Download Chapter 06 Momentum

Chapter 6 Momentum
Section 6.1 Momentum
One part of the definition of inertia is that it is the tendency of a moving object to
keep moving. The greater the tendency of the object to maintain its motion, the greater
the inertia. Objects with greater mass tend to have greater inertia. A larger velocity also
tends to cause the inertia to be greater. Momentum is a measure of inertia in motion.
Just as we might expect from the description of the factors that change inertia,
momentum is equal to mass multiplied by velocity. This is simply expressed as:
p = mv.
Example 1: A 1000 kg gator has a velocity of 15 m/s east. What is the
momentum of the truck?
Answer:
p = mv
p = (1000 kg)(15 m/s) = 15,000 kgm/s east
Momentum is a vector, so we must indicate the direction. Therefore, the
momentum can be negative (unlike kinetic energy). The unit is the kilogram x
meter/second, or kgm/s.
A moving object can have a large momentum if it has a large mass, a high
speed, or both.
Example 2: Is there a situation where a baseball and a car would have the same
momentum?
Answer: If the base ball were moving fast enough and the car slow enough such
that the mass of each multiplied by the velocity of each were equal results, then
the magnitude would be the same. They would also have to be moving in the
same direction.
Another situation would be if both the ball and the car were sitting still, so that
both velocities were zero. If the velocity is zero, the momentum is zero.
If this sounds too simple to be true, you will be happy to know that momentum is
actually this simple. It is just mass multiplied by velocity. Not any more complicated than
that.
Section 6.2 Change of Momentum
Momentum is mass x velocity. Therefore, if momentum changes, either the mass
of the object, or its velocity, or both change. In our studies it is not very common for the
mass of an object to change. In most cases the velocity changes, not the mass.
We have studied changes in velocity. A change in velocity is acceleration, and
we know that acceleration is caused by only one thing: an unbalanced force. An
unbalanced force MUST cause acceleration and a change in momentum. And if the
velocity and the momentum change, it MUST be as a result of an unbalanced force.
There are no exceptions to the statements in this paragraph.
Section 6.3 Impulse and Change of Momentum
The greater the force, the greater the acceleration and the greater the change in
momentum. The length of time the force is applied is also important. The longer the
force is applied the greater the change in momentum.
Force multiplied by time interval is called impulse.
Impulse = Ft
Impulse is related to momentum. Newton’s second law states:
F = ma, and
we know the formula for acceleration: a = v/t. If we replace a in F = ma with v/t,
F = ma becomes: F = mv/t.
We can rearrange this equation to form: Ft = mv, which becomes: Ft = mv.
Impulse is equal to change in momentum. This equation can be very useful. It can be
very difficult to find the force applied to an object, maybe more difficult to find the time
interval over which the force is applied. But while these values, which are necessary to
find impulse, may be hard to find, it is usually very easy to find the change in
momentum. Since the mass seldom changes, the change in momentum is usually the
change in velocity multiplied by the mass.
Example 3: A 2000 kg truck is going north at 15 m/s. It hits a brick wall and is
brought to rest over a time period of 0.42 s. What is the magnitude of the force
exerted on the truck during the collision?
Answer: The truck comes to rest, so the change in velocity is -15 m/s.
Ft = mv
F x 0.42 = (2000)(15)
F = 71,400 N
This seems like a large answer, but notice that the force is applied for less than
one second.
To increase the momentum of an object most affectively, one should apply a
large force for a long period of time.
Decreasing momentum requires an impulse. Te equation mv = Ft shows that a
great mv can be accomplished with little force (F) if the force is applied over a large
time (t).
Example 4: If the truck driver in the previous example has the chance, should he
run into a stack of hay rather than the brick wall?
Answer: The stack of hay will cause the truck to take a longer time to stop;
therefore, the force needed to decrease the momentum to zero will be less.
There are many other examples where the time of impact is extended. The more
the time is increased, the lower the force needed to change the momentum. The
padded dashboard in your car, automobile airbags, bending your knees to absorb the
landing when you jump a long distance, cushioned mats for gymnasts, a glass dish that
doesn’t break because it is dropped on a carpeted floor rather then tile: these are all
examples of time of impact being longer so that force of impact can be smaller.
