47 / 133

12 Slide 48 / Using the F-t graph shown, what is the change in momentum during the time interval from 0 to 6 s? Slide 48 () / Using the F-t graph shown, what is the change in momentum during the time interval from 0 to 6 s? p = area under the graph from 0 to 6 seconds. = area of the triangle from 0 to 3 seconds + area of the rectangle from 3 to 6 seconds = ½ (30 N) (3 s) + (30 N)(3 s) =135 N s or 135 kg m/s Slide 49 / A 5 kg object with an initial velocity of 3 m/s experiences the force shown in the graph. What is its velocity at 6 s? Slide 49 () / A 5 kg object with an initial velocity of 3 m/s experiences the force shown in the graph. What is its velocity at 6 s? From the last question p = 135 kg m/s m = 5 kg, v0 = 3 m/s p0 = mv0 = (5 kg)(3 m/s) =15 kg m/s pf = p0 + p = ( ) kg m/s = 150 kg m/s pf = mvf or vf=pf/m = (150 kg m/s) / (5kg) = 30 m/s 22 Using the F-t graph shown, what is the change in momentum during the time interval from 6 to 10 s? Slide 50 () / Using the F-t graph shown, what is the change in momentum during the time interval from 6 to 10 s? p = area under the graph from 6-10 s Slide 50 / 133 = area of the triangle from 6 to 7 s + area of the rectangle from 6 to 7 s + area of the rectangle from 7 to 10 s = ½ (20 N) (1 s)+(10 N)(1 s)+(10 N)(3 s) =50 N s or 50 kg m/s

13 Slide 51 / 133 Slide 52 / 133 The Momentum of a System of Objects If a system contains more than one object, its total momentum is the vector sum of the momenta of those objects. The Momentum of a System of Objects p syste m = # p p s y s t e m = p 1 + p 2 + p p s y s t e m = m 1 v 1 + m 2 v 2 + m 3 v Return to Table of Contents It's critically important to note that momenta add as vectors, not as scalars. Slide 53 / 133 Slide 54 / 133 The Momentum of a System of Objects Example Determine the momentum of a system of two objects:, has a mass of 6 kg and a velocity of 13 m/s towards the east and, has a mass of 14 kg and a velocity of 7 m/s towards the west. p s y s t e m = m 1 v 1 + m 2 v 2 + m 3 v In order to determine the total momentum of a system: First: Determine a direction to be considered positive. Assign positive values to momenta in that direction. Assign negative values to momenta in the opposite direction. 6 kg + Then: Add the momenta to get a total momentum of the system. - = 6 kg v1 = 13 m/s psystem = v1 + v2 23 Determine the momentum of a system of two objects:, has a mass of 6 kg and a velocity of 13 m/s towards the east and, has a mass of 14 kg and a velocity of 7 m/s towards the west. = 6 kg v1 = 13 m/s 7 m/s 14 kg East (+) = 14 kg v2 = 7 m/s psystem = p1 + p2 psystem = v1 + v2 14 kg East (+) = 14 kg v2 = 7 m/s psystem = v1 + v2 = (6 kg)(13 m/s) + (14 kg)( 7 m/s) = (78 kg m/s) + ( 98 kg m/s) = 20 kg m/s Slide 55 / 133 Example 13 m/s 7 m/s psystem = p1 + p2 Slide 54 () / kg 13 m/s psystem = v1 + v2 = (6 kg)(13 m/s) + (14 kg)( 7 m/s) = (78 kg m/s) + ( 98 kg m/s) = 20 kg m/s Determine the momentum of a system of two objects:, has a mass of 6.0 kg and a velocity of 20 m/s north and, has a mass of 3 kg and a velocity 20 m/s south.

