Lecture Outline Chapter 9 Physics, 4th Edition James S. Walker

Lecture Outline Chapter 9 Physics, 4th Edition James S. Walker
Copyright © 2010 Pearson Education, Inc.

Linear Momentum and Collisions
Chapter 9 Linear Momentum and Collisions

Units of Chapter 9 Linear Momentum Momentum and Newton’s Second Law
Impulse Conservation of Linear Momentum Inelastic Collisions Elastic Collisions

Units of Chapter 9 Center of Mass
Systems with Changing Mass: Rocket Propulsion

9-1 Linear Momentum We have now dealt with the conservation of energy for a single object in translational motion. Now we will deal with the conservation of momentum for 2 or more interacting bodies.

9-1 Linear Momentum Momentum is a vector; its direction is the same as the direction of the velocity.

9-1 Linear Momentum Force is required to change the momentum of an object (to change the velocity in either magnitude or direction)

9-2 Momentum and Newton’s Second Law
Newton’s second law, as we wrote it before: is only valid for objects that have constant mass. Here is a more general form, also useful when the mass is changing:

9-2 Momentum and Newton’s Second Law
This can be shown using Newton’s second law.

9-3 Impulse Having a net force is not enough to cause a change in the motion of an object. A net force must actually be present for some instant of time. A huge force acting for zero seconds accomplishes nothing. A small force acting for a long time can be as effective as a huge force acting for a short time. Impulse Δ momentum

9-3 Impulse Therefore, the same change in momentum may be produced by a large force acting for a short time, or by a smaller force acting for a longer time.

Why use an average force?
9-3 Impulse Why use an average force? At impact, the force jumps from 0 at the moment of contact to a very large value and then drops again in a very short period of time.

Why use an average force?
9-3 Impulse Why use an average force? 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.

9-3 Impulse During a collision, ordinary objects are deformed due to the large forces involved.

Use Impulse for short time forces (ie, hitting a baseball).
Impulse is a vector, in the same direction as the average force. Use Impulse for short time forces (ie, hitting a baseball).

Question Impulse A small beanbag and a bouncy rubber ball are dropped from the same height above the floor. They both have the same mass. Which one will impart the greater impulse to the floor when it hits? a) the beanbag b) the rubber ball c) both the same Answer: b

Question Impulse A small beanbag and a bouncy rubber ball are dropped from the same height above the floor. They both have the same mass. Which one will impart the greater impulse to the floor when it hits? a) the beanbag b) the rubber ball c) both the same Both objects reach the same speed at the floor. However, while the beanbag comes to rest on the floor, the ball bounces back up with nearly the same speed as it hit. Thus, the change in momentum for the ball is greater, because of the rebound. The impulse delivered by the ball is twice that of the beanbag. For the beanbag: Dp = pf – pi = 0 – (–mv ) = mv For the rubber ball: Dp = pf – pi = mv – (–mv ) = 2mv Follow-up: Which one imparts the larger force to the floor?

Question 9.8 Singing in the Rain
A person stands under an umbrella during a rainstorm. Later the rain turns to hail, although the number of “drops” hitting the umbrella per time and their speed remains the same. Which case requires more force to hold the umbrella? a) when it is hailing b) when it is raining c) same in both cases Answer: a

Question 9.8 Singing in the Rain
A person stands under an umbrella during a rainstorm. Later the rain turns to hail, although the number of “drops” hitting the umbrella per time and their speed remains the same. Which case requires more force to hold the umbrella? a) when it is hailing b) when it is raining c) same in both cases When the raindrops hit the umbrella, they tend to splatter and run off, whereas the hailstones hit the umbrella and bounce back upward. Thus, the change in momentum (impulse) is greater for the hail. Because Dp = F Dt, more force is required in the hailstorm. This is similar to the situation with the bouncy rubber ball in the previous question.

Question 9.5a Two Boxes I heavy light F
Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each one for exactly 1 second. Which box has more momentum after the force acts ? a) the heavier one b) the lighter one c) both the same F light heavy Answer: c

Question 9.5a Two Boxes I heavy light Dp = F Dt , F
Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each one for exactly 1 second. Which box has more momentum after the force acts ? a) the heavier one b) the lighter one c) both the same av Dt Dp F = , We know: F light heavy so impulse Dp = Fav Dt. In this case F and Dt are the same for both boxes! Both boxes will have the same final momentum.

