Momentum and Collisions

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 We will usually refer to this as “momentum”, omitting the “linear”. Momentum is a VECTOR. It has components. Don’t forget that.

Linear Momentum The dimensions of momentum are ML/T The SI units of momentum are kg · m / s Momentum can be expressed in component form:  p x = m v x p y = m v y p z = m v z

A 3.00-kg particle has a velocity of. (a) Find its x and y components of momentum. (b) Find the magnitude and direction of its momentum.

Newton and Momentum Newton called the product mv the quantity of motion of the particle 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’s Second Law The time rate of change of the linear momentum of a particle is equal to the net force acting on the particle  This is the form in which Newton presented the Second Law  It is a more general form than the one we used previously  This form also allows for mass changes Applications to systems of particles are particularly powerful

Two Particles (1 dimension for now)

Continuing

NO EXTERNAL FORCES: p total is constant Force on P 1 is from particle 2 and is F 2  1 Force on P 2 is from particle 1 and is F 1  2 F 2  1 =-F 1  2

Summary The momentum of a SYSTEM of particles that are isolated from external forces remains a constant of the motion.

Conservation of Linear Momentum Textbook Statement 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

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 particle of mass m moves with momentum p. Show that the kinetic energy of the particle is K = p2/2m. (b) Express the magnitude of the particle’s momentum in terms of its kinetic energy and mass. An Easy One…

Consider a particle roaming around that is suddenly subjected to some kind of FORCE that looks something like the last slide’s graph.

Let’s drop the vector notation and stick to one dimension. Change of Momentum Impulse

NEW LAW IMPULSE = CHANGE IN MOMENTUM NEW LAW

A 3.00-kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle (Fig. P9.9). If the ball is in contact with the wall for s, what is the average force exerted on the ball by the wall?

LETS TALK ABOUT

Consider two particles: 1 2 m 1 m 2 v1v1 v2v2 V 1 V

What goes on during the collision? N3 Force on m 2 =F(1  2) Force on m 1 =F(2  1)

The Forces Equal and Opposite

More About Impulse: F-t The Graph Impulse is a vector quantity The magnitude of the impulse is equal to the area under the force-time curve 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 I =  t This would give the same impulse as the time-varying force does

SKATEBOARD DEMO

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, 2 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 Total momentum before releasing the arrow is 0 The total momentum after releasing the arrow is p 1f + p 2f = 0

Archer Example, final The archer will move in the opposite direction of the arrow after the release  Agrees with Newton’s Third Law Because the archer is much more massive than the arrow, his acceleration and velocity will be much smaller than those of the arrow

An estimated force-time curve for a baseball struck by a bat is shown in Figure P9.7. 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.

Overview: 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 The time interval during which the velocity changes from its initial to final values is assumed to be short The interaction force is 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 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, cont 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 m 1 v 1i + m 2 v 2i = (m 1 + m 2 ) v f

A 10.0-g bullet is fired into a stationary block of wood (m = 5.00 kg). The relative motion of the bullet stops inside the block. The speed of the bullet-plus-wood combination immediately after the collision is m/s. What was the original speed of the bullet?

. Velcro couplers make the carts stick together after colliding. Find the final velocity of the train of three carts. (b) What If? Does your answer require that all the carts collide and stick together at the same time? What if they collide in a different order?

Most of us know intuitively that in a head-on collision between a large dump truck and a subcompact car, you are better off being in the truck than in the car. Why is this? Many people imagine that the collision force exerted on the car is much greater than that experienced by the truck. To substantiate this view, they point out that the car is crushed, whereas the truck is only dented. This idea of unequal forces, of course, is false. Newton’s third law tells us that both objects experience forces of the same magnitude. The truck suffers less damage because it is made of stronger metal. But what about the two drivers? Do they experience the same forces? To answer this question, suppose that each vehicle is initially moving at 8.00 m/s and that they undergo a perfectly inelastic head-on collision. Each driver has mass 80.0 kg. Including the drivers, the total vehicle masses are 800 kg for the car and kg for the truck. If the collision time is s, what force does the seatbelt exert on each driver?

Elastic Collisions Both momentum and kinetic energy are conserved

Elastic Collisions, cont 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 v 1i – v 2i = v 1f + v 2f 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 The solution is shown on pages in the textbook. (Lots of algebra but nothing all that difficult. We will look at a special case.

Let’s look at the case of equal masses.

Second Particle initially at rest Explains the demo!

Elastic Collisions, final Example of some special cases  m 1 = m 2 – 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

Collision Example – Ballistic Pendulum 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