14 Slide 55 () / Determine the momentum of a system of two objects:, has a mass of 6.0 kg and a velocity of 20 m/s north and, has a mass of 3 kg and a velocity 20 m/s south. Slide 56 / Determine the momentum of a system of two objects: the first has a mass of 8.0 kg and a velocity of 8.0 m/s to the east while the second has a mass of 5.0 kg and a velocity of 15 m/s to the west. = 6 kg v1 = +20 m/s (North) = 3 kg v2 = 20 m/s (South) psystem = v1 + v2 = (6 kg)(20 m/s) + (3 kg)( 20 m/s) = (120 kg m/s) + ( 60 kg m/s) = 60 kg m/s (magnitude) direction is North Slide 56 () / Determine the momentum of a system of two objects: the first has a mass of 8.0 kg and a velocity of 8.0 m/s to the east while the second has a mass of 5.0 kg and a velocity of 15 m/s to the west. Slide 57 / Determine the momentum of a system of three objects: The first has a mass of 7.0 kg and a velocity of 23 m/s north; the second has a mass of 9.0 kg and a velocity of 7 m/s north; and the third has a mass of 5.0 kg and a velocity of 42 m/s south. = 8 kg v1 = +8 m/s (East) = 5 kg v2 = 15 m/s (West) psystem = v1 + v2 = (8 kg)(8 m/s) + (5 kg)( 15 m/s) = (64 kg m/s) + ( 75 kg m/s) = 11 kg m/s (West) Slide 57 () / Slide 58 / 133 Determine the momentum of a system of three objects: The first has a mass of 7.0 kg and a velocity of 23 m/s north; the second has a mass of 9.0 kg and a velocity of 7 m/s north; and the third has a mass of 5.0 kg and a velocity of 42 m/s south. Conservation of Momentum Return to Table of Contents

15 Slide 59 / 133 Slide 60 / 133 Conservation Laws Momentum is Conserved Some of the most powerful concepts in science are called "conservation laws". This was covered in detail in the Energy unit - please refer back to that for more detail. Here is a summary. In the last unit we learned that energy is conserved. Like energy, momentum is a conserved property of nature. This means: Conservation laws: Momentum is not created or destroyed. apply to closed systems - where the objects only interact with each other and nothing else. The total momentum in a closed system is always the same. enable us to solve problems without worrying about the details of an event. The only way the momentum of a system can change is if momentum is added or taken away by an outside force. Slide 61 / 133 Slide 62 / 133 Conservation of Momentum We will use the Conservation of Momentum to explain and predict the motion of a system of objects. As with energy, it will only be necessary to compare the system at two times: just before and just after an event. Slide 63 / 133 Slide 64 / Why don't the internal forces on a system change the momentum of the system? Students type their answers here

16 Slide 64 () / Why don't the internal forces on a system change the momentum of the system? Students type their answers here Slide 65 / An external, positive force acts on a system of objects. Which of the following are true? Select two answers. A The velocity of the system remains the same. B The velocity of the system increases. The internal forces result from the interaction of the objects within the system. By Newton's Third Law, these are all action-reaction pairs, and viewed from outside the system, all cancel out. Thus there is no net force acting on the system and the momentum is conserved. C The momentum of the system decreases. D The momentum of the system increases. Slide 65 () / Slide 66 / 133 An external, positive force acts on a system of objects. Which of the following are true? Select two answers. A The velocity of the system remains the same. B The velocity of the system increases. Types of Collisions C The momentum of the system decreases. B, D D The momentum of the system increases. Return to Table of Contents Slide 67 / 133 Types of Collisions Objects in an isolated system can interact with each other in two basic ways: They can collide. If they are stuck together, they can explode (push apart). In an isolated system both momentum and total energy are conserved. But the energy can change from one form to another. Conservation of momentum and change in kinetic energy can help predict what will happen in these events. Slide 68 / 133 Types of Collisions We differentiate collisions and explosions by the way the energy changes or does not change form. inelastic collisions: two objects collide, converting some kinetic energy into other forms of energy such as potential energy, heat or sound. elastic collisions: two objects collide and bounce off each other while conserving kinetic energy - energy is not transformed into any other type. explosions: an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy.