Question 9.5b Two Boxes II In the previous question, which box has the larger velocity after the force acts? a) the heavier one b) the lighter one c) both the same Answer: b

Question 9.5b Two Boxes II In the previous question, which box has the larger velocity after the force acts? a) the heavier one b) the lighter one c) both the same The force is related to the acceleration by Newton’s Second Law (F = ma). The lighter box therefore has the greater acceleration and will reach a higher speed after the 1-second time interval. Follow-up: Which box has gone a larger distance after the force acts? Follow-up: Which box has gained more KE after the force acts?

Equations Try using it! Calculate the impulse experienced when a 70 kg person lands on firm ground after jumping from a height of 3.0 m. Estimate the average force exerted on the person’s feet by the ground if the landing is stiff-legged, and again With bent legs. With stiff legs, assume the body moves 1.0 cm during impact, and when the legs are bent, about 50 cm.

9-4 Conservation of Linear Momentum
The net force acting on an object is the rate of change of its momentum: If the net force is zero, the momentum does not change:

During a collision, measurements show that the total momentum does not change. Consider a head-on collision of 2 pool balls in an isolated system.

Each ball approaches with its own velocity. After collision, they each have different velocities and momentum. Total momentum, however, remains unchanged

Internal Versus External Forces: Internal forces act between objects within the system. As with all forces, they occur in action-reaction pairs. As all pairs act between objects in the system, the internal forces always sum to zero: Therefore, the net force acting on a system is the sum of the external forces acting on it.

Furthermore, internal forces cannot change the momentum of a system. However, the momenta of components of the system may change.

An example of internal forces moving components of a system:

Equations Try using it! Calculate the recoil velocity of a 5.0 kg rifle that shoots a kg bullet at a speed of 620 m/s.

Question 9.2a Momentum and KE I
A system of particles is known to have a total kinetic energy of zero. What can you say about the total momentum of the system? a) momentum of the system is positive b) momentum of the system is negative c) momentum of the system is zero d) you cannot say anything about the momentum of the system Answer: c

Question 9.2a Momentum and KE I
A system of particles is known to have a total kinetic energy of zero. What can you say about the total momentum of the system? a) momentum of the system is positive b) momentum of the system is negative c) momentum of the system is zero d) you cannot say anything about the momentum of the system Because the total kinetic energy is zero, this means that all of the particles are at rest (v = 0). Therefore, because nothing is moving, the total momentum of the system must also be zero.

Collisions Collision: two objects striking one another
Time of collision is short enough that external forces may be ignored 3 types of collisions: Elastic Inelastic Perfectly Inelastic

Collisions In elastic collisions: kinetic energy is conserved
Since KE is not lost, no energy can go into heat or sound. No damage caused to either colliding object. momentum is conserved Examples: Atomic or nuclear particles with similar charges. Magnets with similar poles facing each other. //

Energy is lost to damage, heat, and sound.
Collisions Inelastic collision: momentum is conserved but kinetic energy is not // Energy is lost to damage, heat, and sound.

Collisions Completely inelastic collision: Examples:
objects stick together afterwards Greatest damage and loss of energy Examples: Car accident where the cars stick together Train crash where the cars stay linked Bullet that stays in the target (ballistic pendulum)

9-6 Elastic Collisions In elastic collisions, both kinetic energy and momentum are conserved. One-dimensional elastic collision:

9-6 Elastic Collisions When dealing with an elastic collision, you will have 2 equations to use: Conservation of momentum Conservation of kinetic energy You could also write this as: Please remember

9-6 Elastic Collisions This equation is a bit of a pain… let’s make it easier. Rearrange REM difference of 2 squares Assumes initial and final speeds of the objects are NOT equal. Divide by the momentum equation. A MUCH easier equation for conservation of KE.

Equations Try using it! A proton of mass 1.01 u (unified atomic mass units) traveling with a speed of 3.60 x 104 m/s has an elastic head-on collision with a helium nucleus (mHe = 4.00 u) initially at rest. What are the velocities of the proton and the helium nucleus after the collision? Assume the collision takes place in nearly empty space. (1 u = 1.66 x kg) V He -final = 1.45 x 104 m/s V p -final = x 104 m/s

Question 9.10a Elastic Collisions I
Consider two elastic collisions: 1) a golf ball with speed v hits a stationary bowling ball head-on. 2) a bowling ball with speed v hits a stationary golf ball head-on. In which case does the golf ball have the greater speed after the collision? a) situation 1 b) situation 2 c) both the same v 2 1 at rest Answer: b

Question 9.10a Elastic Collisions I
Consider two elastic collisions: 1) a golf ball with speed v hits a stationary bowling ball head-on. 2) a bowling ball with speed v hits a stationary golf ball head-on. In which case does the golf ball have the greater speed after the collision? a) situation 1 b) situation 2 c) both the same v 1 Remember that the magnitude of the relative velocity has to be equal before and after the collision! In case 1 the bowling ball will almost remain at rest, and the golf ball will bounce back with speed close to v. v 2 2v In case 2 the bowling ball will keep going with speed close to v, hence the golf ball will rebound with speed close to 2v.

