Chapter 7 Linear Momentum © 2014 Pearson Education, Inc.

Contents of Chapter 7 Momentum and Its Relation to Force Conservation of Momentum Collisions and Impulse Conservation of Energy and Momentum in Collisions Elastic Collisions in One Dimension © 2014 Pearson Education, Inc.

Contents of Chapter 7 Inelastic Collisions Collisions in Two or Three Dimensions Center of Mass (CM) © 2014 Pearson Education, Inc.

Let’s start with everyday language What do you say when a sports team is on a roll? They may not have the lead but they may have ___________ MOMENTUM A team that has momentum is hard to stop.

What is Momentum? An object with a lot of momentum is also hard to stop Momentum = p = mv Units: kg∙m/s m=mass v=velocity Momentum is also a vector (it has direction)

Let’s practice A 1200 kg car drives west at 25 m/s for 3 hours. What is the car’s momentum? Identify the variables: 1200 kg = mass 25m/s, west = velocity 3 hours = time P = mv = 1200 x 25 = 30000 kg m/s, west

How hard is it to stop a moving object? To stop an object, we have to apply a force over a period of time. This is called Impulse Impulse = FΔt Units: N∙s F = force (N) Δt = time elapsed (s)

How hard is it to stop a moving object? Using Newton’s 2nd Law we get FΔt= mΔv Which means Impulse = change in momentum

Why does an egg break or not break? An egg dropped on a tile floor breaks, but an egg dropped on a pillow does not. Why? FΔt= mΔv In both cases, m and Δv are the same. If Δt goes up, what happens to F, the force? Right! Force goes down. When dropped on a pillow, the egg starts to slow down as soon as it touches it. A pillow increases the time the egg takes to stops.

Practice Problem A 57 gram tennis ball falls on a tile floor. The ball changes velocity from -1.2 m/s to +1.2 m/s in 0.02 s. What is the average force on the ball? Identify the variables: Mass = 57 g = 0.057 kg Δvelocity = +1.2 – (-1.2) = 2.4 m/s Time = 0.02 s using FΔt= mΔv F x (0.02 s) = (0.057 kg)(2.4 m/s) F= 6.8 N

Car Crash Would you rather be in a head on collision with an identical car, traveling at the same speed as you, or a brick wall? Assume in both situations you come to a complete stop. Take a guess http://techdigestuk.typepad.com/photos/uncategorized/car_crash.JPG

Car Crash (cont.) Everyone should vote now Raise one finger if you think it is better to hit another car, two if it’s better to hit a wall and three if it doesn’t matter. And the answer is…..

Car Crash (cont.) The answer is… It Does Not Matter! Look at FΔt= mΔv In both situations, Δt, m, and Δv are the same! The time it takes you to stop depends on your car, m is the mass of your car, and Δv depends on how fast you were initially traveling.

FΔt= mΔv Egg Drop connection How are you going to use this in your egg drop? Which of these variables can you control? FΔt= mΔv Which of them do you want to maximize, which do you want to minimize (note: we are looking at the force on the egg. Therefore, m represents the egg mass, not the entire mass of the project)

7-1 Momentum and Its Relation to Force Momentum is a vector symbolized by the symbol p, and is defined as The rate of change of momentum is equal to the net force: This can be shown using Newton’s second law. (7-1) (7-2) © 2014 Pearson Education, Inc.

Can a small sports car ever have the same momentum as a large SUV with three times the mass of the sports car?

Example (Not in your packet) For a top player, a tennis ball may leave the racket on the serve with a speed of 55m/s (about 120 mi/h). If the ball has a mass of 0.060kg and is in contact with the racket for about 4ms(4x10-3s), estimate the average force on the ball. Would this force be large enough to lift a 60kg person? Given: m=0.060kg; t= 4x10-3s; v1=55m/s Formula: p=Ft; mv=Ft Substitution: (0.060kg)(55m/s)=F(4x10-3s) Answer w/unit: 825N This force will be enough because the force required for a 60kg person is 588N (60kg x 9.8m/s2) © 2014 Pearson Education, Inc.

Example 1 Water leaves a hose at a rate of 1.5kg/s with a speed of 20m/s and is aimed at the side of the car, which stops it. (That is we ignore any splashing back.) What is the force exerted by the car? Given: m=1.5kg per second; t= 1s; v1=20m/s; v2=0m/s Formula: ∆p=Ft; p=mv Substitution: (1.5kg)(0-20m/s)=F(1s) Answer w/unit: -30N Why is the force negative? © 2014 Pearson Education, Inc.

