Download - Lecture Notes for Sections 15-2-15-3
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7/28/2019 Lecture Notes for Sections 15-2-15-3
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PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM
AND CONSERVATION OF LINEAR MOMENTUM FOR
SYSTEMS OF PARTICLES
Todays Objectives:
Students will be able to:
1. Apply the principle of linear
impulse and momentum to a
system of particles.
2. Understand the conditions for
conservation of momentum.
In-Class Activities:
Check HomeworkReading Quiz
ApplicationsLinear Impulse and
Momentum for a System of
Particles
Conservation of LinearMomentum
Concept QuizGroup Problem SolvingAttention Quiz
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READING QUIZ
1. The internal impulses acting on a system of particles
always __________A) equal the external impulses. B) sum to zero.
C) equal the impulse of weight. D) None of the above.
2. If an impulse-momentum analysis is considered during the
very short time of interaction, as shown in the picture, weight
is a/an __________
A) impulsive force.B) explosive force.
C) non-impulsive force.
D) internal force.
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APPLICATIONS
Does the release velocity of the ball
depend on the mass of the ball?
As the wheels of this pitching machinerotate, they apply frictional impulses to
the ball, thereby giving it linear
momentum in the direction ofFdt and
Fdt.
The weight impulse, Wt is very smallsince the time the ball is in contact
with the wheels is very small.
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APPLICATIONS (continued)
This large crane-mounted hammer isused to drive piles into the ground.
Conservation of momentum can be
used to find the velocity of the pile
just after impact, assuming thehammer does not rebound off the pile.
If the hammer rebounds, does the pile velocity change from
the case when the hammer doesnt rebound ? Why ?
In the impulse-momentum analysis, do we have to consider
the impulses of the weights of the hammer and pile and the
resistance force ? Why or why not ?
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PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM
FOR A SYSTEM OF PARTICLES
(Section 15.2)
The linear impulse and momentum equation for this system
only includes the impulse ofexternal forces.
mi(v
i)2dtFimi(vi)1
t2
t1=+
For the system of particles shown,
the internal forces fibetween
particles always occur in pairs with
equal magnitude and oppositedirections. Thus the internal
impulses sum to zero.
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For a system of particles, we can define a fictitious centerof mass of an aggregate particle of mass mtot, where mtot is
the sum ( mi) of all the particles. This system of particles
then has an aggregate velocity ofvG = ( mivi) / mtot.
MOTION OF THE CENTER OF MASS
The motion of this fictitious mass is based on motion of the
center of mass for the system.
The position vectorrG = ( miri) / mtot describes the motionof the center of mass.
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CONSERVATION OF LINEAR MOMENTUM FOR A
SYSTEM OF PARTICLES (Section 15.3)
When the sum of external impulses acting on a systemof objects is zero, the linear impulse-momentum
equation simplifies to
mi(v
i)1 = mi(vi)2
This equation is referred to as the conservation oflinear momentum. Conservation of linear momentum
is often applied when particles collide or interact.
When particles impact, only impulsive forces cause a
change of linear momentum.
The sledgehammer applies an impulsive force to the stake. The weight
of the stake is considered negligible, or non-impulsive, as compared to
the force of the sledgehammer. Also, provided the stake is driven into
soft ground with little resistance, the impulse of the ground acting on the
stake is considered non-impulsive.
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EXAMPLE I
Given: M = 100 kg, vi = 20j(m/s)mA = 20 kg, vA = 50i+ 50j(m/s)
mB = 30 kg, vB = -30i50k(m/s)
An explosion has broken the
mass m into 3 smaller particles,
a, b and c.
Find: The velocity of fragment C after
the explosion.
Plan: Since the internal forces of the explosion cancel out, we can
apply the conservation of linear momentum to the SYSTEM.
z
x
y
M
vi
vC
C
vB
B
vA
A
=
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mvi= mAvA + mBvB + mCvC
100(20j) = 20(50i+ 50j) + 30(-30i-50k) + 50(vcx i+ vcyj+ vcz k)
Equating the components on the left and right side yields:
0 = 1000900 + 50(vcx) vcx = -2 m/s
2000 = 1000 + 50 (vcy) vcy = 20 m/s
0 = -1500 + 50 (vcz) vcz = 30 m/s
So vc = (-2i+ 20j+ 30k) m/s immediately after the explosion.