9-6 Elastic Collisions Two-dimensional collisions can only be solved if some of the final information is known, such as the final velocity of one object:

9-5 Inelastic Collisions
Inelastic collision: momentum is conserved but kinetic energy is not. Therefore, you can use the conservation of momentum equation, but you may be asked to solve for the KE lost. DO NOT CONSERVE KE!

Question 9.2b Momentum and KE II
A system of particles is known to have a total momentum of zero. Does it necessarily follow that the total kinetic energy of the system is also zero? a) yes b) no Answer: b

Question 9.2b Momentum and KE II
A system of particles is known to have a total momentum of zero. Does it necessarily follow that the total kinetic energy of the system is also zero? a) yes b) no Momentum is a vector, so the fact that ptot = 0 does not mean that the particles are at rest! They could be moving such that their momenta cancel out when you add up all of the vectors. In that case, because they are moving, the particles would have non-zero KE.

Question 9.2c Momentum and KE III
Two objects are known to have the same momentum. Do these two objects necessarily have the same kinetic energy? a) yes b) no Answer: b

Question 9.2c Momentum and KE III
Two objects are known to have the same momentum. Do these two objects necessarily have the same kinetic energy? a) yes b) no If object #1 has mass m and speed v and object #2 has mass m and speed 2v, they will both have the same momentum. However, because KE = mv2, we see that object #2 has twice the kinetic energy of object #1, due to the fact that the velocity is squared.

A completely inelastic collision:

Ballistic pendulum: the height h can be found using conservation of mechanical energy after the object is embedded in the block.

Equations Try using it! A 10,000 kg railroad car traveling at 24.0 m/s strikes an identical car at rest. If the cars lock together as a result of the collision, what is their common speed just afterward? Calculate how much of the initial KE is transformed to thermal or other forms of energy.

Question 9.4 Collision Course
a) the car b) the truck c) they both have the same momentum change d) can’t tell without knowing the final velocities A small car and a large truck collide head-on and stick together. Which one has the larger momentum change? Answer: c

Question 9.4 Collision Course
a) the car b) the truck c) they both have the same momentum change d) can’t tell without knowing the final velocities A small car and a large truck collide head-on and stick together. Which one has the larger momentum change? Because the total momentum of the because is conserved, that means that Dp = 0 for the car and truck combined. Therefore, Dpcar must be equal and opposite to that of the truck (–Dptruck) in order for the total momentum change to be zero. Note that this conclusion also follows from Newton’s Third Law. Follow-up: Which one feels the larger acceleration?

Question Watch Out! You drive around a curve in a narrow one-way street at 30 mph when you see an identical car heading straight toward you at 30 mph. You have two options: hit the car head-on or swerve into a massive concrete wall (also head-on). What should you do? a) hit the other car b) hit the wall c) makes no difference d) call your physics prof!! e) get insurance! Answer: c

Question Watch Out! You drive around a curve in a narrow one-way street at 30 mph when you see an identical car heading straight toward you at 30 mph. You have two options: hit the car head-on or swerve into a massive concrete wall (also head-on). What should you do? a) hit the other car b) hit the wall c) makes no difference d) call your physics prof!! e) get insurance! In both cases your momentum will decrease to zero in the collision. Given that the time Dt of the collision is the same, then the force exerted on YOU will be the same!! If a truck is approaching at 30 mph, then you’d be better off hitting the wall in that case. On the other hand, if it’s only a mosquito, well, you’d be better off running him down...

Question 9.16a Crash Cars I a) I
b) II c) I and II d) II and III e) all three If all three collisions below are totally inelastic, which one(s) will bring the car on the left to a complete halt? Answer: e

Question 9.16a Crash Cars I a) I
b) II c) I and II d) II and III e) all three If all three collisions below are totally inelastic, which one(s) will bring the car on the left to a complete halt? In case I, the solid wall clearly stops the car. In cases II and III, because ptot = 0 before the collision, then ptot must also be zero after the collision, which means that the car comes to a halt in all three cases.

