Download Since the cars have equal mass and approach each other

1. Provide short answers to the questions (a) through (f). To receive full credit you must justify your reasoning;
yes and no answers without supporting explanation are not sufficient.
a) What is “impulse” and how is it related to an object’s momentum and forces exerted upon it? Why is a wine
glass less likely to break if dropped on a carpet rather than a tile floor? (5)
Impulse is the action of a force over a specified time or I = Ft. By Newton’s Second Law the change in
momentum of a body is numerically equal to the impulse delivered to it (p = Ft).
A wine glass is less likely to break when dropped on a rug because of the rug prolonging the time over which
the glass is brought to rest. Given that the glass falls from rest at a specified height the change in momentum of
the glass is the same regardless of whether it hits tile or rug. However, the force F = p/t is larger for a
shorter collision time with the floor. Since the rug has “give” while the tile does not the collision force is
correspondingly larger when the glass hits the tile.
b) What distinguishes potential energy from kinetic energy? Discuss what effect doing work on a body can have
on each of these quantities. (5)
Potential Energy is energy associated with the configuration or position of a body. One can generally associate
potential energy with stored energy. For example, a mass at a height h above the ground has a potential energy
relative to its position on the ground. Similarly, a stretched rubber band or spring has stored potential energy.
Kinetic energy is energy associated with a body’s motion. For a given speed more massive bodies have greater
kinetic energy. For a body of given mass increasing its speed increases its kinetic energy.
Upon doing work on a system it is possible to change a body’s kinetic energy, potential energy, or both. For
example, pushing a mass on a frictionless horizontal surface over some distance (work) will cause the mass to
increase its kinetic energy. Lifting a block vertically upward with a force equal to its weight will not accelerate
the block but rather increase its gravitational potential energy.
c) State in words how “work” is defined in physics. Is it possible to exert a force on a body without doing work
on it? Give an example to support your answer. (5)
Work is the action of a force over a distance. An important qualification is that one must only consider the
component of force along the direction of motion. Symbolically this can be written W = F|| d.
Yes, it is possible to exert a force on an object and not do any work on it. There are essentially two cases to
distinguish: either the body is not moving (d = 0) or the force acts perpendicular to the body’s motion (F|| = 0).
An example of each would be: holding a book with one’s hand a some fixed height above the ground, and
moving the book horizontally at constant speed.
d) What does it mean to say that a physical quantity is “conserved”? Under what circumstances is the total
momentum of a system conserved? (5)
A physical quantity is said to be “conserved” when its value remains the same before and after some physical
process. For example, provided there are no losses to heat the total mechanical energy of a system remains
fixed.
The total momentum of a system is conserved provided no net external force acts on the system. Momentum is a
vector and thus one can specialize to specific directions (horizontal, vertical) where no net external force acts,
in such case the total momentum in that particular direction is conserved.
Because the forces that act during collisions are typically very large it is usually an extremely good
approximation to ignore external forces that may be present (during the collision) and assume momentum is
conserved. Thus, even when external forces act, so long as one considers events just before and just after a
collision, the total momentum is very nearly conserved.
e) While dealing with falling objects in the absence of air resistance we stated that all objects fall toward the
earth with the same acceleration. How can this be in light of the fact that more massive objects have a greater
force of attraction (weight) toward the earth? That is, why do these more massive bodies not fall at a greater
rate? (5)
Though the more massive object has more weight (force toward the earth) it requires more force to accelerate
the body at the same rate because of its greater mass.
f) Discuss how normal forces, contact area, speed, and sliding/sticking influence the frictional force between
two bodies in contact. Can you give a general rule that determines the direction friction will act? (5)
The frictional force between two surfaces always acts in a direction tangent to the surfaces. The magnitude of
the frictional force is empirically found to be proportional to the normal force that exists between the surfaces,
and is independent of the surfaces’ relative velocity and contact area.
