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Physics 101 Exam 2 Solutions
7 November 2003 Prof L. Weinstein

There are 16 problems. Please give a short explanation for all multiple choice questions. Show your work for all numerical answers.

G = $6.67\cdot10^{-11}$ N$\cdot$m$^2$/kg$^2$

Earth's mass $M_e = 6\cdot 10^{24}$ kg

Earth's radius $R_e = 6.4\cdot 10^6$ m

Moon's mass $M_m = 7\cdot 10^{22}$ kg

Moon's radius $R_m = 1.7\cdot 10^6$ m

Earth-Moon distance $d_{E-m} = 3.8\cdot 10^8$ m

Earth-Sun distance $d_{E-S} = 1.5\cdot10^{11}$ m

closest Earth-Jupiter distance $d_{E-J} = 6.3\cdot10^{11}$ m

Jupiter's mass $M_J = 1.9\cdot10^{27}$ kg

1. You throw a 5 kg block vertically upward with an initial kinetic energy of 30 J. What is its potential energy at the peak of its trajectory (in Joules)?
   1. 0
   2. 10
   3. 30
   4. 50
   5. need more information

   30 Joules. It starts with PE = 0 and KE = 30 J. At the peak of the trajectory, it has PE = 30 J and KE = 0 J. This is because a) energy is conserved and b) at the peak the block is not moving so it has zero velocity and zero KE.

   
   

2. In order to accelerate your 1000-kg Ford Escort from 0 to 30 m/s (about 60 mph), you need to supply the energy contained in about 1 ounce (about 25 g) of gasoline. If instead you want to accelerate a 2000-kg Explorer from 0 to 30 m/s, how much gasoline will you need?
   1. 1/4 as much
   2. 1/2 as much
   3. the same
   4. twice as much
   5. four times as much
   6. need more information

   $KE = 1/2 mv^2$ so that when you double the mass, you double the KE needed to go 30 m/s.

   
   

3. A car is moving at 30 mph, loses control, and hits the highway crash barrels (the barrels of sand that are placed near dangerous objects on the sides of highways). It stops in a distance of 2 m. If the same car was moving at a speed of 60 mph, how much distance would it need to stop? Assume that the crash barrels exert a constant force on the car.
   1. 1/2 m
   2. 1 m
   3. 2 m
   4. 4 m
   5. 8 m
   6. need more information

   There are two parts to this solution: 1) when you double $v$, KE quadruples since $KE = 1/2 mv^2$, and 2) to stop the car, you need to do work $W = Fd$ equal to the change in KE ( $W = \Delta KE = 0 - 1/2mv^2$. Since you need to do 4 times the work and the force exerted is the same, you will need four times the distance.)

   
   

4. 75 people are all walking around. They are each walking with constant velocity at a speed that is between 1 and 5 mph. Can the group have zero total momentum? Can the group have zero total kinetic energy? Explain.

   
   Momentum has a direction. This means that if half the group is walking to the left and the other half is walking to the right, their total momentum can be zero. More technically, momentum is a vector: $\vec p = m \vec v$.

   KE does not have a direction. It is always positive. (Since KE = 1/2mv^2. When you square the $v$, minus signs become positive and there is no direction anymore.)

   Therefore total momentum can be zero but total kinetic energy cannot be zero.

   Many of you wrote that if velocity was constant then momentum could be zero. This is wrong. If velocity is constant, then momentum is not changing and therefore the change in momentum is zero. This does not mean that momentum itself is zero.

   
   

5. Old fashioned record players could play records that rotated at different speeds. An LP (Long Playing record) rotated at approximately once every two seconds. It had a radius of 15 cm. If you doubled the rotational speed, then the linear speed of a point on the edge of the LP would
   1. quarter
   2. halve
   3. stay the same
   4. double
   5. quadruple
   6. need more information

   Rotational speed measures the number of rotations made per second. Linear speed measures the number of meters travelled in that time. You start out covering a distance of $2\pi r$ in 2 s. Then it speeds up and you cover the same distance in half the time. You have doubled your speed.