Bouncing objects require a greater force than an object that doesn’t bounce. The
change in momentum is greater when an object bounces, therefore a greater impulse is
needed.
Example 5: Why is hail more damaging than rain?
Answer: One factor is that the hailstones are ice, so they are harder than liquid
water. But another fact that must be considered is that hailstones bounce when
they hit an object, while water droplets stop when they hit the object. That means
the change of momentum is greater for the hailstones. If the time of impact is the
same, and the change of momentum is greater, then force must be greater.
Hailstones hit harder than raindrops.
Example 6: If you have ever played paintball, you know that a paintball that
doesn’t break when it hits you hurts more than one that breaks. Why?
Answer: This is the same as the hailstones. A paintball that breaks has less of a
momentum change than one that bounces off of its target. More momentum
change means more force. More force, more pain, ouch!
When a kicker kicks a football, the force is not constant throughout the time the
force is applied. At the point the foot just touches the football, the force is zero. As the
ball is compressed by the foot, the force increases to a maximum, then decreases as
the ball rebounds off the foot. The forces involved in impulse are not constant. If a graph
of force vs time is drawn, it follows a bell type curve. The force starts small, increases to
a maximum, then decreases. In calculations we use the average force over the interval
of impulse.
Section 6.4 The Law of Conservation of Momentum
We have already established that in order to change the momentum of an object,
an unbalanced force must be applied. If you are riding in a car, and you push against
the inside of the door; does the car accelerate in the direction of your force? No,
Newton’s third law says that your force on the door is balanced by an equal and
opposite force that the door applies to you. You and the car are a system and, since the
net force is zero, the net force on that system is zero. Since there is no net force, there
is no change in momentum. Internal forces are always in balanced pairs and have no
effect on the total momentum of the system. An external unbalanced force is needed to
change the motion of an object.
If no net force or impulse acts on a system, the momentum of that system cannot
change. This is the law of conservation of momentum: In the absence of an external
force, the momentum of a system remains unchanged. The law of conservation of
momentum is one of the three great conservation laws in physics: the law of
conservation of energy and the law of conservation of angular momentum are the other
two.
Example 7: An 80 kg thrasher is at rest on his 5 kg skateboard. If he steps
forward at a velocity of 2 m/s, what is the final velocity of the skateboard?
Answer: The system of the skateboard and his rider are at rest so the momentum
of the system is zero. The momentum of the system must be conserved, so it
must remain zero. The rider’s momentum forward must be balanced by the equal
(but negative) momentum of the skateboard in the opposite direction.
mv (thrasher) = mv (skateboard)
(80)(2) = (5)v
v = 32 m/s
This 32 m/s must be in the opposite direction of the rider, so if we are calling his
velocity 2 m/s, the velocity of the skateboard must more correctly be -32 m/s.
Example 8: Newton’s Second Law states that if no net force is exerted on a
system, no acceleration occurs. Does it follow that no change in momentum
occurs?
Answer: If an object does not accelerate, its velocity remains constant. That
means momentum remains constant, so it does follow that no change in
momentum occurs.
Example 9: Newton’s Third Law states that the force a cannon exerts on a
cannonball is equal and opposite to the force the cannonball exerts on the
cannon. Does it follow that the impulse the cannon exerts on the cannonball is
equal and opposite to the impulse the cannonball exerts on the cannon?
Answer: Impulse is force multiplied by time. The force on each is equal by
Newton’s third law. The only question is whether the time the forces are applied
is the same. Newton’s third law also says that an action force cannot be applied
without its reaction force also being present. Therefore, the time of application
must be the same. Yes, the impulse is the same.
Section 6.5 Elastic and Inelastic Collisions
When objects collide without being permanently deformed and without
generating heat, it is an elastic collision. Billiard balls colliding is one example of elastic
collisions. (Sort of. Some heat is generated when billiard balls collide. The only real
example of elastic collisions is collisions of subatomic particles.). When perfectly elastic
collisions occur, kinetic energy is conserved. Momentum is conserved in elastic
collisions.
When the objects are distorted in a collision and the collision generates heat, it is
said to be an inelastic collision. Kinetic energy is not conserved in an inelastic collision.