17 Slide 69 / 133 Slide 70 / 133 Inelastic Collisions Elastic Collisions There are two types of Inelastic Collisions. perfect inelastic collisions: two objects collide, stick together and move as one mass after the collision, transferring kinetic energy into other forms of energy. general inelastic collisions: two objects collide and bounce off each other, transferring kinetic energy into other forms of energy. There is really no such thing as a perfect elastic collision. During all collisions, some kinetic energy is always transformed into other forms of energy. But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions. In chemistry, the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law. Other examples include a steel ball bearing dropping on a steel plate, a rubber "superball" bouncing on the ground, and billiard balls bouncing off each other. Slide 71 / 133 Slide 72 / 133 Explosions Explosions A firecracker is an example of an explosion. The chemical potential energy inside the firecracker is transformed into kinetic energy, light and sound. In both an inelastic collision and an explosion, kinetic energy is transformed into other forms of energy - such as potential energy. But they are time reversed! A cart with a compressed spring is a good example. When the spring is against a wall, and it is released, the cart starts moving - converting elastic potential energy into kinetic energy and sound. An inelastic collision transforms kinetic energy into other forms of energy, such as potential energy. Think for a moment - can you see a resemblance between this phenomenon and either an elastic or inelastic collision? Thus, the equations to predict their motion will be inverted. An explosion changes potential energy into kinetic energy. The next slide summarizes the four types of collisions and explosions. Slide 73 / 133 Slide 74 / 133 Collisions and Explosions 28 Momentum Conserved? Kinetic Energy Conserved? Objects bounce off each other Yes No. Kinetic energy is converted to other forms of energy Perfect Inelastic Collision Objects stick together Yes No. Kinetic energy is converted to other forms of energy Elastic Collision Objects bounce off each other Yes Yes Explosion Object or objects break apart Yes No. Release of potential energy increases kinetic energy Event Description General Inelastic Collision In collisions momentum is conserved. A Elastic B Inelastic C All

18 Slide 74 () / 133 In collisions momentum is conserved. 29 In collisions kinetic energy is conserved. A Elastic A Elastic B Inelastic B Inelastic C All C All 28 Slide 75 / 133 C Slide 75 () / In collisions kinetic energy is conserved. Slide 76 / In an inelastic collision, kinetic energy is transformed into what after the collision? Select two answers. A Elastic A Nothing, kinetic energy is conserved. C All B More kinetic energy. B Inelastic C Thermal energy. A D Light energy. Slide 76 () / 133 In an inelastic collision, kinetic energy is transformed into what after the collision? Select two answers. A Nothing, kinetic energy is conserved. B More kinetic energy. C Thermal energy. D Light energy. 30 Slide 77 / 133 Conservation of Momentum During a collision or an explosion, measurements show that the total momentum of a closed system does not change. The diagram below shows the objects approaching, colliding and then separating. A m Av A m Bv B B +x the prime means "after" C, D A B m Bv B' m Av A' A B If the measurements don't show that the momentum is conserved, then this would not be a valid law. Fortunately they do, and it is!

19 Slide 78 / 133 Slide 79 / Slide 79 () / A 13,500 kg railroad freight car travels on a level track at a speed of 4.5 m/s. It collides and couples with a 25,000 kg second car, initially at rest and with brakes released. No external force acts on the system. What is the speed of the two cars after colliding? A 13,500 kg railroad freight car travels on a level track at a speed of 4.5 m/s. It collides and couples with a 25,000 kg second car, initially at rest and with brakes released. No external force acts on the system. What is the speed of the two cars after colliding? Slide 80 / A cannon ball with a mass of kg flies in horizontal direction with a speed of 250 m/s and strikes a ship initially at rest. The mass of the ship is 15,000 kg. Find the speed of the ship after the ball becomes embedded in it. v1'+v2' +) v' +) 0kg)(4.5m/s) / (13,500+25,000)kg irection as the first car's initial velocity Slide 80 () / A cannon ball with a mass of kg flies in horizontal direction with a speed of 250 m/s and strikes a ship initially at rest. The mass of the ship is 15,000 kg. Find the speed of the ship after the ball becomes embedded in it. v1'+v2' +) v' +) )(250m/s) / (100+15,000)kg direction as cannon ball's initial velocity Slide 81 / A 40 kg girl skates at 5.5 m/s on ice toward her 70 kg friend who is standing still, with open arms. As they collide and hold each other, what is their speed after the collision?