Recall that friction can exist between two surfaces even when there is no relative motion between them (e.g.
static friction). A rule that covers both moving and fixed surfaces might read: friction always acts to oppose the
relative motion (perhaps virtual) between the two surfaces. “Virtual” here pertains to the motion that would
occur if no friction were to act.
2. Consider a roller coaster on a frictionless track. Assume the car starts from rest at point A of the figure.
a) As the car rolls along the track does the normal force from the track do any work on the car? Explain your
reasoning. (5)
D
A
The normal force of the track on the car does not
do any work on the car because it always acts
perpendicular to the car’s motion (velocity). The
C
situation is analogous to the tension a cord exerts
?
H
on a swinging pendulum. Also see the answer to
question 1c.
0.5H
B
b) From an energy point of view what causes the car to speed up as it moves from point A to B? As the car
moves from point B to C is the work done by gravity positive or negative? (5)
From an energy point of view, as the car rolls down the track it loses potential energy and gains kinetic energy.
Because of energy conservation the amount of potential energy lost is equal to the kinetic energy gained.
As the car rolls upward the work done by gravity is negative. This is because the force of gravity is in the
opposite direction to the car’s motion, and thus hinders the car’s motion.
c) For points A, B, and C state how the total energy of the car is distributed between kinetic and potential
energy. At which of these points is the potential energy greatest? Where is the kinetic energy greatest? (5)
At point A where the car is at rest the kinetic energy is zero and the potential energy is a maximum. At point B
the car is at its lowest point and therefore its potential energy is a minimum and its kinetic energy a maximum.
Finally, at point C since the height is halfway between the lowest and highest points the potential and kinetic
energies are numerically equal. That is, at this point the total energy is distributed equally among potential and
kinetic energy.
d) At point D the car comes momentarily to rest. What is the height of this point? How would your answer
change if at point A the car has a kinetic energy that is half of its potential energy? (5)
Since the car comes to rest at point D its kinetic energy is zero; all the car’s energy is in the form of potential
energy. By conservation of energy this energy must be the same as when it started at point A. Since the car was
also at rest at point A, the potential energies of points A and D must be the same. This implies D must be the
same height as A.
The maximum height the car attains can be found by considering its total energy. At the beginning, the car has
50% more total energy. Energy-wise this is the same as the car starting from a point 50% higher (recall
gravitational PE is proportional to height). Thus, by reasoning along the lines of the first part of this question
the car will attain a 50% higher maximum height or 1.5H.
3. The following questions concern two railroad cars colliding with one another.
a) Suppose the cars have equal mass and car two is at rest while car one is heading toward it at speed v as
shown in the figure. If the cars collide and stick together what will be their final speed? Is the collision elastic or
inelastic? (5)
Since there is no net external force acting in the horizontal
direction the total momentum in this direction is conserved.
When the cars stick together the mass of the moving object is
double that before the collision. The final speed must
therefore be (1/2) the initial speed to compensate for this.
Since the cars stick together the collision is inelastic.
v
1
2
b) In part (a) how does the final speed change if the second car is twice as massive as the first? In this case how
are the impulses each car delivers to the other related? (5)
Reasoning along the lines of part (a) in the event that the second car is twice as massive the final speed will be
(1/3) the initial speed.
The impulses the cars deliver to each other are equal and opposite. This is because the forces each car exerts
on the other are equal and opposite and the time duration that those forces act is the same.
c) Assume the cars have equal mass and suppose instead of sticking the cars bounce off from one another
elastically. What now will be each car’s final speed? Explain your result in terms of impulses and the forces
each car exert on one another. (5)
As we have seen and discussed in class when the cars have equal mass and the collision is elastic the first car
will stop and the second car will continue with the first car’s initial speed. By the Third Law the forces acting
on each car are equal and opposite. Since the forces act for the same time the momentum changes are equal
and opposite – thus implying that the total momentum is conserved in the collision. Also note that since the cars
have the same mass, for every bit the second speeds up the first slows down. The collision is, however, also
elastic and to conserve energy the final situation should look similar to the initial – i.e. have one car moving
and the other at rest. The only possibility is to have the first car slow to a stop and the second go off with the
initial speed of the first.
