Linear momentum and Collisions Chapter 9

Center of mass and linear momentum I.The center of mass - System of particles / - Solid body II.Newton’s Second law for a system of particles III.Linear Momentum - System of particles / - Conservation - System of particles / - Conservation IV. Collision and impulse - Single collision / - Series of collisions - Single collision / - Series of collisions V.Momentum and kinetic energy in collisions VI. Inelastic collisions in 1D -Completely inelastic collision/ Velocity of COM

VII. Elastic collisions in 1D VII. Elastic collisions in 1D VIII. Collisions in 2D IX. Systems with varying mass X. External forces and internal energy changes

I. Center of mass The center of mass of a body or a system of bodies is the point that moves as though all the mass were concentrated there and all external forces were applied there.

System of particles: Origin of reference system coincides with m 1 Two particles of masses m 1 and m 2 separated by a distance d

System of particles: General: The center of mass lies somewhere between the two particles. Choice of the reference origin is arbitrary  Shift of the coordinate system but center of mass is still at the same distance from each particle. M = total mass of the system

3D: System of particles: We can extend this equation to a general situation for n particles that strung along x-axis. The total mass of the system M=m 1 +m 2 +m 3 +……+m n The location of center of the mass:

3D: The vector form Position of the particle: M = mass of the object object System of particles: Position COM:

Solid bodies: Continuous distribution of matter. Particles = dm (differential mass elements). 3D: M = mass of the object Assumption: Uniform objects  uniform density

The center of mass of an object with a point, line or plane of symmetry lies on that point, line or plane. The center of mass of an object does not need to lie within the object (Examples: doughnut, horseshoe )

Problem solving tactics: (1) Use object’s symmetry. (2) If possible, divide object in several parts. Treat each of these parts as a particle located at its own center of mass. (3) Chose your axes wisely. Use one particle of the system as origin of your reference system or let the symmetry lines be your axis.

A water molecule consists of an oxygen atom with two hydrogen atoms bound to it. The angle between the two bonds is 106 . If the bonds are nm long, where is the center of mass of the molecule?

x y

A uniform piece of sheet steel is shaped as in Figure. Compute the x and y coordinates of the center of mass of the piece.

A1A1 A3A3 A2A2

A1A1 A3A3 A2A2

II. Newton’s second law for a system of particles Center of the mass of the system moves as a particle whose mass is equal to the total mass of the system. Motion of the center of mass: F net is the net of all external forces that act on the system. Internal forces (from one part of the system to another) are not included. The system is closed: no mass enters or leaves the system during the movement. (M=total system mass). a com is the acceleration of the system’s center of mass. a com is the acceleration of the system’s center of mass.

Prove: (*) includes forces that the particles of the system exert on each other (internal forces) and forces exerted on the particles from outside the system (external). Newton’s third law  internal forces from third-law force pairs cancel out in the sum (*)  Only external forces.

Linear Momentum The linear momentum of a particle or an object that can be modeled as a particle of mass m moving with a velocity v is defined to be the product of the mass and velocity: p = m v The terms momentum and linear momentum will be used interchangeably in the text

III. Linear momentum The linear momentum of a particle is a vector p defined as: Momentum is a vector with magnitude equal mv and have direction of v. SI unit of the momentum is kg-meter/second Momentum can be expressed in component form: p x = m v x p y = m v y p z = m v z

Newton II law in terms of momentum: The time rate of change of the momentum of a particle is equal to the net force acting on the particle and is in the direction of the force. Newton called the product mv the quantity of motion of the particle

System of particles: The total linear moment P is the vector sum of the individual particle’s linear momentum. The linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity of the center of mass. Net external force acting on the system.

Conservation: If no external force acts on a closed, isolated system of particles, the total linear momentum P of the system cannot change. Closed: no matter passes through the system boundary in any direction. boundary in any direction.

If the component of the net external force on a closed system is zero along an axis  component of the linear momentum along that axis cannot change. The momentum is constant if no external forces act on a closed particle system. internal forces can change the linear momentum of portions of the system, but they cannot change the total linear momentum of the entire system. If no net external force acts on the system of particles the total linear momentum P of the system cannot change. Each component of the linear momentum is conserved separately if the corresponding component of the net external force is zero. Conservation of Linear Momentum

IV. Collision and impulse Collision: isolated event in which two or more bodies exert relatively strong forces on each other for a relatively short time. Single collision Measures the strength and duration of the collision force Third law force pair F R = - F L Impulse:

Impulse-linear momentum theorem The change in the linear momentum of a body in a collision is equal to the impulse that acts on that body. Units: kg m/s F avg such that: Area under F(t) vs Δt curve = Area under F avg vs t

An estimated force-time curve for a baseball struck by a bat is shown in Figure. From this curve, determine (a) the impulse delivered to the ball, (b) the average force exerted on the ball, and (c) the peak force exerted on the ball.

