Transcript kg· m/s - University of Redlands

Section 2.4 – 2.6
Momentum Principle
& the Universal Speed limit
Predictions of Motion under the influence of Forces
Meet Gravitation – constant force
Meet Springs – changing force
Q 2.2 a
An object is moving in the +x direction.
Which of the following statements about the net force acting on the
object could be true?
a)
b)
c)
d)
e)
f)
g)
The net force is in the +x direction
The net force is in the –x direction
The net force is zero
A and B
B and C
A and C
A, B and C
Q 2.2 b
Three carts are moving on three tracks. Which cart(s) experience a net
force to the left?
A.
B.
C.
D.
E.
F.
Green Cart which moves to the left at constant speed.
Red Cart which moves to the left, gradually speeding up.
Blue Cart which moves to the left, gradually slowing down.
Both Green and Red carts
Both Blue and Red carts
Both Green and Blue carts
Question
A car is moving at 3 m/s in the positive x
direction. Which of these describes that fact:
a) V = <-3,0,0> m/s
b) V = 3 m/s x^
^
c) V = 3 m/s -x
d) V = <0,3,0> m/s
Question
A car experiences a force in the -x^ direction. Which
of these is a possible change in momentum it would
experience:
a) Dp = <-10,0,0> kg m/s
b) Dp = 10 kg m/s x^
c) Dp = 10 kg m/s -x^
d) Dp = <0,-10,0> kg m/s
e) Answers (b) and (d)
f) Answers (a) and (c)
Question
A ball has a momentum in the y^ direction at one
moment. At the next moment has a motion in the –y^
direction. What direction does the force on it point?
^
a) Dp = 10 kg m/s y
^
b) Dp = -10 kg m/s y
c) Dp = 10 kg m/s x^
^
d) Dp = -10 kg m/s x
e) Not enough information to tell.
Q 2.3 d: You practice on paper like homework
Inside a spaceship in outer space there is a small steel ball. At a
particular instant, the ball has momentum < -8, 3, 0 > kg· m/s and is
pulled by a string, which exerts a force < 20, -10, 0 > N on the ball.
What is the ball’s (vector) momentum 2 seconds later?
A.
B.
C.
D.
E.
< -28, 23, 0 > kg· m/s
< 12, -7, 0 > kg· m/s
36.2 kg· m/s
< 32, -17, 0 > kg· m/s
< 40, -20, 0 > kg· m/s
Students collide
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The initial momentum of the one-student system is nonzero.
The initial momentum of the two-student system is zero.
We're interested in the force exerted on one student, so we apply the Momentum
Principle to a system consisting of one student.
We can estimate Δt from the compression distance in the collision.
The final momentum of the one-student system is nonzero.
The net force on the two-student system is zero because the interatomic force
exerted by student 1 on student 2 is equal in magnitude but opposite in direction
to the force exerted by student 2 on student 1.
A good estimate of the initial speed of one of the running students is 0.1 m/s.
The Momentum Principle can't be applied in this analysis because we have no way
of estimating Δt.
The final momentum of the two-student system is nonzero.
The Momentum Principle when applied to the two-student system is simply 0 = 0
and doesn't give us any information.
Updating Position in Computer
Simulations
Position Update (ch 1)

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
robject new  robjectold  vobjectave Dt
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vobjectave 
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pobjectave
mobject
Dt
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Momentum Update (ch 2) pobjectnew  pobjectold  Fnet object ave Dt
Note: If Dt is small enough, old speed is close enough to average, old force is
close enough to average to approximate in simulation.
A “while loop”
while t<tmax
t  t  Dt
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pobject  pobject  Fnet objectDt
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p
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object
robject  robject 
Dt
mobject
1-D Motion fan cart (1-D). Suppose you have a fan cart whose mass is 400 gram
and it’s on a huge, 8m long track. Initially you set the cart moving down the track in
the x-direction at location <0.5,0,0>m with velocity <1.2,0,0>m/s. The force due to the
fan is <0.2,0,0>N; comparatively, friction is negligible. What are the momentum and
position of the cart 3 seconds later?
Calculate this in a series of 1 second intervals.
Gravitational force
• Constant force over small regions. Works well
for an integrated solution.

Magnitude: FEarth  mg
9.8 m/s2
Mass
Direction: toward earth
(sign and component depend on your choice of coordinate systems)
Smoothly-Varying Force: Spring


Fsp  k s s  k s sLˆ
Example: With the spring over here, I hang 4.9 N (0.5 kg) weight from it, and it stretches by
______m. So, what’s it’s stiffness?
Example: A spring is 0.14 m long when it is relaxed. When a force of magnitude 325 N is
applied, the spring becomes 0.22 m long.
(a) What is the stiffness of this spring?
(b) Next, this spring is compressed so that its length is 0.08 m. What magnitude of force is
required to do this?
Q 2.5 c
A spring is 12 cm (0.12 m) long when relaxed. Its stiffness is 30 N/m.
You push on the spring, compressing it so its length is now 10 cm
(0.10 m).
What is the magnitude of the force the spring now exerts on your
hand?
a)
0.6 N
b)
3N
c)
3.6 N
d)
30 N