d) Suppose again that the cars collide elastically but now that car one is twice as massive as car two. How are
the forces each car exerts on the other related? How are the changes in each car’s velocity related? Describe in
as much detail as you can the final velocities of each car after the collision. (5)
By the Third Law the forces each car exerts on the other are equal and opposite. With this in mind, since the
first car’s mass is twice that of the second, its acceleration will be half that of the second. Since the cars are
subject to interaction forces for the same time duration the velocity change of car one must be one half that of
car two. Thus, car one slows down half as much as car two speeds up. The collision is elastic and there is an
instant (collision halfway point) where the cars have the same speed. Since the acceleration of each car
continues awhile after this instant it follows that both cars must be moving forward with car two moving faster
than car one.
For a more detailed answer assume that car one has an initial speed of 12 units then when car one has speed 8
so does car two (why?) – this is the collision halfway point. Hence, car one loses 4 more units of speed and car
two gains 8 more. Hence if car one approaches at speed v then the final velocities of car one and two are
respectively v/3, (4/3)v.
e) Suppose the cars have equal mass and approach one another with the same speed and stick together. What
will be their final velocity after they stick together? Is the kinetic energy conserved in this process? (3)
Since the cars have equal mass and approach each other
with equal speed the total momentum before the collision
is zero (the momentum vectors cancel one another).
v
v
1
2
By conservation of momentum the total momentum after
the collision must also be zero. But the cars are stuck together.
The only possibility is that the single body composed of both cars is not moving. Thus, after the collision the
system is at rest.
As far as macroscopic objects are concerned kinetic energy is not conserved in this process. This is clear in
light of the fact that objects are moving before the collision with nothing in motion after the collision. As we
mentioned in class, the missing energy serves to deform the bodies and increase their molecular kinetic energy
(heat).
f) What is the final velocity if instead car two is twice as massive as car one? (2)
If car two is twice as massive as car one there is a net momentum to the left in the figure; thus, when the cars
stick they will move to the left. To find their speed, note that if car one has 1 unit of momentum then there is a
net momentum of one unit to the left before the collision. Since the moving object is three times as massive as
car one it must be moving with 1/3 car one’s initial speed (to the left).
4. A person pushes on a book of mass m = 2kg as it slides down a wall as shown in the figure. The force the
person exerts on the book is perpendicular to the wall and has magnitude F = 15N. The frictional force between
the book and wall has magnitude f = 8N.
a) Draw a free-body diagram for the book. (5)
f
F
N
W
b) Determine the weight of the book and the normal force the wall exerts on the book. (5)
W = mg = (2)(10) = 20N
Since there is no acceleration in the horizontal direction: F – N = 0  N = F = 15N
c) What is the acceleration of the book down the wall? (5)
Applying Newton’s Second Law (F = ma) in the vertical direction: F(net) = 20 – 8 = 12N (down).
But F(net) = ma or a = F(net)/m = 6m/s2
d) For each force acting on the book determine its respective reaction force. Make sure to state clearly on what
body this reaction force acts. (5)
N (normal force of wall on book) – reaction: Normal force of book on wall
f (frictional force of wall on book) – reaction: Frictional force of book on wall
F (force of person on book) – reaction: Force of book on person
W (gravitational force of earth on book) – reaction: Gravitational force of book on earth
e) Suppose that instead of taking place on earth the book is being pushed against the wall of a spaceship in deep
outer space. In such an instance does the book have mass or weight? Would friction still be present? (5)
In deep outer space, assuming the ship to be very far from any large massive body, the book would not have
weight. However, the book still has mass – this aspect of matter being independent (so far as we know) of a
body’s environment.
Since the book does not have weight and the person is pushing horizontally on the book there is no friction force
acting on the book – there is no virtual movement for friction to oppose. This is analogous to a book resting on
a horizontal table on earth. Though there is a normal force between the bodies no friction acts. Of course, if the
person pushed up or down on the book while pushing it against the wall a force of friction would be present.