Linear momentum and Collisions

Linear momentum and Collisions Chapter 9

Linear Momentum and its Conservation • Impulse and Momentum • Collisions in One Dimension • Collisions in Two Dimensions • V. The Center of Mass • Motion of a System of Particles • Deformable Systems • Rocket Propulsion

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 is defined to be the product of the mass and velocity: The terms momentum and linear momentum will be used interchangeably in the text

III. Linear momentum The linear momentum of a particle is a vectorp defined as: Momentum is a vector with magnitude equalmvand has direction of v. The dimensions of momentum are ML/T SI unitof the momentum iskg-meter/second • Momentum can be expressed in component form: • px = m vxpy= m vypz = m vz

Newton and Momentum Newton’s Second Law can be used to relate the momentum of a particle to the resultant force acting on it with constant mass

Newton called the product mv the quantity of motion of the particle 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.

System of particles: The total linear momentPis 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 momentumP of the system cannot change. Closed:no matter passes through the system boundary in any direction.

Conservation of Linear Momentum If no net external force acts on the system of particles the total linear momentumPof the system cannot change. Each component of the linear momentum is conserved separately if the corresponding component of the net external force is zero. 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.

Conservation of Linear Momentum • Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant • The momentum of the system is conserved, not necessarily the momentum of an individual particle • This also tells us that the total momentum of an isolated system equals its initial momentum

Conservation of Momentum • Conservation of momentum can be expressed mathematically in various ways • In component form, the total momenta in each direction are independently conserved pix = pfxpiy = pfypiz = pfz • Conservation of momentum can be applied to systems with any number of particles • This law is the mathematical representation of the momentum version of the isolated system model

Conservation of Momentum, Archer Example • The archer is standing on a frictionless surface (ice) • Approaches: • Newton’s Second Law – no, no information about F or a • Energy approach – no, no information about work or energy • Momentum – yes

Archer Example • Conceptualize • The arrow is fired one way and the archer recoils in the opposite direction • Categorize • Momentum Let the system be the archer with bow (particle 1) and the arrow (particle 2) There are no external forces in the x-direction, so it is isolated in terms of momentum in the x-direction • Analyze • Total momentum before releasing the arrow is 0

Archer Example • Analyze. • The total momentum after releasing the arrow is • Finalize • The final velocity of the archer is negative • Indicates he moves in a direction opposite the arrow • Archer has much higher mass than arrow, so velocity is much lower

Impulse and Momentum • From Newton’s Second Law, • Solving for gives • Integrating to find the change in momentum over some time interval • The integral is called the impulse, , of the force acting on an object over Δt

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. Impulse: Measures the strength and duration of the collision force Third law force pair FR = - FL Single collision

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 Favgsuch that: Area underF(t)vsΔtcurve = Area under Favgvs 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.

More About Impulse • Impulse is a vector quantity • The magnitude of the impulse is equal to the area under the force-time curve • The force may vary with time • Dimensions of impulse are M L / T • Impulse is not a property of the particle, but a measure of the change in momentum of the particle

Impulse • The impulse can also be found by using the time averaged force • This would give the same impulse as the time-varying force does

Impulse Approximation • In many cases, one force acting on a particle acts for a short time, but is much greater than any other force present • When using the Impulse Approximation, we will assume this is true • Especially useful in analyzing collisions • The force will be called the impulsive force • The particle is assumed to move very little during the collision • represent the momenta immediately before and after the collision

Impulse-Momentum: Crash Test Example • Categorize • Assume force exerted by wall is large compared with other forces • Gravitational and normal forces are perpendicular and so do not effect the horizontal momentum • Can apply impulse approximation

Crash Test Example • Analyze • The momenta before and after the collision between the car and the wall can be determined • Find • Initial momentum • Final momentum • Impulse • Average force • Finalize • Check signs on velocities to be sure they are reasonable

Collisions – Characteristics • We use the term collision to represent an event during which two particles come close to each other and interact by means of forces • May involve physical contact, but must be generalized to include cases with interaction without physical contact • The time interval during which the velocity changes from its initial to final values is assumed to be short • The interaction forces are assumed to be much greater than any external forces present • This means the impulse approximation can be used

Collisions – Example 1 • Collisions may be the result of direct contact • The impulsive forces may vary in time in complicated ways • This force is internal to the system • Momentum is conserved

Collisions – Example 2 • The collision need not include physical contact between the objects • There are still forces between the particles • This type of collision can be analyzed in the same way as those that include physical contact

