Download Physics 170 Week 11, Lecture 2

Physics 170 Week 11, Lecture 2
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Physics 170 203 Week 11, Lecture 2
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Textbook Chapter 15:
Section 15.2-4
Physics 170 203 Week 11, Lecture 2
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Learning Goals:
• We will review the concept of work due to a force. the law of
work and energy and the law of impulse and linear momentum.
• We will discuss conservation of momentum in a system where
no forces act.
• We will study the law of impulse and momentum for a system
of many particles and the concept of conservation of
momentum for systems with only mutual interactions.
• We will solve an example which uses these ideas.
• Students should learn how to recognize problems where they
can apply conservation of momentum and then to use
conservation of momentum to help solve mechanical problems.
• We will begin analyzing impacts.
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Example
Two men, one having a weight of 160 lb and the other a weight of
180 lb, stand on the 200 lb cart. Each runs with a speed of 3 f t/s
measured relative to the cart. Determine the final speed of the cart
if (a) the lighter runs and jumps off, then the heavier runs and
jumps off the same end, (b) the heavier runs and jumps off, then
the lighter runs and jumps off the same end, and (c) both run at
the same time and jump off at the same time. Neglect the mass of
the wheels and assume the jumps are made horizontally.
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Review
• Principle of work and energy
Z
t2
1
1
d~r(t)
dtF~ (t) ·
m~v 2 (t2 ) = m~v 2 (t1 ) +
dt
2
2
dt
t1
R ~r2
~
• If F is conservative, ~r1 dtF~ (t) · d~r = −V (~r2 ) + V (~r1 )
1
1
m~v 2 (t2 ) + V (~r2 ) = m~v 2 (t1 ) + V (~r1 )
2
2
• Law of impulse and linear momentum
Z t2
m~v (t2 ) = m~v (t1 ) +
F~ (t) dt
t1
All of these assume that F~ is the resultant of all forces acting
on the particle and that its motion obeys F~ = m~a.
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Law of impulse and linear momentum in a system of
particles
Apply it to each particle:
Z t2
mi~vi (t2 ) = mi~vi (t1 ) +
F~i (t) dt
t1
where ~vi (t) is the velocity of the i0 th particle and F~i is the resultant
of all forces acting on the i0 th particle.
If forces are only internal. i.e. if F~ij is the force on particle i
caused by particle j, then F~ji is the force on particle j caused by
particle i then, Newton’s third law tell us
F~ij = −F~ji
For particle i:
Z
t2
mi~vi (t2 ) = mi~vi (t1 ) +
t1
Physics 170 203 Week 11, Lecture 2
X
F~ij (t) dt
j:j6=i
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Law of impulse and linear momentum in a system of
particles cont’d
For each particle
Z t2 X
F~ij (t) dt
mi~vi (t2 ) = mi~vi (t1 ) +
t1
j:j6=i
P
~
Total momentum P = i mi~vi (t2 ),
X
mi~vi (t2 ) =
i
But
P
i,j:j6=i
X
Z
mi~vi (t1 ) +
t1
i
F~ij =
t2
X
F~ij (t) dt
i,j:j6=i
´
³
~
~
i,j:i<j Fij − Fji = 0 and
P
X
i
mi~vi (t2 ) =
X
mi~vi (t1 )
i
The total momentum is conserved (three components).
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Example
Two men, one having a weight of 160 lb and the other a weight of
180 lb, stand on the 200 lb cart. Each runs with a speed of 3 f t/s
measured relative to the cart. Determine the final speed of the cart
if (a) the lighter runs and jumps off, then the heavier runs and
jumps off the same end, (b) the heavier runs and jumps off, then
the lighter runs and jumps off the same end, and (c) both run at
the same time and jump off at the same time. Neglect the mass of
the wheels and assume the jumps are made horizontally.
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Example cont’d
If we consider the “particles” to be the two men and the cart, the
only forces which are acting in the horizontal direction are internal,
mutual interactions of the particles..
The horizontal momentum of the total system is therefore
conserved.