Example 2 What if the water splashed back from the car? Would the force on the car be greater or less? The force would be greater.

7-2 Conservation of Momentum During a collision, measurements show that the total momentum does not change: (7-3) © 2014 Pearson Education, Inc.

7-2 Conservation of Momentum More formally, the law of conservation of momentum states: The total momentum of an isolated system of objects remains constant. © 2014 Pearson Education, Inc.

7-2 Conservation of Momentum Momentum conservation works for a rocket as long as we consider the rocket and its fuel to be one system, and account for the mass loss of the rocket. © 2014 Pearson Education, Inc.

Example 3 A 10,000kg railroad car traveling at a speed of 24.0m/s strikes an identical car at rest. If the cars lock together as a result of the collision, what is their common speed afterward? Given: ma&b=10000kg; v1a=24m/s; v1b=0m/s; v2a&b = ? Formula: mav1a + mbv1b = mav2a + mbv2b mav1a + mbv1b = (ma+ mb)v2a&b Substitution: (10000kg)(24m/s)= (10000kg+ 10000kg)v2a&b Answer w/unit: 12.0m/s This is an example of inelastic collision which we will discuss further later. © 2014 Pearson Education, Inc.

Example 4 Calculate the recoil velocity of a 5.0 kg rifle that shoots a 0.050 kg bullet at a speed of 120 m/s. Given: ma=5kg; mb= 0.050kg; v2b=120m/s; v1a&b=0m/s; v2a=? Formula: mav1a + mbv1b = mav2a + mbv2b Substitution: 0=(5kg)v2a + (0.050kg)(120m/s) Answer w/unit: -1.20m/s It is negative because recoil is backwards. © 2014 Pearson Education, Inc.

Conceptual Example (Not in notes) An empty sled is sliding on frictionless ice when Susan drops vertically from a tree down onto the sled. When she lands, does the sled speed up, slow down, or keep the same speed? b) Later: Susan falls sideways off the sled. When she drops off, does the sled speed up, slow down, or keep the same speed? Because Susan falls vertically onto the sled, she has no initial horizontal momentum. Thus the total horizontal momentum afterward equals the momentum of the sled initially. Since the mass of the system (sled + person) has increased, the speed must decrease. At the instant Susan falls off, she is moving with the same horizontal speed as she was while on the sled. At the moment she leaves the sled, she has the same momentum she had an instant before. Because her momentum does not change, neither does the sled’s (total momentum conserved); the sled keep the same speed. © 2014 Pearson Education, Inc.

Example (not in notes) A gun is fired vertically into a 1.40 kg block of wood at rest directly above it. If the bullet has a mass of 21.0 g and a speed of 210 m/s, how high will the block rise into the air after the bullet becomes embedded in it? Find the final speed. Given: ma=1.40kg; mb= 0.021kg; v1b=210m/s; v1a=0m/s; v2a&b=? Formula: mav1b + mav1b = mav2b + mav2b mav1b + mav1b = (ma+ mb)v2a&b Substitution: (0.021kg)(210m/s)= (1.40kg+0.021kg)v2a&b Answer w/unit:3.10m/s Find the height using conservation of energy (all KE is now PEG) 1/2v2 = gh ½(3.10m/s)2= 9.8m/s2h 0.490m = h

7-3 Collisions and Impulse During a collision, objects are deformed due to the large forces involved. Since the force is equal to the change in momentum divided by time, we can write: The definition of impulse: (7-4) (7-5) © 2014 Pearson Education, Inc.

7-3 Collisions and Impulse Since the time of the collision is very short, we need not worry about the exact time dependence of the force, and can use the average force. © 2014 Pearson Education, Inc.

7-3 Collisions and Impulse The impulse tells us that we can get the same change in momentum with a large force acting for a short time, or a small force acting for a longer time. This is why you should bend your knees when you land; why airbags work; and why landing on a pillow hurts less than landing on concrete.

