Please enable JavaScript.
Coggle requires JavaScript to display documents.
CHAPTER 8: LINEAR MOMENTUM, NUR ADIBAH BINTI ROZAINI - Coggle Diagram
CHAPTER 8:
LINEAR MOMENTUM
8.3 - COLLISIONS AND IMPULSE
During a collision, objects are deformed due to the large forces involved
Such an example, tennis racket strikig a ball. Both the ball and the racket strings are deformed due to the large force each exerts on the other.
F = dp/dt
Impulse is equal to the change in momentum:
J = ∆p = pf - pi
Since the time of the collision is often very short, we may be able to use the average force, which would produce the same impulse over the same time interval.
Force as a function of time during a typical collision: F can become very large; Δt is typically milliseconds for macroscopic collisions.
The average force acting over a very brief time interval gives the same impulse (FavgΔt) as the actual force.
8.5 - ELASTIC COLLISIONS IN ONE DIMENSION
Two small objects of masses mA and mB, (a) before the collision and (b) after the collision.
There 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.
8.4 - 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.
Two equal-mass objects (a) approach each other with equal speeds, (b) collide, and then (c) bounce off with equal speeds in the opposite directions if the collision is elastic, or (d) bounce back much less or not at all if the collision is inelastic.
8.6 - INELASTIC COLLISIONS
With inelastic collisions, some of the initial kinetic energy is lost to thermal or potential energy. Kinetic energy may also be gained during explosions, as there is the addition of chemical or nuclear energy.
A completely inelastic collision is one in which the objects stick together afterward, so there is only one final velocity.
8.7 - 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.
Object A, the projectile, collides with object B, the target. After the collision, they move off with momenta pA’ and pB’ at angles θA’ and θB’. The objects are shown here as particles, as we would visualize them in atomic or nuclear physics. But they could also be macroscopic pool balls.
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.
If the collision is elastic, apply conservation of kinetic energy as well.
Solve
Apply momentum conservation; there will be one equation for each dimension.
Check units and magnitudes of result.
8.10 - SYSTEMS OF VARIABLE MASS; ROCKET REPULSION
Applying Newton’s second law to the system shown gives:
(a) At time t, a mass dM is about to be added to our system M. (b) At time t + dt, the mass dM has been added to our system.
8.8 - CENTER OF MASS (CM)
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)
b)
The motion of the diver is pure translation in (a), but is translation plus rotation in (b). The black dot represents the diver’s CM at each moment.
Particle would take if subjected to the same force as the diver. This point is called the center of mass (CM).
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.
Translation plus rotation: a wrench moving over a horizontal surface. The CM, marked with a red cross, moves in a straight line.
For two particles, the center of mass lies closer to the one with the most mass:
M is total mass.
The center of mass of a two-particle system lies on the line joining the two masses. Here mA > mB, so the CM is closer to mA than to mB .
The center of mass of a two-particle system lies on the line joining the two masses. Here mA > mB, so the CM is closer to mA than to mB .
For an extended object, we imagine making it up of tiny particles, each of tiny mass, and adding up the product of each particle’s mass with its position and dividing by the total mass. In the limit that the particles become infinitely small, this gives:
An extended object, here shown in only two dimensions, can be considered to be made up of many tiny particles (n), each having a mass Δmi. One such particle is shown located at a point ri = xii + yij + zik. We take the limit of n →∞ so Δmi becomes the infinitesimal dm.
The center of gravity is the point at which 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.
Determining the CM of a flat uniform body.
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.
Finding the CG
8.9 - CENTER OF MASS AND TRANSLATIONAL MOTION
The total momentum of a system of particles is equal to the product of the total mass and the velocity of the center of mass.
The sum of all the forces acting on a system is equal to the total mass of the system multiplied by the acceleration of the center of mass:
Therefore, the center of mass of a system of particles (or objects) with total mass M moves like a single particle of mass M acted upon by the same net external force.
8.1 - MOMENTUM AND ITS RELATION TO FORCE
Momentum is a vector symbolized by the symbol p, and defined as
p = mv
The rate of change of momentum is equal to the net force.
F = dp/dt
This can be shown using Newton's
second
law
8.2 - CONSERVATION OF MOMENTUM
During collision, measurements show that the total momentum does not change:
m1v1 + m2v2 = m1v'1 + M2v'2
For more than two objects,
dP/dt = ΣdPi/dt = ΣFi
Or, if internal forces is canceled,
dP/dt = ΣFext
Collision of two objects. Their momenta before collision are pA and pB, and after collision are pA’ and pB’. At any moment during the collision each exerts a force on the other of equal magnitude but opposite direction.
Conservation of momentum can also be derived from Newton's laws. A collision takes a short enough time that we can ignore external forces. Since the internal forces are equal and opposite, the total momentum is constant.
This is the law of conservation of linear momentum:
when the net external force on a system of object is zero, the total momentum of the system remains constant.
Equivalently,
the total momentum of an isolated system remains constant.
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
(a) A rocket, containing fuel, at rest in some reference frame. (b) In the same reference frame, the rocket fires, and gases are expelled at high speed out the rear. The total vector momentum, P = pgas + procket, remains zero.
NUR ADIBAH BINTI ROZAINI