Collisions where objects stick together are said to be perfectly inelastic. Some energy is
lost in the deformation of the objects and some energy is lost in the production of heat
and sound. Momentum is still conserved in inelastic collisions. Momentum is always
conserved in all collisions.
Example 10: A great white shark (mass = 3000 kg) swims at a velocity of 2 m/s
and swallows an unlucky fish (200 kg) in one gulp. If the unlucky fish is
motionless before becoming lunch, what is the final velocity of the great white +
his meal?
Answer: The momentum before the collision must equal that after the collision.
mv = mv
(3000)(2) + (200)(0) = (3000 + 200) v
v = 1.88 m/s
(The velocity didn’t change much because the mass of the smaller fish was a
small percentage of the shark’s mass.)
Example 11: How would the situation be different if the smaller fish were
swimming toward the shark at 10 m/s?
Answer: The momentum must still be conserved.
mv = mv
(3000)(2) + (200)(-10) = (3000 + 200) v v = 1.25 m/s
Most collisions are neither perfectly elastic nor perfectly inelastic. Most collisions
are partly elastic and partly inelastic.
Example 12: A 4.00 kg ball moving east at 4.00 m/s makes an elastic head-on
collision with a 5.00 kg ball moving 3.00 m/s west. After the collision, the 4.00 kg
ball moves west at 3.78 m/s. What is the final velocity of the 5.00 kg ball?
Answer: This is solved using the conservation of momentum. We define east as
the positive direction.
pinitial = pfinal
(4 kg)(4 m/s) + (5 kg)(-3 m/s) = (4 kg)(-3.78 m/s) + (5 kg)(v)
v = 3.22 m/s The positive answer indicates the 5 kg ball is now moving east.
Example 13: Using the velocity just calculated for the 5 kg ball, determine if
kinetic energy is conserved. (Is the collision elastic?)
½ (4 kg)(4 m/s)2 + ½ (5 kg)(-3 m/s) 2 = ½ (4 kg)(-3.78 m/s) 2 + ½ (5 kg)(3.22) 2
32 j + 22.5 j = 28.6 j + 25.9 j
54.5 j = 54.5 j
The kinetic energy is the same before and after the collision, so this is an elastic
collision.
(Note: We must assume there is no friction involved in the movement of these
two balls as that would use some of the kinetic energy. We must also assume the
balls are not rolling. There is a form of kinetic energy involved in rotation of
masses.)
Example 14: Two toy railroad cars meet in a collision and couple together. The
first car has a mass of 2 kg and an initial velocity of 0.4 m/s to the right. The
second car has a mass of 1.5 kg and is moving to the left at 0.3 m/s. What is the
final velocity of the two coupled cars?
Answer: Momentum must be conserved.
pinitial = pfinal
(2 kg)(0.4 m/s) + (1.5 kg)(-0.3 m/s) = (2 kg +1.5 kg) (v)
v = 0.1 m/s
The positive answer indicates that the cars are moving to the right.
Example 15: The collision in Example 14 is inelastic. How much kinetic energy is
lost in the collision?
Answer:
K E = ½ mv2
The KE before the collision is ½ (2 kg)(0.4 m/s)2 + ½ (1.5 kg)(-0.3 m/s)2 =
0.16 j + 0.068 j = 0.228 j
The KE after the collision is ½ (3.5 kg)(0.1 m/s)2 = 0.018 j.
So the KE lost is 0.228 j – 0.018 j = 0.210 j.
When objects collide that are not moving parallel to each other, vector analysis is
used to find the resulting momentum. When two dimensional motion is calculated, the
momentum along the x-axis must be conserved, and the momentum along the y-axis
must be conserved.
Momentum of a system in the absence of unbalanced external forces must be
conserved even if the system splits apart into fragments. If a spaceship moving at
constant velocity in space explodes, the net momentum of all the particles combined will
have the same momentum as the spacecraft had before the explosion, even though
each individual particle might have its own unique velocity. The explosion results in a
set of Newton’s third law pairs of internal forces, and these can’t change the net
momentum of the system. This is true for an exploding firecracker as well. The sum of
the individual fragments will be equal to the original momentum of the firecracker as a
whole. If thrown on earth, the firecracker will follow a parabola as it is in projectile
motion. After the firecracker explodes, the net motion of all the pieces combined (the
center of gravity) will continue to follow the parabola.