Types of Collisions • In an elastic collision, momentum and kinetic energy are conserved • Perfectly elastic collisions occur on a microscopic level • In macroscopic collisions, only approximately elastic collisions actually occur • Generally some energy is lost to deformation, sound, etc. • In an inelastic collision, kinetic energy is not conserved, although momentum is still conserved • If the objects stick together after the collision, it is a perfectly inelastic collision

Collisions • In an inelastic collision, some kinetic energy is lost, but the objects do not stick together • Elastic and perfectly inelastic collisions are limiting cases, most actual collisions fall in between these two types • Momentum is conserved in all collisions

Perfectly Inelastic Collisions • Since the objects stick together, they share the same velocity after the collision

Elastic Collisions • Both momentum and kinetic energy are conserved

Elastic Collisions • Typically, there are two unknowns to solve for and so you need two equations • The kinetic energy equation can be difficult to use • With some algebraic manipulation, a different equation can be used v1i – v2i = v1f + v2f • This equation, along with conservation of momentum, can be used to solve for the two unknowns • It can only be used with a one-dimensional, elastic collision between two objects

Elastic Collisions • Example of some special cases m1 = m2 – the particles exchange velocities • When a very heavy particle collides head-on with a very light one initially at rest, the heavy particle continues in motion unaltered and the light particle rebounds with a speed of about twice the initial speed of the heavy particle • When a very light particle collides head-on with a very heavy particle initially at rest, the light particle has its velocity reversed and the heavy particle remains approximately at rest

Problem-Solving Strategy: One-Dimensional Collisions • Conceptualize • Image the collision occurring in your mind • Draw simple diagrams of the particles before and after the collision • Include appropriate velocity vectors • Categorize • Is the system of particles isolated? • Is the collision elastic, inelastic or perfectly inelastic?

Problem-Solving Strategy: One-Dimensional Collisions • Analyze • Set up the mathematical representation of the problem • Solve for the unknown(s) • Finalize • Check to see if the answers are consistent with the mental and pictorial representations • Check to be sure your results are realistic

Example: Stress Reliever • Conceptualize • Imagine one ball coming in from the left and two balls exiting from the right • Is this possible? • Categorize • Due to shortness of time, the impulse approximation can be used • Isolated system • Elastic collisions

Example: Stress Reliever • Analyze • Check to see if momentum is conserved • It is • Check to see if kinetic energy is conserved • It is not • Therefore, the collision couldn’t be elastic • Finalize • Having two balls exit was not possible if only one ball is released

Example: Stress Reliever • What collision is possible • Need to conserve both momentum and kinetic energy • Only way to do so is with equal numbers of balls released and exiting

Collision Example – Ballistic Pendulum • Conceptualize • Observe diagram • Categorize • Isolated system of projectile and block • Perfectly inelastic collision – the bullet is embedded in the block of wood • Momentum equation will have two unknowns • Use conservation of energy from the pendulum to find the velocity just after the collision • Then you can find the speed of the bullet

Ballistic Pendulum • A multi-flash photograph of a ballistic pendulum • Analyze • Solve resulting system of equations • Finalize • Note different systems involved • Some energy was transferred during the perfectly inelastic collision

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) Elastic collision: If the total kinetic energy of the system of two colliding bodies is unchanged (conserved) by the collision. Example:Ball into hard floor. Inelastic collision: The kinetic energy of the system is not conserved  some goes into thermal energy, sound, etc. After the collision the bodies lose energy and stick together. Completely inelastic collision: Example:Ball of wet putty into floor

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

Stationary target: v2f>0 always v1f>0 if m1>m2  forward mov. v1f<0 if m1<m2  rebounds Equal masses:m1=m2  v1f=0 and v2f = v1i In head-on collisions bodies of equal masses simply exchange velocities.

Massive target:m2>>m1 v1f ≈ -v1iand v2f ≈ (2m1/m2)v1i Body 1 bounces back with approximately same speed. Body 2 moves forward at low speed. Massive projectile:m1>>m2 v1f ≈ v1iand v2f ≈ 2v1i 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 Moving target: Closed, isolated system  Linear momentum Kinetic energy

VIII. Collisions in 2D Closed, isolated system  Linear momentum conserved Elastic collision  Kinetic energy conserved 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 3Mmoves 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 = 0.350 kg.

A tennis player receives a shot with the ball (0.0600 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?

Two blocks are free to slide along the frictionless wooden track ABC shown in Figure. A block of mass m1 = 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 m2 = 10.0 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic 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 lengthℓand negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle?

A small block of mass m1 = 0.500 kg is released from rest at the top of a curve-shaped frictionless wedge of mass m2 = 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 hof the wedge?

Linear momentum and Collisions