Transcript 2.5 kg m/s - Purdue Physics

This Week
Momentum
Is momentum in basketball physics?
Rockets and guns
How do spaceships work?
Collisions of objects
They get impulses!
Practical Propulsion
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Momentum
What happens when a force acts on an object
We know
So that
v = v0 + at
mv – mvo = mat
mv – mvo = Ft
F
The quantity mv is known as the momentum p and it is a vector quantity in the
same direction as the velocity. The result of applying a force for a time t results
in a change of momentum.
So a body moving with velocity v has kinetic energy = 1/2mv2 and
momentum = mv
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Conservation of Momentum
If we have a collision between two objects then in general the
velocity of each object changes and kinetic energy is lost in the
form of heat, sound or in a permanent deformation of the bodies.
For a short interval of time t
we have mv – mvo = Ft and
p = F t
This is the impulse equation
F2
F1
F1 = -F2 and Δp1 + Δp2 = 0
Total momentum is conserved
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Isolated systems
In each case
m1v1 = - m2v2
http://www.physics.purdue.edu/academic_programs/courses/phys214/lectures/anim0010.mov
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Momentum conservation
The conservation of momentum also is
connected to the fundamental physical laws
The laws of physics do not change under
translation or rotation in space
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Impulse
We have seen that a force F acting for a time Δt changes
momentum
FΔt = Δp.
FΔt is called an impulse
There are many situations where it is more useful to use the
impulse when two objects collide than equating the change in
momentum of each. In the case shown below the momentum of
the earth changes but that is too difficult to calculate. It is better to
use the fact that the earth exerts a force for a short time.
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Collisions
In a closed, isolated system containing a
collision, the linear momentum of each
colliding object can change but the total
momentum of the system is a constant.
This statement is true even if energy is lost by
the colliding bodies
*
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Types of collision
Elastic - no energy is lost
Inelastic - Energy is lost (transformed)
Perfectly inelastic – objects stick together
In two dimensions momentum is conserved along
the x and y axes separately
y
x
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Perfectly Inelastic
Let initial velocity be vi and the mass of a car
be m.Then
mvi = 3mvfinal and vfinal = vi/3
Kinetic Energy before = 1/2mvi2
Kinetic energy after = ½ x 3mv2final
So KEbefore/KEfinal
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Head on Elastic collisions
Pool ball collisions are close to being elastic
v
v
Pool balls
v
Bowling ball hits tennis ball
Tennis ball hits bowling ball
http://www.physics.purdue.edu/class/applets/phe/collision.htm
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Collisions of Particles
Both energy and momentum and charge are conserved
in elementary particle interactions and are a powerful
tool in analyzing the fundamental physics.
There are also other important conservation laws that
for example prevent the proton from decaying.
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Summary of Chapter 7
The action of a force changes the momentum of an object
mv – mvo = Ft p = mv and is a vector
When two bodies interact they feel equal
and opposite forces
F1 = -F2 and Δp1 - Δp2 = 0
Total momentum is conserved
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Summary: Impulse
FΔt = Δp
is the impulse equation and is
used to determine the momentum
change when a force acts
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Practical Propulsion

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A space vessel has to have a propulsion system and this requires an engine
and fuel or an external force. It has to be able to maneuver and be able to
escape from the gravitational attraction of all objects that affect it’s path. Or,
for example, to leave the moon after landing.
Conventional rockets are ~90% fuel by weight most of which is used
escaping from the earth.
Very small satellites might be put into orbit using a powerful laser beam
Nuclear engines are used in deep space probes. In this radioactive decay of
the fuel emits particles which eject “backwards” and give the probe
momentum. In this case it takes a long time to achieve high velocities
Solar sails have been tested which use enormous sails pushed by the solar
wind of particles. This technique has limited application
 Remember the nearest star is 4 light years away that is ~ 2 x 1013 miles or
~200,000 times the distance to the sun (ONE WAY!!)
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1N-02 Collision of Two Large Balls
What happens when two large balls of equal mass collide
one is at rest at the other has velocity v1?
Conservation of momentum
mv1 + mv2 = mv1A + mv2A
 v1 = v1A + v2A
Conservation of Energy (Elastic)
Can we predict the
velocities of each
ball after a
collision ?
½ mv12 + ½ mv22 = ½ mv12 + ½ mv22
 v12 = v1A2 + v2A2 v1A = 0 & v2A = v1
Completely Inelastic collision (stick together)
mv1 + mv2 = (m + m)vA
(v2 =0)
 v1 = 2vA vA = ½v1
In practice some energy is always lost. You can hear
the noise when they hit and there will be some heat
generated at impact
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1N-04 Conservation of Linear Momentum
Two carts move under tension of weight on frictionless track
Is momentum
conserved in
this system ?