20 Slide 81 () / Slide 82 / 133 A 40 kg girl skates at 5.5 m/s on ice toward her 70 kg friend who is standing still, with open arms. As they collide and hold each other, what is their speed after the collision? Explosions In an explosion, one object breaks apart into two or more pieces (or coupled objects break apart), moving afterwards as separate objects. To make the problems solvable at this math level, we will assume: the object (or a coupled pair of objects) breaks into two pieces. v1'+v2' 1 the explosion is along the same line as the initial velocity. +) v' +) (5.5m/s) / (40+70)kg ction as the 40kg girls's initial This object is a pull tab] Slide 83 / 133 Slide 84 / Slide 84 () / 133 A 5 kg cannon ball is loaded into a 300 kg cannon. When the cannon is fired, it recoils at 5 m/s. What is the cannon ball's velocity after the explosion? = 5kg = 300kg 34 v1 = v2 = v = 0 v2' = 5m/s (+) v = v1'+v2' 0 = v1'+v2' v1' = -v2' v1' = -v2'/ = -(300kg)(5m/s) / (5kg) = - 300m/s A 5 kg cannon ball is loaded into a 300 kg cannon. When the cannon is fired, it recoils at 5 m/s. What is the cannon ball's velocity after the explosion? Slide 85 / Two railcars, one with a mass of 4000 kg and the other with a mass of 6000 kg, are at rest and stuck together. To separate them a small explosive is set off between them. The 4000 kg car is measured travelling at 6 m/s. How fast is the 6000 kg car going?

21 Slide 85 () / 133 Slide 86 / 133 Two railcars, one with a mass of 4000 kg and the other with a mass of 6000 kg, are at rest and stuck together. To separate them a small explosive is set off between them. The 4000 kg car is measured travelling at 6 m/s. How fast is the 6000 kg car going? 35 Elastic Collisions In an elastic collision, two objects collide and bounce off each other, as shown below, and both momentum and kinetic energy are conserved. This will give us two simultaneous equations to solve to predict their motion after the collision. (+) v = v1'+v2' Before (moving towards) 0 = v1'+v2' v1' = -v2' pa=mava v1' = -v2'/ = -(4000kg)(6m/s) / (6000kg) A = -4m/s pb=mbvb After (moving apart) pa'=mava' B A pb'=mbvb' B Slide 87 / 133 Slide 88 / 133 Slide 89 / 133 Slide 90 / 133 Properties of Elastic Collisions Elastic Collision Simultaneous Equations Conservation of Momentum Conservation of Kinetic Energy m 1v 1 + m 2v 2 = m 1v 1' + m 2v 2'½ m 1v 1 + ½ m 2v 2 = ½ m 1v 1' + ½ m 2v 2' m 1 v 1 - m 1 v 1 ' = m 2 v 2 ' - m 2 v 2 m 1 v m 2 v 2 2 = m 1 v 1 '2 + m 2 v 2 '2 m 1 v m 1 v 1 '2 = m 2 v 2 '2 - m 2 v 2 2 m 1 (v 1 - v 1 ') = m 2 (v 2 ' - v 2 ) m 1 (v v 1 '2 ) = m 2 (v 2 '2 - v 2 2 ) By solving the conservation of momentum and constant kinetic energy equations simultaneously, the following result appeared: v 1 - v 2 = -(v 1 ' - v '2 ) Do you recognize the terms on the left and right of the above equation? And, what does it mean? m 1 (v 1 + v 1 ')(v 1 - v 1 ') = m 2 (v 2 ' + v 2 )(v 2 ' - v 2 ) m 1 (v 1 + v 1 ')(v 1 - v 1 ') = m 2 (v 2 ' + v 2 )(v 2 ' - v 2 ) m 1 (v 1 - v 1 ') = m 2 (v 2 ' - v 2 ) v 1 + v 1' = v 2' + v 2 v 1 - v 2 = -(v 1 ' - v '2 ) The terms are the relative velocities of the two objects before and after the collision. It means that for all elastic collisions - regardless of mass - the relative velocity of the objects is the same before and after the collision.