EXAMPLE I
(continued)Solution:
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EXAMPLE II
Find: The speed of the car A after collision if the cars
collide and rebound such that B moves to the rightwith a speed of 2 m/s. Also find the average
impulsive force between the cars if the collision
place in 0.5 s.
Plan: Use conservation of linear momentum to find thevelocity of the car A after collision (all internal
impulses cancel). Then use theprinciple of impulse
and momentum to find the impulsive force by looking
at only one car.
Given: Two rail cars with masses
of mA = 20 Mg and mB =15 Mg and velocities as
shown.
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EXAMPLE II
(continued)
Conservation of linear momentum (x-dir):mA(vA1) + mB(vB1) = mA(vA2)+ mB(vB2)
20,000 (3) + 15,000 (-1.5)
= (20,000) vA2 + 15,000 (2)
vA2 = 0.375 m/s
The average force is
F dt = 52,500 Ns = Favg(0.5 sec); Favg = 105 kN
Solution:
Impulse and momentum on car A (x-dir):
mA (vA1)+ F dt = mA (vA2)
20,000 (3) - F dt = 20,000 (0.375)
F dt = 52,500 Ns
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CONCEPT QUIZ
2) A drill rod is used with a air hammer for making holes in
hard rock so explosives can be placed in them. How many
impulsive forces act on the drill rod during the drilling?
A) None B) One
C) Two D) Three
1) Over the short time span of a tennis ball hitting the racket
during a players serve, the balls weight can beconsidered _____________
A) nonimpulsive.
B) impulsive.
C) not subject to Newtons second law.D) Both A and C.
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GROUP PROBLEM SOLVING
Find: The ramps speed when the crate reachesB.
Plan: Use the energy conservation equation as well as
conservation of linear momentum and the relative
velocity equation (you thought you could safely forget
it?) to find the velocity of the ramp.
Given:The free-rolling ramp has a
weight of 120 lb. The 80 lb crateslides from rest at A, 15 ft down
the ramp to B.
Assume that the ramp is smooth,
and neglect the mass of thewheels.
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GROUP PROBLEM SOLVING
(continued)
Energy conservation equation:0 + 80 (3/5) (15)
= 0.5 (80/32.2)(vB)2 + 0.5 (120/32.2)(vr)
2
Solution:
Since vB = vr+ vB/r -vBx i+ vByj= vri+vB/r(4/5 i3/5 j)
-vBx = vr (4/5)vB/r (2)
vBy = (3/5)vB/r (3)
+
To find the relations between vB andvr, use
conservation of linear momentum: 0 = (120/32.2) vr (80/32.2) vBx vBx =1.5 vr (1)
Eliminating vB/rfrom Eqs. (2) and (3) and Substituting Eq.
(1) results in vBy
=1.875 vr
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GROUP PROBLEM SOLVING
(continued)
Then, energy conservation equation can be rewritten ;0 + 80 (3/5) (15) = 0.5 (80/32.2)(vB)
2 + 0.5 (120/32.2)(vr)2
0 + 80 (3/5) (15) = 0.5 (80/32.2) [(1.5 vr)2 +(1.875 vr)
2]
+ 0.5 (120/32.2) (vr)2
720 = 9.023 (vr)2
vr
= 8.93 ft/s
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ATTENTION QUIZ
2. The 200-g baseball has a horizontal velocity of 30 m/s when it
is struck by the bat, B, weighing 900-g, moving at 47 m/s.
During the impact with the bat, how many impulses of
importance are used to find the final velocity of the ball?
A) Zero B) One
C) Two D) Three
1. The 20 g bullet is fired horizontally at 1200 m/s into the
300 g block resting on a smooth surface. If the bulletbecomes embedded in the block, what is the velocity of the
block immediately after impact.
A) 1125 m/s B) 80 m/s
C) 1200 m/s D) 75 m/s
1200 m/s
BAT
vball
vbat
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