Question 9.16b Crash Cars II
If all three collisions below are totally inelastic, which one(s) will cause the most damage (in terms of lost energy)? a) I b) II c) III d) II and III e) all three Answer: c

Question 9.16b Crash Cars II
If all three collisions below are totally inelastic, which one(s) will cause the most damage (in terms of lost energy)? a) I b) II c) III d) II and III e) all three The car on the left loses the same KE in all three cases, but in case III, the car on the right loses the most KE because KE = mv2 and the car in case III has the largest velocity.

For collisions in two dimensions, conservation of momentum is applied separately along each axis:

Collisions in 2D x y In the case shown above

Collisions in 2- or 3-D Problem solving:
Choose the system. If it is complex, subsystems may be chosen where one or more conservation laws apply. Is there an external force? If so, is the collision time short enough that you can ignore it? Draw diagrams of the initial and final situations, with momentum vectors labeled. Choose a coordinate system. Apply momentum conservation; there will be one equation for each dimension. If the collision is elastic, apply conservation of kinetic energy as well. Solve. Check units and magnitudes of result.

Equations Try using it! Billiard ball A moving with speed vA = 3.0 m/s in the +x direction strikes an equal-mass ball B initially at rest. The two balls are observed to move off at 45° to the x axis, ball A above the x axis and ball B below. What are the speeds of the two balls after the collision? 2.1 m/s

Question 9.13a Nuclear Fission I
A uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater momentum? a) the heavy one b) the light one c) both have the same momentum d) impossible to say 1 2 Answer: c

Question 9.13a Nuclear Fission I
A uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater momentum? a) the heavy one b) the light one c) both have the same momentum d) impossible to say 1 2 The initial momentum of the uranium was zero, so the final total momentum of the two fragments must also be zero. Thus the individual momenta are equal in magnitude and opposite in direction.

Question 9.13b Nuclear Fission II
A uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater speed? a) the heavy one b) the light one c) both have the same speed d) impossible to say 1 2 Answer: b

Question 9.13b Nuclear Fission II
A uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater speed? a) the heavy one b) the light one c) both have the same speed d) impossible to say We have already seen that the individual momenta are equal and opposite. In order to keep the magnitude of momentum mv the same, the heavy fragment has the lower speed and the light fragment has the greater speed. 1 2

9-7 Center of Mass The center of mass of a system is the point where the system can be balanced in a uniform gravitational field.

9-7 Center of Mass Imagine any body being made of many tiny particles. (In this simple case, only 2 and in 1D.) The center of mass can be found by: The center of mass is closer to the more massive object.

9-7 Center of Mass The center of mass need not be within the object:

Question 9.18 Baseball Bat a) at the midpoint
b) closer to the thick end c) closer to the thin end (near handle) d) it depends on how heavy the bat is Where is the center of mass of a baseball bat located? Answer: b

Question 9.18 Baseball Bat a) at the midpoint
b) closer to the thick end c) closer to the thin end (near handle) d) it depends on how heavy the bat is Where is the center of mass of a baseball bat located? Because most of the mass of the bat is at the thick end, this is where the center of mass is located. Only if the bat were like a uniform rod would its center of mass be in the middle.

Question 9.20 Center of Mass
The disk shown below in (1) clearly has its center of mass at the center. Suppose the disk is cut in half and the pieces arranged as shown in (2). Where is the center of mass of (2) as compared to (1) ? a) higher b) lower c) at the same place d) there is no definable CM in this case (1) X CM (2) Answer: a

Question 9.20 Center of Mass
The disk shown below in (1) clearly has its center of mass at the center. Suppose the disk is cut in half and the pieces arranged as shown in (2). Where is the center of mass of (2) as compared to (1) ? a) higher b) lower c) at the same place d) there is no definable CM in this case The CM of each half is closer to the top of the semicircle than the bottom. The CM of the whole system is located at the midpoint of the two semicircle CMs, which is higher than the yellow line. (1) X CM (2) CM

Equations Try using it! Three people of roughly equal masses m on a lightweight (air-filled) banana boat sit along the x axis at positions xA = 1.0 m, xB = 5.0 m and xC = 6.0 m, measured from the left-hand end. Find the position of the CM. Ignore the mass of the boat. 4.0 m

Question 11.12a Tipping Over I
a) all b) 1 only c) 2 only d) 3 only e) 2 and 3 A box is placed on a ramp in the configurations shown below. Friction prevents it from sliding. The center of mass of the box is indicated by a blue dot in each case. In which case(s) does the box tip over? 1 2 3 Answer: d