22 Slide 91 / Two objects have an elastic collision. Before they collide, they are approaching with a velocity of 4 m/s relative to each other. With what velocity do they move apart from one another after the collision? Slide 91 () / Two objects have an elastic collision. Before they collide, they are approaching with a velocity of 4 m/s relative to each other. With what velocity do they move apart from one another after the collision? 4 m/s away from each other v1 -v2 = -(v1 '-v2 ') The difference in the initial velocities is the same as the negative difference of the final velocities. Or, the relative velocity between the objects before the collision is equal to the negative of the relative velocity between the objects after the collision. Slide 92 / 133 Two objects have an elastic collision. Object, has an initial velocity of +4.0 m/s and has a velocity of -3.0 m/s. After the collision, has a velocity of 1.0 m/s. What is the velocity of? 37 Two objects have an elastic collision. Object, has an initial velocity of +4.0 m/s and has a velocity of -3.0 m/s. After the collision, has a velocity of 1.0 m/s. What is the velocity of? 37 Slide 92 () / 133 v1 = 4 m/s v2 = -3 m/s v1' = 1 m/s v2 ' =? v1-v2 = -(v1'-v2') v2' = v1+v1'-v2 v2' = 4m/s + 1m/s - (-3m/s) v2' = 8 m/s Slide 93 / 133 Two objects have an elastic collision. Object, has an initial velocity of +6.0 m/s and has a velocity of 2.0 m/s. After the collision, has a velocity of 1.0 m/s. What is the velocity of? 38 Two objects have an elastic collision. Object, has an initial velocity of +6.0 m/s and has a velocity of 2.0 m/s. After the collision, has a velocity of 1.0 m/s. What is the velocity of? v1 = 6 m/s v2 = 2 m/s v1 ' = 1 m/s v2 ' =? 38 Slide 93 () / 133 v1-v2 = -(v1'-v2') v2 ' = v1 +v1 '-v2 v2 ' = 6m/s + 1m/s - (2m/s) v2 ' = 5 m/s

23 Slide 94 / 133 Slide 95 / 133 Slide 96 / 133 Slide 97 / 133 Slide 98 / 133 Slide 99 / 133

24 Slide 101 / 133 Slide 102 / 133 Slide 103 / 133 Slide 104 / 133 Slide 104 () / 133 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest. The velocity of the bowling ball is virtually unaffected by the collision. What will be the speed of the ping pong ball? 39 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest. The velocity of the bowling ball is virtually unaffected by the collision. What will be the speed of the ping pong ball? 39 Slide 100 / 133 v1 = +v v2 = 0 v1' = +v v2' =? v1-v2 = -(v1'-v2') v2' = v1+v1'-v2 v2' = v + v - 0 v2' = 2 v (ping pong ball's speed is twice that of the bowling ball)

25 Slide 105 / 133 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of -2v. The bat barely changes velocity during the collision. How fast is the baseball going after it's hit? 40 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of -2v. The bat barely changes velocity during the collision. How fast is the baseball going after it's hit? v1 = +v v2 = -2v v1' = +v v2 ' =? 40 Slide 105 () / 133 v1-v2 = -(v1'-v2') v2 ' = v1 +v1 '-v2 v2 ' = v + v - (-2v) v2 ' = 4 v Slide 106 / Two objects with identical masses have an elastic collision: the initial velocity of is +6.0 m/s and is -3.0 m/s. What is the velocity of after the collision? Slide 106 () / Two objects with identical masses have an elastic collision: the initial velocity of is +6.0 m/s and is -3.0 m/s. What is the velocity of after the collision? v1 = +6m/s v2 = -3m/s v1' =? v2' =? When identical mass objects experience an elastic collision, they swap their initial velocities: v1' = v2 = -3.0 m/s v2' = v1 = 6.0 m/s So the velocity of is -3.0 m/s. 42 Two objects with identical masses have an elastic collision: the initial velocity of is +6.0 m/s and is -3.0 m/s. What is the velocity of after the collision? Slide 107 () / Two objects with identical masses have an elastic collision: the initial velocity of is +6.0 m/s and is -3.0 m/s. What is the velocity of after the collision? v1 = +6m/s v2 = -3m/s v1' =? v2' =? Slide 107 / 133 When identical mass objects experience an elastic collision, they swap their initial velocities: v1' = v2 = -3.0 m/s v2' = v1 = 6.0 m/s So the velocity of is 6.0 m/s.

