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Chapter 9

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Chapter 9

Linear Momentum and Collisions

- Consider bowling:
- Bowling ball collides with initial pin
- Force on/Acceleration of the Pin
- Force on/Acceleration of the ball

- Bowling ball collides with initial pin
- Momentum- simplified way to study these moving objects.

Consider two particles (isolated) m1 and m2 moving at v1 and v2

From Newton’s 3rd Law

- And if the the masses are constant
- If the derivative of a function is 0 it is constant (conserved)

- Linear Momentum- the product of the mass and velocity of a moving particle.
- Momentum is a vector quantity
- Has dimensions MLT-1 and SI units kg.m/s
- All momentum is conserved
- Three Components of Momentum

- Directly related to Newton’s 2nd Law
- Instead of Net Force equals mass times accel.
- Can be described as Net Force equals time rate of change of momentum

- Conservation of Momentum
- For isolated systems the time derivative of the total momentum is 0
- The total momentum is therefore constant or conserved.
Law of Conservation of Momentum

- And in 3 components

- Quick Quizzes p 254-255
- Examples 9.1-9.2

- The momentum of an object changes when a net force acts on it
or

- Integrating this gives

- Impulse-Momentum Theorem- the impulse of the force F acting on a particle equals the change in momentum of the particle
- Force Varies, impulse time is short impulse can generally be calculated with the average force.

- Quick Quizzes p. 258
- Examples 9.3-9.4

- Momentum is conserved
- Three types of Collisions
- Inelastic
- Perfectly Inelastic
- Elastic

- Inelastic Collision- Momentum is conserved but kinetic energy is not.
- Inelastic- objects collide and separate, some K is lost
- Perfectly Inelastic- objects collide and stick together (moving as one), some K is lost

- Elastic Collisions-A collision in which no energy is lost to (surroundings / internal / potential)
- Both momentum and kinetic energy are conserved

- By combining the Conservation of p and K equations
- In all collision types careful attention to the direction (and sign) of velocities must be paid.

- Quick Quizzes p. 262
- Examples 9.5 - 9.9

- Momentum is conserved on each axis
- Examples 9.10 – 9.12

- We can describe the overall motion of a mechanical system by tracking its center of mass
- System could be a group of particles
- System could be a large extended object

- A force applied to the center of mass will cause no rotation to the system

- To find the center of mass in 3-D space for a number (i) particles
- Or in terms of the position vector of each particle

- For extended objects that have a continuous mass distribution
- Consider them an infinite number of closely spaced particles
- The sum becomes an integral

- Or in terms of the position vector
- For symmetrical objects, the center of mass lies on the axis/plane of symmetry
- Examples: uniform rod,
sphere,

cube,

donut?

- Examples: uniform rod,

- For extended objects, the force of gravity acts individually on each small piece of mass (dm)
- The net effect of all these forces is equivalent to the single force Mg, through a point called the center of gravity.
- If the gravitational field is uniform across all dm, the center of gravity and center of mass are one and the same.

- Quick quiz p 272
- Examples 9.13, 9.14, 9.15

- If the mass of a system remains constant (no particles entering/leaving) then we can track the motion of the center of mass, rather than the individual particles.
- Also assumes any forces on the system are internal (isolated)

- Velocity of the center of mass
- Acceleration of the center of mass

- If there is a net force on the system, it will move equivalent to the way a single M with the same net force would move.
- And if the net force is zero

- Quick Quizzes p. 276
- Examples 9.17, 9.18

- Most forms of vehicular motion result from action/reaction friction.
- A rocket has nothing to push against so its motion/control depend on conservation of motion of the system.
- The system includes the rocket body (and payload) plus the ejected fuel

- The rocket burns fuel and oxidizer creating expanding gases that are directed through the nozzle.
- Each gas molecule has a mass (that was once part of the rockets total mass) and velocity, therefore a downward momentum.
- The rocket receives the same compensating momentum upward.

- Looking a rocket initially with mass M + Δm, moving with velocity v…

- And some time, Δt, later...
- The rocket now has mass, M and velocity v + Δv, compensating the momentum of the exhausted mass, Δm.

- The conservation of momentum expression for this change…
- Can be simplified to…

- A rocket motor produces a continuous flow of exhaust gas a fairly constant speed, through the burn
- For continually changing values…
Δv dv

Δm dm

So…

- Because the increase in exhaust mass = the decrease in rocket mass…
- Then integrate this expression

- Discuss integral of M-1
- Evaluating from vi to vf gives the basic expression for rocket propulsion.

- Mi is the total mass of the rocket/payload plus fuel
- Mf is the mass of the rocket/payload
- Mi – Mf is the mass of fuel needed to achieve a certain speed (eg. Escape speed to power down rocket)

- Thrust- the actual force on the rocket at any given time is
- Thrust is proportional to exhaust speed and also the rate of change of mass (burn rate).
- Examples 9.19 p. 279