Due Monday, February 6, 1995, In Class
Please Answer Each Question As Briefly As Possible
You May Work Together, But Write Up Your Answers Separately
Question 1:
Two cars are driving at 88 km/h (55 miles-per-hour) toward an intersection. A 1500 kg touring sedan is heading east and a 700 kg subcompact is heading north. The subcompact has a green light and drives into the intersection. The tired driver of the touring sedan runs the red light and crashes directly into the driver's side of the subcompact. The subcompact experiences a sudden, strong force to its right (toward the east).
(A) Which car experiences the largest force due to the collision?
Answer: They experience identical forces, oppositely-directed, so that neither force is largest.
Why: Because Newton's Third Law states that for every force exerted by one object on a second object, there is an equal but oppositely-directed force exerted by the second object on the first object. The first car exerts a force on the second car and the second car exerts an equal but oppositely-directed force on the first car.
(B) Which car experiences the largest acceleration due to the collision?
Answer: The subcompact experiences the largest acceleration.
Why: Because the subcompact has less mass than the touring sedan, the subcompact accelerates more when both cars experience forces of equal strengths.
(C) Which car is most likely to be thrown off the roadway because of the collision?
Answer: The subcompact is more likely to be thrown off the road than the touring sedan.
Why: Because the subcompact accelerates more than the touring sedan, the subcompact's velocity changes more as the result of the collision. The subcompact's new velocity directs it toward the northeast corner of the intersection, where it will probably drive off the road. The touring sedan's new velocity will still be directly toward the east, along the road itself, and that car will probably remain on the road.
(D) If the force on the subcompact is directly toward that car's right (directly toward the east), what happens to the north/south component of the subcompact's velocity?
Answer: The north/south velocity component is unchanged by the collision.
Why: Because the force that the subcompact experiences is directly toward the east, there is no component of force in the north/south direction. The subcompact does not accelerate in the north/south direction. It thus retains its original forward component of velocity. The subcompact's new velocity after the collision is toward the northeast. It will still have a northward velocity component of 88 kilometers-per-hour, but it will now have an eastward velocity component, too. The subcompact will probably drive off the road and into the grass on the northeast corner of the intersection.
(E) If the sedan loses 10,000 kg-m/s of momentum to the east due to the collision, what is the subcompact's change in momentum due to the collision? (Note: the kg-m/s is the SI unit of momentum)
Answer: The subcompact receives 10,000 kg-m/s of momentum to the east.
Why: Momentum is a conserved quantity so it is only transferred. The subcompact receives what the sedan lost.
Question 2:
Cable cars have ridden the hills of San Francisco for generations. As they ascend the hills, the cable cars are pulled up by moving, underground cables. As they descend those hills, the cable cars are held back by those same cables.
(A) As the cable pulls the car uphill, which does work on which? As the cable lowers the car downhill, which does work on which?
Answer: As the cable pulls the car uphill, the cable does (positive) work on the car. As the cable lowers the car downhill, the car does (positive) work on the cable.
Why: As the cable pulls the cable car uphill, the cable exerts an uphill force on the cable car and moves that car uphill. Since the force it exerts is in the same direction as the cable car's motion, the cable does work on the cable car. As the cable car descends the hill, the cable still applies an uphill force to the car, but now the cable car moves in the direction opposite the force applied by the cable. The cable car does work on the cable.
(B) On a gentle hill, the car rises 1 m in height as it travels 10 m along the road. On a second, steeper hill, the car rises 1 m in height as it travels 5 m along the road. If the cable must exert 3,000 N of force on the car to pull it up the gentle hill at constant velocity, how much force will be needed to pull the car up the steeper hill at constant velocity?
Answer: 6,000 newtons.
Why: On the gentle hill, the car will rise 1 meter in height by traveling 10 meters along the road. Since the force needed to keep the cable car at constant velocity on the gentle hill is 3,000 newtons, the work done in lifting the cable car 1 meter is 10 meters times 3,000 newtons. The product is 30,000 newton-meters or 30,000 joules. The cable did 30,000 newton-meters of work by exerting a force of 3,000 newtons to the cable car and moving that cable car 10 meters in the direction of the applied force. On the steeper hill, the cable car is lifted 1 meter when the car travels only 5 meters along the road. Since the same amount of work is done on the car, no matter how you raise it 1 meter in height, the cable must do 30,000 newton-meters of work on the cable car by pulling it 5 meters along the road. Since the product of 5 meters times the force must be 30,000 newton-meters, that force must be 6,000 newtons. Thus, the cable applies an uphill force of 6,000 newtons to keep the cable car at constant velocity on the steeper hill.
(C) The car now begins to descend the gentle hill. How much force is needed to keep the car from accelerating downhill and in which direction does the cable exert its force on the car?
Answer: 3,000 newtons uphill.