26 Slide 108 / 133 Slide 108 () / A golf ball is hit against a solid cement wall, and experiences an elastic collsion. The golf ball strikes the wall with a velocity of +35 m/s. What velocity does it rebound with? 43 A golf ball is hit against a solid cement wall, and experiences an elastic collsion. The golf ball strikes the wall with a velocity of +35 m/s. What velocity does it rebound with? When a light object strikes a very massive object, the light object rebounds with the opposite velocity in this case, the golf ball will leave the wall with a velocity of -35 m/s. Slide 109 / 133 Slide 110 / 133 Conservation of Momentum in Two Dimensions Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame Collisions in Two Dimensions This, of course also applies to three dimensions, but we'll stick with two for this chapter! Return to Table of Contents Slide 111 / 133 This means that the momentum conservation equation p = p ' can be solved independently for each component: Slide 112 / 133 Example: Collision with a Wall m Consider the case of a golf ball colliding elastically with a hard wall, rebounding with the same velocity, where its angle of incidence equals its angle of reflection. p y' Is momentum conserved in this problem? py p' p x' θ px θ m p The solid lines represent the momentum of the ball (blue - prior to collision, red - after the collision). The dashed lines are the x and y components of the momentum vectors. Example: Collision with a Wall m Momentum is not conserved! An external force from the wall is being applied to the ball in order to reverse its direction in the x axis. However, since we have an elastic collision, the ball bounces off the wall with the same speed that it struck the wall. Hence, the magnitude of the initial momentum and the final momentum is equal: p y' p' p x' θ px θ py m p Now it's time to resolve momentum into components along the x and y axis.

27 Slide 114 / 133 Slide 115 / 133 Slide 115 () / 133 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially. What was the change in momentum of the ball? m A -mv B -2mv C -mv cosθ p y' m A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially. What was the change in momentum of the ball? m A -mv p' p x' θ p y' D -2mv cosθ 44 B -2mv p x' θ C -mv cosθ p 44 Slide 113 / 133 p y' p y' m D -2mv cosθ p' p x' θ D p x' θ p Slide 116 / 133 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially. What was the change in momentum of the ball in the y direction? m A -mv B 0 C mv D 2mv p y' p' p x' θ p y' m p x' θ p 45 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially. What was the change in momentum of the ball in the y direction? m A -mv B 0 C mv D 2mv 45 Slide 116 () / 133 p y' p y' m p' p x' θ B p x' θ p

28 Slide 117 / 133 Slide 118 / 133 General Two Dimensional Collisions General Two Dimensional Collisions Before After p2 = 0 p2 =? This is not a head on collision - note how heads off with a y component of velocity after it strikes. Also, did you see how we rotated the coordinate system so the x axis is horizontal? We'll now consider the more general case of two objects moving in random directions in the x-y plane and colliding. Since there is no absolute reference frame, we'll line up the x-axis with the velocity of one of the objects. To reduce the amount of variables that must be specified, we will also assume that mass 2 is at rest. The problem now is to find the momentum of after the collision. Slide 119 / 133 Slide 120 / 133 General Two Dimensional Collisions Before After General Two Dimensional Collisions Here is the momentum vector breakdown of mass 1 after the collision: p2 = 0 p2 =? This will be done by looking at the vectors first - momentum must be conserved in both the x and the y directions.? needs to have a component in the y direction to sum to zero with 's final y momentum. And it needs a component in the x direction to add to 's final x momentum to equal the initial x momentum of : Since the momentum in the y direction is zero before the collision, it must be zero after the collision. and this is the final momentum for mass 2 by vectorially adding the final px and py. And, the value that has for momentum in the x direction must be shared between both objects after the collision - and not equally - it will depend on the masses and the separation angle. Slide 121 / 133 After the collision shown below, which of the following is the most likely momentum vector for the blue ball? A before 46 After the collision shown below, which of the following is the most likely momentum vector for the blue ball? after? before after? A B B C C D D E E 46 Slide 121 () / 133 D