(a) area under curve (b) (c) (c)

V. Momentum and kinetic energy in collisions Assumptions: Closed systems (no mass enters or leaves them) Isolated systems (no external forces act on the bodies within the system) Isolated systems (no external forces act on the bodies within the system) Elastic collision: If the total kinetic energy of the system of two colliding bodies is unchanged (conserved) by the collision. Inelastic collision: The kinetic energy of the system is not conserved  some goes into thermal energy, sound, etc. Example: Ball into hard floor. Completely inelastic collision: After the collision the bodies lose energy and stick together. Example: Ball of wet putty into floor

Velocity of the center of mass: In a closed, isolated system, the velocity of COM of the system cannot be changed by a collision. (No net external force). Completely inelastic collision  V = v com

VII. Elastic collisions in 1D Closed, isolated system  In an elastic collision, the kinetic energy of each colliding body may change, but the total kinetic energy of the system does not change. Stationary target: Linear momentum Kinetic energy

v 2f >0 always v 1f >0 if m 1 >m 2  forward mov. v 1f <0 if m 1 <m 2  rebounds Stationary target: Equal masses: m 1 =m 2  v 1f =0 and v 2f = v 1i  In head-on collisions bodies of equal masses simply exchange velocities.

Massive target: m 2 >>m 1  v 1f ≈ -v 1i and v 2f ≈ (2m 1 /m 2 )v 1i  Body 1 bounces back with approximately same speed. Body 2 moves forward at low speed. Massive projectile: m 1 >>m 2  v 1f ≈ v 1i and v 2f ≈ 2v 1i  Body 1 keeps on going scarcely lowed by the collision. Body 2 charges ahead at twice the initial speed of the projectile.

VII. Elastic collisions in 1D Closed, isolated system  Moving target: Linear momentum Kinetic energy

VIII. Collisions in 2D Closed, isolated system  Linear momentum conserved Kinetic energy conserved Elastic collision  Example: If the collision is elastic 

Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them. A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2.00 m/s. (a) What is the speed of the block of mass M? (b) Find the original elastic potential energy in the spring if M = kg.

(a) (motion toward the left).

Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them. A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2.00 m/s. (a) What is the speed of the block of mass M? (b) Find the original elastic potential energy in the spring if M = kg. (b)

A tennis player receives a shot with the ball ( kg) traveling horizontally at 50.0 m/s and returns the shot with the ball traveling horizontally at 40.0 m/s in the opposite direction. (a) What is the impulse delivered to the ball by the racquet? (b) What work does the racquet do on the ball?

Assume the initial direction of the ball in the –x direction. (a) (b) W

Two blocks are free to slide along the frictionless wooden track ABC shown in Figure. A block of mass m 1 = 5.00 kg is released from A. Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet embedded in the back end of the block of mass m 2 = 10.0 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m 1 rises after the elastic collision.

- speed ofat B before collision.

Two blocks are free to slide along the frictionless wooden track ABC shown in Figure. A block of mass m 1 = 5.00 kg is released from A. Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet embedded in the back end of the block of mass m 2 = 10.0 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m 1 rises after the elastic collision., speed of at B just after collision.

As shown in Figure, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length ℓ and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle?

Energy is conserved for the bob-Earth system between bottom and top of swing. At the top the stiff rod is in compression and the bob nearly at rest.

As shown in Figure, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length ℓ and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? Momentum of the bob-bullet system is conserved in the collision:

A small block of mass m 1 = kg is released from rest at the top of a curve-shaped frictionless wedge of mass m 2 = 3.00 kg, which sits on a frictionless horizontal surface as in Figure (a). When the block leaves the wedge, its velocity is measured to be 4.00 m/s to the right, as in Figure (b). (a) What is the velocity of the wedge after the block reaches the horizontal surface? (b) What is the height h of the wedge?