Example 5 a) Calculate the impulse experienced when a 70. kg person lands on firm ground after jumping from a height of 3.0 m. b) Then estimate the average force exerted on the person’s feet by the ground, if the landing is stiff-legged and the body only moves 1.0 cm during impact, and c) if the person bends their legs and the body moves 50. cm during impact. Find the final speed using conservation equation Given: m=70kg; h1=3m; v1=m?; v2=0m/s; h2=0 Formula: ½mv21 + mgh1= ½mv22 + mgh2 (m’s cancel out (same mass) and since v2&h2=0, those parts of the equation cancels) ½v21 + gh1= 0 + 0 Substitution: ½v22 = -(9.8m/s2)(3.0m) Answer w/unit: -7.67m/s Find the impulse using the impulse-momentum theorem. J = m∆v J= 70kg (0- (-7.67m/s)) J=537 kgm/s

Example 5 a) Calculate the impulse experienced when a 70. kg person lands on firm ground after jumping from a height of 3.0 m. b) Then estimate the average force exerted on the person’s feet by the ground, if the landing is stiff-legged and the body only moves 1.0 cm during impact, and c) if the person bends their legs and the body moves 50. cm during impact. b) Find the average force if stiff-legged. Given: m=70kg; v1=-7.67m/s; v2=0m/s Formula: Fnet = Fground – Fgravity; Fground=ma + mg= m(a+ g) (We have to use a kinematic equation to solve for a since we don’t have it.) Fground = m[(v22 – v12) + g] 2d Substitution: (70kg)[(0 – (-7.67)2 + 9.8m/s2] 2(0.01m) Answer w/unit: 2.07 x 105 N

Example 5 a) Calculate the impulse experienced when a 70. kg person lands on firm ground after jumping from a height of 3.0 m. b) Then estimate the average force exerted on the person’s feet by the ground, if the landing is stiff-legged and the body only moves 1.0 cm during impact, and c) if the person bends their legs and the body moves 50. cm during impact. c) Find the average force if legs bent. Given: m=70kg; v1=-7.67m/s; v2=0m/s Formula: Fnet = Fground – Fgravity; Fground=ma + mg= m(a+ g) (We have to use a kinematic equation to solve for a since we don’t have it.) Fground = m[(v22 – v12) + g] 2d Substitution: (70kg)[(0 – (-7.67)2 + 9.8m/s2] 2(0.5m) Answer w/unit: 4.80 x 103 N

7-3 Collisions and Impulse The impulse tells us that we can get the same change in momentum with a large force acting for a short time, or a small force acting for a longer time. This is why you should bend your knees when you land; why airbags work; and why landing on a pillow hurts less than landing on concrete. © 2014 Pearson Education, Inc.

7-4 Conservation of Energy and Momentum in Collisions Momentum is conserved in all collisions. Collisions in which kinetic energy is conserved as well are called elastic collisions, and those in which it is not are called inelastic. © 2014 Pearson Education, Inc.

7-5 Elastic Collisions in One Dimension Here we have two objects colliding elastically. We know the masses and the initial speeds. Since both momentum and kinetic energy are conserved, we can write two equations. This allows us to solve for the two unknown final speeds. v1 – v2 = - (v'1 – v'2) This equation works only in 1-D not 2-D. v= initial; v‘= final For any elastic AND head-on collision, the relative speed of 2 particles after collision Has the same magnitude as before (but opposite direction), regardless of mass! © 2014 Pearson Education, Inc.

Example 6 A billiard ball of mass m moving with speed v collides head-on with a second ball of equal mass at rest. What are the speeds of the two balls after the collision, assuming it is elastic? Given: ma=mb; v2b=?; v1b=0m/s; v2a=? Formula: v1a + v1b = v2a + v2b v1a – v1b = v2b - v2a 2v1a = 2v2b Substitution: Get rid of the 2 and you will see that v1a =v2b Answer w/unit: Since v1a =v2b, then v2a must now be 0m/s since momentum is conserved. (We have two unknowns so I will use the KE conservation eqn. and add them to make one equation. © 2014 Pearson Education, Inc.