29 Slide 122 / 133 Slide 123 / 133 General Two Dimensional Collisions General Two Dimensional Collisions Now that we've seen the vector analysis, let's run through the algebra to find the exact velocity (magnitude and direction) that leaves with after the collision. after before 20.0 kg-m/s 12.0 kg-m/s 60.0 Slide 125 / 133 General Two Dimensional Collisions General Two Dimensional Collisions 12.0 kg-m/s θ y-direction θ ball 53.1 pin kg-m/s A 5.0 kg bowling ball strikes a stationary bowling pin. After the collision, the ball and the pin move in directions as shown and the magnitude of the pin's momentum is 18.0 kg-m/s. What was the velocity of the ball before the collision? 47 before after p =? p before 1x p1y = 0 = 5 kg? 60.0 Slide 126 () / 133 A 5.0 kg bowling ball strikes a stationary bowling pin. After the collision, the ball and the pin move in directions as shown and the magnitude of the pin's momentum is 18.0 kg-m/s. What was the velocity of the ball before the collision? before 12.0 kg-m/s Now that the x and y components of the momentum of mass 2 have been found, the total final momentum is calculated. Slide 126 / after before 20.0 kg-m/s 60.0 Use Conservation of Momentum in the x and y directions. x direction θ Find: Slide 124 / θ after 12.0 kg-m/s Given: There is a bowling ball with momentum 20.0 kg-m/s that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above. What is the final velocity of the pin? before 20.0 kg-m/s after before 20.0 kg-m/s 2x = 0 p2y = 0? after p'2x = 18cos(30 )kg-m/s p'2y = 18sin(30 )kg-m/s p'1x =? p'1y =? x-direction y-direction p1x + p2x = p'1x + p'2x p1y + p2y = p'1y + p'2y = p'1y +18sin(-30 ) tan(53.1 ) = p1y' / p1x' 0 = p'1y - 9 p'1y = 9 kg-m/s p1x = p'1y/tan(53.1 ) + p'2x p1x = 9/tan(53.1 ) + 18cos(30 ) p1x = 9/ = 22.4 kg-m/s v1x = p1x / = 22.4/5 = 4.48 m/s ball 53.1 pin kg-m/s after

30 Slide 127 / 133 Slide 128 / 133 Perfect Inelastic Collisions in Two Dimensions Perfect Inelastic Collisions in Two Dimensions One common kind of inelastic collision is where two cars collide and stick at an intersection. Before In this situation the two objects are traveling along paths that are perpendicular just prior to the collision. Before mm 12 p2 mm 12 p2 After p' p1 p-conservation in x: θ After p' x θ p' y p' in y: final momentum: p1 final velocity: final direction: Slide 129 / Object A with mass 20.0 kg travels to the east at 10.0 m/s and object B with mass 5.00 kg travels south at 20.0 m/s. They collide and stick together. What is the velocity (magnitude and direction) of the objects after the collision? 48 Object A with mass 20.0 kg travels to the east at 10.0 m/s and object B with mass 5.00 kg travels south at 20.0 m/s. They collide and stick together. What is the velocity (magnitude and direction) of the objects after the collision? Slide 129 () / 133 south of east Slide 130 / 133 Slide 131 / 133 Explosions in Two Dimensions Explosions in Two Dimensions The Black object explodes into 3 pieces (blue, red and green). before: p x = p y = 0 We want to determine the momentum of the third piece. p'2 p'1 p '3 During an explosion, the total momentum is unchanged, since no EXTERNAL force acts on the system. By Newton's Third Law, the forces that occur between the particles within the object will add up to zero, so they don't affect the momentum. if the initial momentum is zero, the final momentum is zero. the third piece must have equal and opposite momentum to the sum of the other two. p'2 p'1 p'3 # after: p '1 x + p '2 x + p '3 x = 0 p '1 y + p '2 y + p '3 y = 0 The Black object explodes into 3 pieces (blue, red and green). We want to determine the momentum of the third piece. In this case the blue and red pieces are moving perpendicularly to each other, so:

31 Slide 132 / Slide 133 / 133 A stationary cannon ball explodes in three pieces. The momenta of two of the pieces is shown below. What is the direction of the momentum of the third piece? 50 A B C D Slide 133 () / 133 A stationary 10.0 kg bomb explodes into three pieces. A 2.00 kg piece moves west at m/s. Another piece with a mass of 3.00 kg moves north with a velocity of m/s. What is the velocity (speed and direction) of the third piece? 50 A stationary 10.0 kg bomb explodes into three pieces. A 2.00 kg piece moves west at m/s. Another piece with a mass of 3.00 kg moves north with a velocity of m/s. What is the velocity (speed and direction) of the third piece?