Why: The fact that the cable car is at constant velocity means that it is not accelerating. For the cable car to not accelerate, it must be experiencing a net force of zero. Since we know that an uphill force of 3,000 newtons will allow it to move uphill at constant velocity, that means that 3,000 newtons is just the amount of uphill force needed to give the cable car a net force of zero. Regardless of the direction the cable car is heading along the road, that same 3,000 newton uphill force is what you need to keep the car at constant velocity.
(D) What is the weight of the cable car in newtons? (Hint: how hard would the cable have to pull if the hill were straight up?)
Answer: 30,000 newtons.
Why: Imagine a hill so steep that it is vertical. As it rides on this hill, the cable car will rise 1 meter as it travels 1 meter along the road. To keep the cable car at constant velocity, the cable must exert an upward force that exactly balances the downward weight of the cable car. The road will no longer be applying any force at all. The work needed to raise the cable car 1 meter in height is still 30,000 newton-meters (from part B). Therefore the product of 1 meter times the force needed to oppose the cable car's weight is 30,000 newton-meters. The cable car's weight is thus 30,000 newtons.
(E) What is the mass of the cable car in kilograms?
Answer: 3,000 kilograms (or 3,061 kilograms, to be somewhat more accurate).
Why: The weight of an object near the surface of the earth is 9.8 newtons-per-kilogram. Since the cable car's weight is 30,000 newtons, its mass must by 30,000 newtons divided by 9.8 newtons-per-kilogram, which is 3,061 kilograms.
Question 3:
Three children are at a playground, playing with a seesaw. Two of the children are 3 years old and each has a mass of 15 kg; the third child is 7 years old and has a mass of 30 kg. The seesaw is made of a high-tech low-mass plastic, so that its mass can be ignored throughout this problem. The seesaw's pivot is in the middle of the board. The two 15 kg children sit together at the east end of the seesaw and the 30 kg child sits at the west end.
(A) The seesaw balances, meaning that it experiences no net torques when the children and the seesaw are not touching the ground. Describe the nature of any possible motion of the seesaw during the periods when the children and the seesaw are not touching the ground.
Answer: The seesaw rotates with constant angular velocity.
Why: A balanced seesaw is one that experiences a net torque of zero. Without any net torque, an object experiences no angular acceleration. If an object has no angular acceleration, than it has a constant angular velocity. That means that, if it is rotating at a certain rate, it will continue to rotate at that same rate indefinitely. If it is initially not rotating at all, meaning that it has an angular velocity of zero, it will not start rotating. While the children and board are not touching the ground, the balanced seesaw is either motionless or turning at a steady rate (constant angular velocity).
(B) What happens to the seesaw if the 30 kg child at the west end pushes downward on the ground?
Answer: The seesaw will begin to rotate so that the west end rises upward.
Why: The balanced seesaw is experiencing a net torque of zero. But as soon as one child pushes against the ground, the ground pushes back and exerts a torque on the seesaw. The seesaw experiences a net torque and undergoes an angular acceleration. The angular acceleration is such that the seesaw acquires an angular velocity that lifts the pushing child upward.
(C) One of the 15 kg children loses interest (as 3 year olds are wont to do) and jumps off the board while the seesaw is exactly horizontal. What happens to the seesaw and the other 2 children?
Answer: The seesaw begins to rotate and 30 kilogram child's side descends rapidly toward the ground. The 15 kilogram child's side rises rapidly.
Why: The balanced seesaw is experiencing no net torque. The removal of one of the 15 kilogram children ruins the balance. Now gravity creates a large net torque on the seesaw and its occupants. Angular acceleration appears and the seesaw begins to acquire an ever increasing angular velocity. The torque is such that the seesaw rotates to make the 30 kilogram child descend and the remaining 15 kilogram child rise.
(D) The two remaining children decide to continue playing on the seesaw. If the 15 kg child stays at the east end of the seesaw, exactly where should the 30 kg child sit in order for the seesaw to balance?
Answer: The 30 kilogram child should sit exactly half way between the west end and the pivot.
Why: The 30 kilogram child weighs twice as much as the 15 kilogram child. For the two children to apply identical torques, but with opposite senses, to the seesaw the light child must sit twice as far from the pivot as the heavy child. Then the torques, which have strengths of force times distance, will be equal but with the opposite senses. Since the two torques will then be equal but opposite, they will cancel one another. The net torque on the seesaw will be zero and it will balance.
(E) The other 15 kilogram child returns and sits on the middle of the board, right above the pivot. The child has essentially no effect on the seesaw. Why not?
Answer: The child's weight is exerted on the axis of rotation. Without a lever arm, it creates no torque.
Why: To affect an object's rotation, you must exert a torque on that object. If you push on the object about its center of rotation (a fixed pivot), you will be exerting no torque because your force has no lever arm.