Example 7 A proton traveling with mass of 1.01u (unified atomic mass units) traveling with a speed of 3.60 x 104m/s has an elastic head-on collision with a He nucleus. (mass of He=4.00u) initially at rest [tough for a gas molecule!]. What are the velocities of the proton and helium nucleus after the collision? Given: ma= 1.01u; mb=4.00u; v1a= 3.60 x 104m/s v2b=?; v1b=0m/s; v2a=? Formula: mav1a + mbv1b = mav2a + mbv2b v2b - v1a = v2a mav1a = mav2b - mav1a + mbv2b 2mav1a = (ma+ mb)v2b Substitution:2(1.01u)(3.60 x 104m/s)=(1.01u+4.00u)v2b Answer w/unit: v2b= 14500m/s Plug in your answer in for v2b to solve for v2a. v2b - v1a = v2a; 14500m/s- 3.60 x 104m/s=-21500m/s (We have two unknowns so I will use the KE conservation eqn. Solve for a variable in the equation to substitute into the main conservation equation.) © 2014 Pearson Education, Inc.

7-6 Inelastic Collisions With inelastic collisions, some of the initial kinetic energy is lost to thermal or potential energy. It may also be gained during explosions, as there is the addition of chemical or nuclear energy. A completely inelastic collision is one where the objects stick together afterwards, so there is only one final velocity. © 2014 Pearson Education, Inc.

Example 8 A 10,000kg railroad car traveling at a speed of 24.0m/s strikes an identical car at rest. Calculate how much of the initial KE is transformed to thermal or other forms of energy. Given: ma&b=10000kg; v1a=24m/s; v1b=0m/s; v2a&b = 12.0m/s Formula: KE1 = KE2+ E? ½mav1a2 =1/2(ma+ mb)v2a&b2 + other energy form Substitution: 1/2(10000kg)(24m/s)2 =1/2(10000kg+10000kg)12.0m/s2 +E? Answer w/unit: 1440000J © 2014 Pearson Education, Inc.

7-7 Collisions in Two or Three Dimensions Conservation of energy and momentum can also be used to analyze collisions in two or three dimensions, but unless the situation is very simple, the math quickly becomes unwieldy. Here, a moving object collides with an object initially at rest. Knowing the masses and initial velocities is not enough; we need to know the angles as well in order to find the final velocities. © 2014 Pearson Education, Inc.

7-8 Center of Mass In (a), the diver’s motion is pure translation; in (b) it is translation plus rotation. There is one point that moves in the same path a particle would take if subjected to the same force as the diver. This point is called the center of mass (CM). © 2014 Pearson Education, Inc.

7-8 Center of Mass The general motion of an object can be considered as the sum of the translational motion of the CM, plus rotational, vibrational, or other forms of motion about the CM. © 2014 Pearson Education, Inc.

7-8 Center of Mass For two particles, the center of mass lies closer to the one with the most mass: where M is the total mass. © 2014 Pearson Education, Inc.

7-8 Center of Mass The center of gravity is the point where the gravitational force can be considered to act. It is the same as the center of mass as long as the gravitational force does not vary among different parts of the object. © 2014 Pearson Education, Inc.

7-8 Center of Mass The center of gravity can be found experimentally by suspending an object from different points. The CM need not be within the actual object—a doughnut’s CM is in the center of the hole. © 2014 Pearson Education, Inc.

7-9 CM for the Human Body The x’s in the small diagram mark the CM of the listed body segments. © 2014 Pearson Education, Inc.

7-9 CM for the Human Body The location of the center of mass of the leg (circled) will depend on the position of the leg. © 2014 Pearson Education, Inc.

7-9 CM for the Human Body High jumpers have developed a technique where their CM actually passes under the bar as they go over it. This allows them to clear higher bars. © 2014 Pearson Education, Inc.

7-10 Center of Mass and Translational Motion This is particularly useful in the analysis of separations and explosions; the center of mass (which may not correspond to the position of any particle) continues to move according to the net force. © 2014 Pearson Education, Inc.

Summary of Chapter 7 Momentum of an object: Newton’s second law: Total momentum of an isolated system of objects is conserved. During a collision, the colliding objects can be considered to be an isolated system even if external forces exist, as long as they are not too large. Momentum will therefore be conserved during collisions. (7-1) (7-2) © 2014 Pearson Education, Inc.

Summary of Chapter 7 In an elastic collision, total kinetic energy is also conserved. In an inelastic collision, some kinetic energy is lost. In a completely inelastic collision, the two objects stick together after the collision. The center of mass of a system is the point at which external forces can be considered to act. (7-4) © 2014 Pearson Education, Inc.

From the California Standards Test Copyright © 2004 California Department of Education.

From the California Standards Test Copyright © 2004 California Department of Education.

From the California Standards Test Copyright © 2004 California Department of Education.