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Example cont’d
Let us first assume that:
Man A: mass mA jumps off with horizontal velocity vA relative to
the cart.
Then
Man B: mass mB jumps off with horizontal velocity vB relative to
the cart.
Cart: mass mC has initial horizontal velocity 0, then has velocity
v1 after A jumps, then has velocity v2 after B jumps.
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Example cont’d
Total momentum of two men and cart is initially zero.
Immediately after A jumps, his velocity relative to the ground is
vA + v1
mA (vA + v1 ) + (mB + mC )v1 = 0
Now, total momentum of man B plus cart is (mB + mC )v1 .
Immediately after B jumps
mB (vB + v2 ) + mC v2 = (mB + mC )v1
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Example cont’d
We have obtained the two equations for conservation of momentum,
mA (vA + v1 ) + (mB + mC )v1 = 0
mB (vB + v2 ) + mC v2 = (mB + mC )v1
From the first equation
v1 = −
Physics 170 203 Week 11, Lecture 2
mA
vA
mA + mB + mC
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Example cont’d
v1 = −
mA
vA
mA + mB + mC
mB (vB + v2 ) + mC v2 = (mB + mC )v1
Combining these,
v2 = v1 −
mB
vB
mB + mC
The final velocity of the cart is
mA
mB
v2 = − (mA +m
v
−
vB
A
+m
)
m
B
C
B +mC
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Example cont’d
vA = vB = 3 f t/s
Lightest jumps first: mA = 160, mB = 180, mC = 200
µ
¶
160
180
+
= −2.31 f t/s
v2 = −(3 f t/s)
180 + 200 160 + 180 + 200
Heaviest jumps first: mA = 180, mB = 160, mC = 200
µ
¶
180
160
v2 = −(3 f t/s)
+
= −2.33 f t/s
160 + 200 160 + 180 + 200
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Example cont’d
Both jump at the same time:
(mA + mB )(vAB + v1 ) + mC v1 = 0
160 + 180
mA + mB
vAB = −(3 f t/s)
= −1.89 f t/s
v1 = −
mA + mB + mC
160 + 180 + 200
The final velocity of the cart is least if both men jump off at the
same time.
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Impact
Impact occurs when two bodies collide with each other during a
very short period of time, causing relatively large (impulsive) forces
to be exerted between the bodies.
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Five stages of Impact
Approach
Deformationion
Z
mA~v = mA~vA1 +
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Z
P~ , mB ~v = mB ~vB1 −
P~
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Maximum compression
Restitution
Z
Z
mA~vA2 = mA~v −
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~ , mB ~vB2 = mB ~v +
R
~
R
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Rebound
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Kinematics of a collision
All forces are internal, so momentum is conserved,
mA~vA2 + mB ~vB2 = mA~vA1 + mB ~vB1
Consider components parallel to the line of collision.
Let v be the velocity at the instant of maximum deformation.
Z
Z
mA v = mA vA1 + P , mA vA2 = mA v − R
Coefficient of restitution
R
R
R
R
v − vA2
−v + vB2
R
R
e=
=
, e=
=
vA1 − v
−vB1 + v
P
P
eliminate v
e=
Physics 170 203 Week 11, Lecture 2
vB2 − vA2
vA1 − vB1
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Coefficient of restitution:
e=
vB2 − vA2
vA1 − vB1
Elastic impact e = 1
Plastic impact e = 0
Later, we will analyze collisions where we use the equation for e
plus conservation of momentum to solve, for example, for the final
velocities of particles in terms if initial velocities.
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Example:
If we drop an object from height h, its velocity when it hits the
ground is (using conservation of energy)
p
vA1 = − 2gh
where h is the initial height. It then has a collision with the earth.
The velocity of the earth is
vB1 = 0
After the collision, the object has a velocity vA2 and the earth has
velocity vB2 = 0.
p
−vA2
→ v2A = −ev1A = e 2gh
e=
vA1
Height to which the object bounces is (by conservation of energy)
q
0
e = hh
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For the next lecture, please read
Textbook Chapter 14:
Section 14.5-6
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