Question 4:
An escalator is essentially a moving staircase. The individual steps are supported by metal tracks that run on either side of the escalator. These steps follow one another in a complete loop, driven by an electric motor. When you step onto an escalator at the ground floor, it soon begins to carry you upward and forward at a constant velocity toward the second floor.
(A) While you are moving toward the second floor at a constant velocity, what is the net force exerted on you by all outside forces (specify the amount and the direction of the net force)?
Answer: Zero net force.
Why: If you are moving with constant velocity, you are not accelerating. You must be experiencing zero net force.
(B) You know that gravity gives you a weight in the downward direction. What force does the escalator exert on you as you move toward the second floor at a constant velocity (specify the amount and the direction of force)?
Answer: A force equal in amount to your weight but pointing directly upward.
Why: If you are experiencing zero net force, all of the forces on you must cancel. You have a downward weight, so the escalator must be exerting an upward force equal in amount to your weight. The two forces are equal but oppositely-directed and cancel one another.
(C) Is the escalator doing work on you as you move toward the second floor?
Answer: Yes
Why: When the escalator pushes upward on you and you move at least partly upward (in the direction of the push), the escalator does work on you. The fact that you experience zero net force does not prevent the escalator from doing work on you as you move uphill. As you rise, gravity is also doing negative work on you, which is why you do not accelerate as the escalator does work on you. Your kinetic energy does not change, but the escalator is transferring work to you in the form of gravitational potential energy.
(D) As you first step onto the escalator, you begin to accelerate toward the second floor. Is the net force exerted on you by all outside forces the same as in part (A)?
Answer: No
Why: When you accelerate, you are experiencing a non-zero net force. Since the net force in parts (A) and (B) is zero, the net force is different during the acceleration process.
(E) If the rapidly-moving escalator suddenly stopped moving, you would be thrown forward and might even fall over. What causes you to be thrown forward?
Answer: Your momentum (or inertia)
Why: As you travel uphill on the escalator, you have momentum in the uphill direction. It would require a force downhill to remove this momentum and stop you from moving. If the escalator stops abruptly, it will not exert enough force on you to bring you to rest with it. You will continue forward, due to your momentum, and might even fall over onto the now stationary steps. Inertia is also an appropriate answer; momentum is simply one of the physical quantities associated with the concept of inertia.
Question 5:
You are seated in a crowded bus that has stopped at a bus stop. Two people board the bus, one wearing rubber-soled shoes and one wearing inline skates (roller skates - i.e. no friction). They both have to stand in the middle of the aisle and neither one holds onto anything.
(A) The bus driver is new and the bus leaps forward from the bus stop. Which way does person wearing inline skates move in relationship to the bus?
Answer: Toward the back of the bus.
Why: The bus is unable to exert any significant force on the person wearing skates. As a result, that person remains at constant, zero velocity and the bus moves out from under that person.
(B) The bus is accelerating forward as it pulls away from the bus stop. Why doesn't the person wearing inline skates accelerate with the bus?
Answer: The bus is unable to exert any significant horizontal force on the person in skates.
Why: If the bus cannot exert horizontal forces on a passenger, the passenger will not accelerate with the bus. The passenger will appear to move around the bus when, in fact, it is the bus that is moving out from under the constant-velocity passenger.
(C) The person wearing rubber soled shoes remains in place as the bus starts moving. What provides the force that caused that person to accelerate with the bus?
Answer: Friction (between the floor of the bus and person's shoes).
Why: As the bus accelerates forward, static friction between the floor of the bus and the shoes exerts a forward force on the shoes (and a backward force on the bus). As a result, the person accelerates forward, along with the bus.
(D) Once the bus has reached a constant velocity (it is traveling along a straight road at a steady pace), the person wearing inline skates is able to stand comfortably in the aisle without rolling anywhere. What is the net force on the person wearing skates?
Answer: Zero net force
Why: Once the bus reaches constant velocity, the net forces on the bus and on every item in the bus is zero. That means that the skater experiences zero net force. That person has a weight downward and a support force upward and the two forces cancel. Skates will convey upward forces just fine, even without friction. It is horizontal forces that skates do not convey.
(E) The driver accidentally runs into a curb and the bus stops so abruptly that everything slides forward, including the person wearing rubber soles shoes. In stopping quickly, the bus experiences an enormous backward acceleration. Why doesn't the person wearing rubber-soled shoes accelerate backward with the bus?
Answer: The maximum force possible from static friction is exceeded.
Why: For the person in rubber soled shoes to accelerate backward at the same rate as the bus requires a large backward force. The only way the bus can exert such a large backward force on the passenger is via static friction between the floor and the shoes. However, static friction has a maximum value, beyond which it cannot prevent sliding. The passenger slides forward in the bus, experience sliding friction, before finally reaching the same velocity as the bus.