ACCOUNTING FOR LINEAR MOMENTUM

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ACCOUNTING FOR LINEAR MOMENTUM Class 28.1 1. Mass is a state quantity. 2. Velocity is a state quantity. 3. Linear Momentum is a state quantity. 4. Linear Momentum is a vector quantity. 5. Linear Momentum is always conserved. Rat 28.1

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### ACCOUNTING FOR LINEAR MOMENTUM

Class 28.1

1. Mass is a state quantity.

2. Velocity is a state quantity.

3. Linear Momentum is a state quantity.

4. Linear Momentum is a vector quantity.

5. Linear Momentum is always conserved.

Rat 28.1

Turn off computer monitors. Take 3 minutes to answer the following (true or false):

Objectives
• Know that linear momentum is conserved
• Know that linear momentum can be added to a system by forces, or by adding mass
• Know the connection between Newton's Laws and accounting for linear momentum
• Be able to do calculations involving accounting for linear momentum
Linear Momentum
• The concept of linear momentum was developed by Newton.
• Linear momentum, p, is defined as the product of the mass, m, and velocity, v

p=mv

Linear Momentum
• Mass is a state quantity.
• Velocity is also a state quantity, but since it also has direction, it is also a vector quantity (thus the boldface notation).
• Since mass is a scalar and velocity is a vector, then momentum is also a vector.
• The algebraic combination of state quantities yields a state quantity, thus momentum is a state quantity (CLOSED SYSTEMS ONLY).
Restatement of Newton’s Laws
• Newton’s laws can be re-stated in terms of linear momentum:
• #1: Linear momentum is constant for a body in its ‘natural’ state.
• #2: Linear momentum changes when net force is applied to a body.
• #3: Forces always exist by the interaction of bodies; the force on one body is equal and opposite to the force on the other body.
Revisiting the UAE
• Before applying the Universal Accounting Equation (UAE):
• Define a system.
• Determine what quantity will be counted.
• Define time interval for counting.
Terms of the UAE
• FINAL AMOUNT: …in system at the END of time period
• INITIAL AMOUNT: …in system at the START of time period
• INPUT: …PASSING through boundary INTO system during time period
• OUTPUT: …PASSING through boundary OUT of system during time period
• GENERATION: …PRODUCED during time period within boundary
• CONSUMPTION: …DESTROYED during time period within boundary
Revisiting the UAE
• The UAE equation states that the net difference in state quantities equals the net difference in path quantities, OR
• Accumulation = Net Input + Net Generation, OR,
• FINAL AMOUNT - INITIAL AMOUNT =

INPUT - OUTPUT + GENERATION - CONSUMPTION

Linear Momentum Is Conserved
• Linear momentum is a conserved quantity; i.e. the total amount of linear momentum in the universe is constant and cannot be changed.
• Thus, generation = consumption = 0, so
• Final Amt - Initial Amt = Input - Output,

or

• Accumulation = Net Input
Conservation of Momentum

from http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/momentum/fcb.html

Steps to Applying the UAE
• To use the UAE, define the system, then compute:

1. INITIAL linear momentum (scalar) for the quantity(ies) of mass in the system,

2. FINAL linear momentum (scalar) for the quantity(ies) in the system,

3. Linear momentum (scalar) of quantity(ies) INPUT into the system,

4. Linear momentum (scalar) of quantity(ies) OUTPUT from the system.

Linear Momentum Change
• Consider the accumulation term of the UAE
• There are two ways that the linear momentum of a system can change:
• mass transfer
• unbalanced forces
• Otherwise, the accumulation of linear momentum is zero
Individual Exercise (5 minutes)
• This is Problem 20.8
• A 5.0 g bullet fired horizontally at 300 m/s passes through a 500 g block of wood initially at rest on a frictionless surface. The bullet emerges with a speed of 100 m/s. What is the final speed of the block?

block

mw = 500 g

block

mw = 500 g

bullet

mb=5g

vb,f=100 m/s

bullet

mb=5g

vb,i=300 m/s

vw,i = 0 m/s

vw,f = ? m/s

Initial

Final

Solution
Linear Momentum Change by Mass Transfer
• Suppose you define a system that contains a single object of mass m1 moving at a velocity of v1 in the positive x direction
• As time passes, two more objects enter the system with masses m2 and m3 and velocities v2 and v3 respectively, also in the x direction.

system

boundary

m2v2

m2v2

Time

Passes

Time

Passes

m1v1

m1v1

m1v1

m3v3

m3v3

Linear Momentum Change by Mass Transfer
• The initial linear momentum is

pi = m1v1

• The final linear momentum is

pf = (m1v1 + m2v2 + m3v3)

• Note that for this case we can write these as scalars if we assume that we are looking at the magnitude of the x component of the vectors.
• Then, accumulation = pf - pi is not zero.
Pairs Exercise (5 minutes)
• Problem 20.2:

A 1.0 ton (including ammo and pilot) military helicopter is flying at 40 mph. It has a machine gun that fires 60 bullets per second in the forward direction. Each 0.5 lbm bullet exits the gun at at speed of 1800 mph. After a 2 second burst of fire, what is the helicopter speed.

i.e. the helicopter is going backwards!!