Question 11.12a Tipping Over I
a) all b) 1 only c) 2 only d) 3 only e) 2 and 3 A box is placed on a ramp in the configurations shown below. Friction prevents it from sliding. The center of mass of the box is indicated by a blue dot in each case. In which case(s) does the box tip over? The torque due to gravity acts like all the mass of an object is concentrated at the CM. Consider the bottom right corner of the box to be a pivot point If the box can rotate such that the CM is lowered, it will!! 1 2 3

Question 11.12b Tipping Over II
Consider the two configurations of books shown below. Which of the following is true? a) case 1 will tip b) case 2 will tip c) both will tip d) neither will tip 1/2 1/4 1 2 Answer: a

Question 11.12b Tipping Over II
Consider the two configurations of books shown below. Which of the following is true? a) case 1 will tip b) case 2 will tip c) both will tip d) neither will tip The CM of the system is midway between the CM of each book. Therefore, the CM of case #1 is not over the table, so it will tip. 1/2 1/4 1 2

9-7 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).

9-7 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.

9-7 Center of Mass Motion of the center of mass:
Say you have 3 particles lie on the x axis at positions x1, x2, and x3. If the particles move with velocities v1, v2, and v3, Then the new X´CM is given by: Total Linear Momentum

9-7 Center of Mass Velocity of the center of mass:

9-7 Center of Mass Motion of the center of mass:
If there are forces acting on the particles… Then: Net Force Acceleration of the center of mass:

9-7 Center of Mass The total mass multiplied by the acceleration of the center of mass is equal to the net external force: The center of mass accelerates just as though it were a point particle of mass M acted on by

9-7 Center of Mass 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.

9-8 Systems with Changing Mass: Rocket Propulsion
If a mass of fuel Δm is ejected from a rocket with speed v, the change in momentum of the rocket is: The force, or thrust, is

Extra Fun - Center of Mass of a Human Body
The x’s in the small diagram mark the CM of the listed body segments.

The location of the center of mass of the leg (circled) will depend on the position of the leg.

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.

Equations Try using it! Determine the position of the CM of a whole leg When stretched out When bent at 90° Assume the person is 1.70 m tall. 0.54 m above the bottom of the foot 0.39 m above the floor and 25 cm to the right of the hip joint

Equations Try using it! Determine the position of the CM of a whole leg When stretched out Assume the person is 1.70 m tall. Put x=0 at the hip. The % distance to the CM of the thigh is then: = 9.6 0.54 m above the bottom of the foot 0.39 m above the floor and 25 cm to the right of the hip joint

Equations Try using it! Determine the position of the CM of a whole leg When stretched out Assume the person is 1.70 m tall. 0.54 m above the bottom of the foot 0.39 m above the floor and 25 cm to the right of the hip joint

Equations Try using it! Determine the position of the CM of a whole leg When stretched out Assume the person is 1.70 m tall. 0.54 m above the bottom of the foot 0.39 m above the floor and 25 cm to the right of the hip joint So, the CM of the leg and foot is 20.4 units from the hip joint. Or… = 31.7 units from the floor. Since the person is 1.70 m tall, this is above the floor

Equations Try using it! Determine the position of the CM of a whole leg When bent at 90° Assume the person is 1.70 m tall. Separate x and y coordinates 0.54 m above the bottom of the foot 0.39 m above the floor and 25 cm to the right of the hip joint

Equations Try using it! Determine the position of the CM of a whole leg When bent at 90° Assume the person is 1.70 m tall. From the hip 0.54 m above the bottom of the foot 0.39 m above the floor and 25 cm to the right of the hip joint Above the floor Note: this places the CM outside the body

Summary of Chapter 9 Linear momentum: Momentum is a vector
Newton’s second law: Impulse: Impulse is a vector The impulse is equal to the change in momentum If the time is short, the force can be quite large

Summary of Chapter 9 Momentum is conserved if the net external force is zero Internal forces within a system always sum to zero In collision, assume external forces can be ignored Inelastic collision: kinetic energy is not conserved Completely inelastic collision: the objects stick together afterward

Summary of Chapter 9 A one-dimensional collision takes place along a line In two dimensions, conservation of momentum is applied separately to each Elastic collision: kinetic energy is conserved Center of mass:

Summary of Chapter 9 Center of mass:

Summary of Chapter 9 Motion of center of mass: Rocket propulsion:
Center of Mass of the Human Body

Lecture Outline Chapter 9 Physics, 4th Edition James S. Walker

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