(a) The initial momentum of the system is zero, which remains constant throughout the motion. Therefore, when m 1 leaves the wedge, we must have:

A small block of mass m 1 = kg is released from rest at the top of a curve-shaped frictionless wedge of mass m 2 = 3.00 kg, which sits on a frictionless horizontal surface as in Figure (a). When the block leaves the wedge, its velocity is measured to be 4.00 m/s to the right, as in Figure (b). (a) What is the velocity of the wedge after the block reaches the horizontal surface? (b) What is the height h of the wedge? (b)Using conservation of energy for the block-wedge-Earth system as the block slides down the smooth (frictionless) wedge, we have:

Rocket Propulsion The operation of a rocket depends upon the law of conservation of linear momentum as applied to a system of particles, where the system is the rocket plus its ejected fuel. IV. Systems with varying mass The initial mass of the rocket plus all its fuel is M + Δm at time t i and velocity v The initial momentum of the system is p i = (M + Δm) v

Rocket Propulsion At some time t + Δt, the rocket’s mass has been reduced to M and an amount of fuel, Δm has been ejected. The rocket’s speed has increased by Δv

Because the gases are given some momentum when they are ejected out of the engine, the rocket receives a compensating momentum in the opposite direction Therefore, the rocket is accelerated as a result of the “push” from the exhaust gases In free space, the center of mass of the system (rocket plus expelled gases) moves uniformly, independent of the propulsion process Rocket Propulsion

IV. Systems with varying mass Example: most of the mass of a rocket on its launching is fuel that gets burned during the travel. System: rocket + exhaust products Closed and isolated  mass of this system does not change as the rocket accelerates. P=const  P i =P f P=const  P i =P f Linear momentum of exhaust products released during the interval dt Linear momentum of rocket at the end of dt Afterdt After dt

R=Rate at which the rocket losses mass= -dM/dt = rate of fuel consumption First rocket equation Velocity of rocket relative to frame = (velocity of rocket relative to products)+ (velocity of products relative to frame) to products)+ (velocity of products relative to frame)

Second rocket equation

The basic equation for rocket propulsion isThe basic equation for rocket propulsion is The increase in rocket speed is proportional to the speed of the escape gases (v e )The increase in rocket speed is proportional to the speed of the escape gases (v e ) –So, the exhaust speed should be very high The increase in rocket speed is also proportional to the natural log of the ratio M i / M fThe increase in rocket speed is also proportional to the natural log of the ratio M i / M f –So, the ratio should be as high as possible, meaning the mass of the rocket should be as small as possible and it should carry as much fuel as possible

Thrust The thrust on the rocket is the force exerted on it by the ejected exhaust gasesThe thrust on the rocket is the force exerted on it by the ejected exhaust gases Thrust = Thrust = The thrust increases as the exhaust speed increasesThe thrust increases as the exhaust speed increases The thrust increases as the rate of change of mass increasesThe thrust increases as the rate of change of mass increases –The rate of change of the mass is called the rate of fuel consumption or burn rate

The first stage of a Saturn V space vehicle consumed fuel and oxidizer at the rate of 1.50  10 4 kg/s, with an exhaust speed of 2.60  10 3 m/s. (a) Calculate the thrust produced by these engines. (b) Find the acceleration of the vehicle just as it lifted off the launch pad on the Earth if the vehicle’s initial mass was 3.00  10 6 kg. Note: You must include the gravitational force to solve part (b).

(a) Thrust (b)

Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocket engine has an average thrust of 5.26 N, a fuel mass of 12.7 grams, and an initial mass of 25.5 grams. The duration of its burn is 1.90 s. (a) What is the average exhaust speed of the engine? (b) If this engine is placed in a rocket body of mass 53.5 grams, what is the final velocity of the rocket if it is fired in outer space? Assume the fuel burns at a constant rate.

(a) The fuel burns at a rate: Trust

Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocket engine has an average thrust of 5.26 N, a fuel mass of 12.7 grams, and an initial mass of 25.5 grams. The duration of its burn is 1.90 s. (a) What is the average exhaust speed of the engine? (b) If this engine is placed in a rocket body of mass 53.5 grams, what is the final velocity of the rocket if it is fired in outer space? Assume the fuel burns at a constant rate. (b)

A rocket for use in deep space is to be capable of boosting a total load (payload plus rocket frame and engine) of 3.00 metric tons to a speed of m/s. (a) It has an engine and fuel designed to produce an exhaust speed of m/s. How much fuel plus oxidizer is required? (b) If a different fuel and engine design could give an exhaust speed of m/s, what amount of fuel and oxidizer would be required for the same task?

(a) The mass of fuel and oxidizer is:

A rocket for use in deep space is to be capable of boosting a total load (payload plus rocket frame and engine) of 3.00 metric tons to a speed of m/s. (a) It has an engine and fuel designed to produce an exhaust speed of m/s. How much fuel plus oxidizer is required? (b) If a different fuel and engine design could give an exhaust speed of m/s, what amount of fuel and oxidizer would be required for the same task? (b) Because of the exponential, a relatively small increase in fuel and/or engine efficiency causes a large change in the amount of fuel and oxidizer required.