Helicopter

w/ ammo

mi = 1 ton

vi = 40 mph

Helicopter

w/ less ammo

mf = ?

vf = ?

Solution

bullets

Initial

Final

Pairs Exercise (2 minutes)
• For the previous exercise, if the final system had been defined to include the fired bullets, would the answer be different?

Helicopter

w/ ammo

mi = 1 ton

vi = 40 mph

Helicopter

w/ less ammo

mf = ?

vf = ?

bullets

Initial

Final

Initial State

-F

F

m1vi

Positive Direction

Time Passes

Final State

F

-F

m3vf

Linear Momentum Change by Unbalanced Forces
Linear Momentum Change by Unbalanced Forces
• Momentum flow INTO or OUT OF a system can result from FORCES.
• Mathematically,
• SI Units of Force = kg*m/s2
• SI Units of Momentum = kg*m/s
• Thus, Units of Force = Units of Momentum/s = Units of Momentum Flow Rate across a boundary, or
Linear Momentum and Newton’s Laws
• Newton’s 2nd Law, F=ma, is equivalent to stating ‘a net Force acting on a body changes the Linear Momentum of the body’.
• Adding Newton’s 3rd Law to this yields CONSERVATION OF LINEAR MOMENTUM (universe) since the linear momentum of one body changes the same as the other body, but in opposite directions.
Designating the System
• When solving these types of problems, the system boundary must be defined.
• This boundary is at the discretion of the engineer.
• There is no requirement that the system contain all bodies involved in the process.
• Thus, a system can have unbalanced forces which will change the linear momentum of the system.
Individual Exercise: (7 minutes)
• This is Problem 20.5
• A 1.000 ton dragster has a jet engine that provides a thrust of 3000. lbf. The dragster starts from a dead stop at the start line and crosses the finish line 9 seconds later. Neglect mass loss.
• What is its final velocity?
• How long was the track?

mi = 1 ton

vf = ?

Dt=9s

final

initial

F

Solution (a)

Burning fuel neglected, so mass is constant

Individual Exercise (7 minutes)
• This is Problem 20.19
• A 0.100 kg hockey puck is stationary on the ice. Then it is hit with a hockey stick and 0.020 s later, the puck is traveling at 100. km/h.
• What is the average force (N) on the puck?
• What is the average force (N) on the hockey stick?

Initial

Final

Dt=0.020

Fave

vo=0 m/s

vf=100 km/hr

Solution (a)
Solution (b)
• By Newton’s 3rd Law of equal and opposite reactions, the force on the hockey stick is the same as that on the puck.
Systems Without Net Linear Momentum Input
• No unbalanced forces & no mass transfer into or out of the system.
• Thus, UAE simplifies to:

ACCUMULATION=0, OR

FINAL AMOUNT - INITIAL AMOUNT = 0, OR

FINAL AMOUNT = INITIAL AMOUNT

• Therefore, there is no change in linear momentum of the system.
Individual Exercise: (7 minutes)
• In Example 20.5, what is the change in linear momentum of the black ball? The white ball? What is the total kinetic energy of the two balls before the impact and after the impact? Is energy conserved in this impact?
Solution

White Ball:

Dpwhite = mwhite(vwhite,final- vwhite,initial)

= 2 lbm * (1.5 - 2) ft/s

= -1.0 lbm*ft/s = -0.0311 lbf.s

Black Ball:

Dpblack = mblack(vblack,final- vblack,initial)

= 1 lbm * (1 - 0) ft/s

= 1.0 lbm*ft/s = 0.0311 lbf.s

Assignment# 12TEAM ASSIGNMENT
• Foundations: #20.4, 20.5, 20.9, 20.11, 20.18, 20.22 plus problem on next slide.
• DUE: 4/22/03
• Include cover sheet with names of team members and % participation
• FORMAT WILL BE GRADED AS WELL AS YOUR SOLUTION (i.e. – Given, Required, etc.)
Assignment # 12 (cont.)TEAM ASSIGNMENT
• A 4.00 oz (0.250 lbm) baseball is thrown by a pitcher towards a batter at a velocity of 80.0 ft/s.The batter impacts the ball with the bat for 0.015 seconds and propels the baseball back at the pitcher at a velocity of 100. ft/s. Assuming the ball travels the same path as it was thrown (but in the opposite direction) compute the force exerted on the ball by the bat.