T
T
The initial momentum of the carts is
zero and they each feel equal and
opposite forces. So at any time the
net momentum will be zero.
0 = mAvA – mBvB  mAvA = mBvB
vA / vB = mB / mA
dA/dB = (vAtA) / (vBtB)
If, tA = tB
dA/dB = mB /
mA
IF THE MASSES ARE IN INVERSE RATIO TO THE INITIAL DISTANCES,
THE CARTS WILL ARRIVE AT THE STOPS SIMULTANEOUSLY.
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1H-01 Action – Reaction
Two Carts, Supporting different masses, are
pulled and pushed together in different ways.
Note where
the carts
come together.
CM
NO MATTER HOW THE PULLING IS DONE, THE CARTS END
UP IN THE SAME PLACE.
NO EXTERNAL FORCES ACT ON SYSTEM. THE CENTER OF
MASS STAYS WHERE IT IS. SO THE CARTS ALWAYS MEET
AT THE CENTER OF MASS.
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1N-05 Elastic Collision (Magnets)
Two Magnets collide with like poles facing each other
We know that the
magnets repel each
other, so they will
not touch. So is this
a collision?
Rest
v
v
S
S
S
S
Rest
What does it mean for
objects to ‘touch’ ?
MOMENTUM TRANSFER AND CONSERVATION REQUIRE
ONLY THAT THERE BE A MUTUAL INTERACTION.
AT THE MICROSCOPIC LEVEL, ALL CONTACT INVOLVES
ELECTROMAGNETIC INTERACTIONS.
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1N-06 Equality of Momentum
Two cylinders are exploded apart by a spring
What happens
when the spring
is released ?
First case: Both identical
Second case: One much heavier
Use Momentum conservation m1v1 = m2v2
The height reached is an indication of the
initial speed since 1/2mv2 = mgh
v = sqrt (2gh)
THE SPRING FORCE THAT DRIVES THEM APART IS
INTERNAL TO THE SYSTEM, SO THE NET MOMENTUM
REMAINS ZERO.
SINCE THE METAL CYLINDER IS HEAVIER IT FLIES AWAY
WITH A SMALLER VELOCITY TO CONSERVE MOMENTUM
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1N-10 Elastic & Inelastic Collisions
Elastic and Inelastic collisions of Two identical Carts on a Frictionless Track
How do the
collisions
compare?
Conservation of momentum
mvA + mvB = mvA’ + mvB’
Conservation of Energy (Elastic)
½ mvA2 + ½ mvB2 = ½ mvA’2 + ½ mvB’2
If vB = 0 then vA’ = 0 and vB’ = vA
Completely inelastic (two carts stick)
if vB = 0 then vAB = ½ vA
WE CAN MEASURE THE SPEED BY TIMING THE CARTS
ACROSS A FIXED DISTANCE. For THE INELASTIC CASE
HALF THE VELOCITY IMPLIES IT SHOULD TAKE TWICE THE
TIME.
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1N-12 Fun Balls
An enlarged version of the Classic Toy - The Array of Steel Balls
First case pull back one ball and release
What
happens if
we use the
big ball?
What happens
when more of the
balls are pulled
back than are left
at rest?
The collision is nearly elastic so we can
use both momentum conservation and
kinetic energy conservation
NO MATTER HOW MANY BALLS ARE PULLED BACK, THE
SAME NUMBER RECOIL AT THE SAME SPEED.
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Questions Chapter 7
Q5 Are impulse and momentum the same thing? Explain.
No impulse changes momentum
Q6 If a ball bounces off a wall so that its velocity coming back has
the same magnitude that it had prior to bouncing:
A. Is there a change in the momentum of the ball? Explain.
B. Is there an impulse acting on the ball during its collision with the
wall? Explain.
A. Yes momentum is a vector
B. Yes a force acts for a short time
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Q9 What is the advantage of an air bag in reducing injuries during
collisions? Explain using impulse and momentum ideas.
It increases the time over which the force acts.
It also spreads the force over a larger area
Q11 If you catch a baseball or softball with your bare hand, will
the force exerted on your hand by the ball be reduced if you pull
your arm back during the catch? Explain.
Yes. The impulse is the same but the impact time is longer.
From a work point of view the kinetic energy = Fd so increasing d
reduces F
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Q17 A compact car and a large truck have a head-on collision.
During the collision, which vehicle, if either, experiences:
A. The greater force of impact? Explain.
B. The greater impulse? Explain.
C. The greater change in momentum? Explain.
D. The greater acceleration? Explain.
A. The forces are equal and opposite
B. The impulse for each is the same
C. The momentum changes are equal and opposite
D. F = ma so a is larger for the compact car
Q22 Is it possible for a rocket to function in empty space (in a
vacuum) where there is nothing to push against except itself?