A 60.0-kg person running at an initial speed of 4.00 m/s jumps onto a 120-kg cart initially at rest. The person slides on the cart’s top surface and finally comes to rest relative to the cart. The coefficient of kinetic friction between the person and the cart is Friction between the cart and ground can be neglected. (a) Find the final velocity of the person and cart relative to the ground. (b) Find the friction force acting on the person while he is sliding across the top surface of the cart. (c) How long does the friction force act on the person? (d) Find the change in momentum of the person and the change in momentum of the cart. (e) Determine the displacement of the person relative to the ground while he is sliding on the cart. (f) Determine the displacement of the cart relative to the ground while the person is sliding. (g) Find the change in kinetic energy of the person. (h) Find the change in kinetic energy of the cart. (What kind of collision is this, and what accounts for the loss of mechanical energy?)

(a)(b)

(c) For the person, (d)person: cart: (e) (f)

(g) (h) (i) The force exerted by the person on the cart must equal in magnitude and opposite in direction to the force exerted by the cart on the person. The changes in momentum of the two objects must be equal in magnitude and must add to zero. Their changes in kinetic energy are different in magnitude and do not add to zero. The following represent two ways of thinking about “way”. The distance the cart moves is different from the distance moved by the point of application of the friction force to the cart. The total change of mechanical energy for both objects together, -320J, becomes +320J of additional internal energy in this perfectly inelastic collision.

Suppose the incline is frictionless for the system in Figure. The block is released from rest with the spring initially unstretched. (a) How far does it move down the incline before coming to rest? (b) What is its acceleration at its lowest point? Is the acceleration constant?

A 2.00-kg block situated on a frictionless incline is connected to a spring of negligible mass having a spring constant of 100 N/m. The pulley is frictionless.The block is released from rest with the spring initially unstretched. (a) How far does it move down the incline before coming to rest? (b) What is its acceleration at its lowest point? Is the acceleration constant? A 2.00-kg block situated on a frictionless incline is connected to a spring of negligible mass having a spring constant of 100 N/m. The pulley is frictionless. The block is released from rest with the spring initially unstretched. (a) How far does it move down the incline before coming to rest? (b) What is its acceleration at its lowest point? Is the acceleration constant? (a)

(b) (from (a)) The negative sign indicates a is up the incline. The.

Example: Tennis Racket A 50 g ball is struck by a racket. If the ball is initially travelling at 5 m/s up and ends up travelling at 5 m/s down after 0.1 s of contact what is the (average) force exerted by the racket on the ball? What is the impulse?

Example: Tennis Racket A 50 g ball is struck by a racket. If the ball is initially travelling at 5 m/s up and ends up travelling at 5 m/s down after 0.1 s of contact what is the (average) force exerted by the racket on the ball? What is the impulse? a = (v f - v i )/  t = (5 - (-5))/0.1 = 100 m/s 2 down F av = ma = 0.05 x 100 = 5 N down  p = F  t = 5 x 0.1 = 0.5 Nsdown  p = F  t = 5 x 0.1 = 0.5 Ns down

Example: Bowling balls Two particles masses m and 3m are moving towards each other at the same speed, but opposite velocity, and collide elastically After collision, m moves off at right angles What are v mf and v 3mf ? At what angle is 3m scattered? m 3m m vovo vovo v fm x y

By conservation of KE: ½mv im 2 + ½3mv i3m 2 = ½mv fm 2 + ½3mv f3m 2 By conservation of momentum: In x direction: mv im + 3mv i3m = mv fm cos  m  3mv f3m cos  3m In y direction: 0 = mv fm sin  m  3mv f3m sin  3m

½mv im 2 + ½3mv i3m 2 = ½mv fm 2 + ½3mv f3m 2 v im 2 =v i3m 2 =v o 2 so 2mv o 2 = ½m(v fm 2 + 3v f3m 2 ) – In x direction: v i3m = -v im  m =90 o – In x direction: v i3m = -v im and  m =90 o mv o - 3mv o = mv fm 0  3mv f3m cos  3m -2mv o = 3mv f3m cos  3m v o = -3/2 v f3m cos  3m –In y direction:  m =90 o 0 = mv fm  3mv f3m sin  3m v f3m = -v fm /(3sin  3m )  3m  35 o, v f3m =  (2/3)v o, v fm =  2v o

Sulfur dioxide (SO 2 ) consist of two oxygen atoms (each of mass 16u) and single sulfur atom (of mass 32u). The center-to-center distance between the atoms is 0.143nm. Angle is given on the picture. Find the x- and y-coordinate of center of the mass of this molecule.