Yes. It ejects material at high velocity and momentum conservation
means the rocket recoils
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Q23 Suppose that you are standing on a surface that is so slick that
you can get no traction at all in order to begin moving across this
surface. Fortunately, you are carrying a bag of oranges. Explain
how you can get yourself moving.
Throw the oranges opposite to the direction you wish to move
Q24 A railroad car collides and couples with a second railroad
car that is standing still. If external forces acting on the system
are ignored, is the velocity of the system after the collision equal
to, greater than, or less than that of the first car before the
collision?
The velocity after is exactly half
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Ch 7 E 2
What is the momentum of a 1200 kg car traveling at
27 m/s?
P= mv = (1200 kg)(27 m/s)
P = 32400 kg m/s
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Ch 7 E 6
A ball experiences a change in momentum of 9.0 kg∙m/s.
a) What is the impulse?
b) If the time of interaction = 0.15 s, what is the magnitude
of the average force on the ball?
a) Impulse = p = 9.0 kg m/s
b) Impulse = Ft, F = 9/0.15 = 60N
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Ch 7 E 8
A ball has an initial momentum = 2.5 kg m/s, it bounces
off a wall and comes back in opposite direction with
momentum = -2.5 kg m/s
a) What is the change in momentum of the ball?
b) What is the impulse?
a) Δp = pf – pi = -2.5 – (+2.5) = - 5kgm/s
Pi = 2.5kgm/s
b) Impulse = Δp = - 5kgm/s
F
Pf = -2.5kgm/s
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Ch 7 E 10
M1 and M2 collide head on
a) Find initial momentum of M1 and M2
b) What is the total momentum of the system before
collision?
c) Ignore external forces, if they stick together after
collision, which way do the masses travel?
west M2 = 80kg
6.0m/s
3.5m/s
a) p1 = -100 x 3.5 = 350kgm/s
M1 = 100kg
east
p2 = 80 x 6 = 480kgm/s
b) Total momentum = 480 – 350 = 130kgm/s east
c) The masses will travel east with p = 130kgm/sec
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Ch 7 E 16
M1 = 4000kg v1 = 10m/s due north
M2 = 1200kg v2 = 20m/s due south
Masses collide and stick.
a) Find initial momentum of each mass.
b) Find size and direction of momentum after
collision.
N
M2
+x
v2
a) Call due north the +x direction
p1 = m1 v1 = 4000 kg (10m/s) = 40000 kg m/s
p2 = m2 v2 = 1200 kg (-20m/s) = - 24000 kg m/s
v1
S
M1
b) p = p1 + p2 = +16000 kg m/s
This is momentum of system before collision, but momentum is
conserved. So after masses stick:
p = 16000 kg m/s due North.
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Ch 7 E 18
A truck of mass 4000kg and speed 10m/s collides at right
angles with a car of mass 1500kg and a speed of 20m/s.
a) Sketch momentum vectors before collision
b) Use vector addition to get total momentum of system
before collision.
a) p1 = 40000kgm/s + y
p2 = 30000kgm/s + x
b)
p
y
p1
p1 = 40000
p  p1  p2
p2
p2 = 30000
x
p2 = p12 + p22
P = 50000kgm/s
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Ch 7 CP 2
A bullet is fired into block sitting on ice. The bullet travels at
500 m/s with mass 0.005 kg. The wooden block is at rest with a
mass of 1.2 kg. Afterwards the bullet is embedded in the block.
a) Find the velocity of the block and bullet after the impact (assume
momentum is conserved).
b) Find the magnitude of the impulse on the block of wood.
c) Does the change in momentum of the bullet equal that of wood?
a) pfinal = pinitial = (0.005 kg)(500 m/s)
m
v
M
pfinal = (Mbullet + Mwood)v = 2.5 kg m/s
No friction (ice)
v = (2.5 kg m/s)/(1.205 kg) = 2.07 m/s
b) Impulse = Δp = pfinal – pinitial
= (1.2 kg)(2.07 m/s) – 0 = 2.50 kg m/s
c) Δp for bullet = (0.005 kg)(500 m/s) – (0.005 kg)(2.07 m/s)
= 2.50 kg m/s
Momentum is conserved, so momentum lost by bullet is gained
by wood.
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Ch 7 CP 4
Car travels 18 m/s and hits concrete wall. Mdriver = 90 kg.
a) Find change in momentum of driver.
b) What impulse produces this change in momentum?
c) Explain difference between wearing and not wearing a
seat belt.
a) When driver comes to a stop his p = 0.
p = 0 – (18 m/s)(90 kg) = -1620 kg m/s
b) Impulse = p = -1620 kg m/s
v
p
c) Impulse = Ft
With seat belt: t is large and F is spread over torso of driver
Without seat belt: t is small. This makes F much larger (F =
Impulse/t).
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