   Note that the radius of the LP did not change.

   
   

6. NASA uses centrifuges to simulate the effects of high gravity on astronauts. The centrifuge, like some amusement park rides, has a passenger cabin at the end of a long (10 m) are. The cabin then swings around in a 10-m radius circle at high speed.

   An astronaut is in the centrifuge as it rotates around. She has a linear speed $v$ and feels a centripetal force from the centrifuge. If her linear speed doubles, then the force on her from the centrifuge will

   1. quarter
   2. halve
   3. stay the same
   4. double
   5. quadruple
   6. need more information

   Centripetal force $F = mv^2/r$. If you double $v$ then $F$ will quadruple.

   
   

7. When an object is raised off the ground to a certain height, it gains a certain amount of potential energy. If instead the same object is raised to four times the height, its potential energy will
   1. quarter
   2. halve
   3. stay the same
   4. double
   5. quadruple
   6. need more information

   $PE = mgh$. Quadruple height and PE will quadruple.

   
   

8. A car travels in a circle with constant speed. The net force on the car is
   1. directed forward, in the direction of travel
   2. directed towards the center of the circle
   3. zero because the car is not accelerating
   4. none of these

   You need a centripetal force to go in a circle. If there is no force, then you will go in a straight line. (Try rounding a curve at high speed on an icy day. You won't make it. On second thought, don't try it.) If the force is directed forward, then you will go faster but you won't change direction.

   
   

9. The reason that the moon does not crash into the Earth is
   1. it has a large tangential velocity
   2. the net force on it is zero
   3. it is beyond the main pull of Earth's gravity
   4. it is being pulled by the sun and the planets as well as by the Earth
   5. all of the above
   6. none of the above

   If the moon was not moving, then Earth's gravity would pull it toward the Earth and there would be a horrible crash. The large tangential velocity makes the moon miss the Earth.

   If there was no net force on the moon, then it would travel in a straight line and leave us forever.

   The Earth's gravity still pulls on the moon. The Earth's pull is weaker that far away, but it is still very strong and keeps the moon in orbit.

   The pull of the sun and the planets does not significantly affect the orbit of the moon around the Earth.

   
   

10. Calculate the gravitational force on a newborn baby (mass = 4 kg) from the planet Jupiter (see numbers at top of exam).

    $F = {GMm\over d^2}$ where $G$ is the gravitational constant, $M$ and $m$ are the masses of the two objects that are pulling on each other (in this case the baby with $m = 4$ kg and Jupiter with $M = 1.9\cdot10^{27}$ kg), and $d$ is the distance between them (in this case $d = 6.3\cdot 10^{11}$ m). Thus:

    
    

    $$\eqalign{F &= \frac{(6.67\cdot 10^{-11} N\cdot m^2/kg^2) (4 k... ...\cdot m^2}{3.97\cdot 10^{23} m^2} \cr &= 1.28\cdot 10^{-6} N}$$
    
    
    Rather a tiny amount. It makes it hard to believe that the planets can influence our lives.

    
    

11. If you moved to another planet with the same mass as the Earth but twice the radius, your new weight would be
    1. four times smaller
    2. two times smaller
    3. the same
    4. two times larger
    5. four times larger
    6. need more information

    $F = {GMm\over d^2}$ You are now twice as far from the center of the planet so $d$ doubles and $F$ gets four times smaller.

    
    

12. You are in a rotating, cylindrical, space habitat far from any solar system. Assume that there are no external forces on it. If you climb 'up' a ladder from the edge of the habitat to the center, the habitat's rotational speed will

    = 1.5in

    1. increase
    2. remain the same
    3. decrease
    4. need more information

    There are no external torques on the space habitat. Therefore angular momentum is conserved. When you move to the center of the habitat, you decrease the total rotational inertia of the habitat. Therefore, the rotational speed of the habitat increases. This is just like the ice skater bringing her arms in and spinning faster or the student on the rotating stool who rotated faster when she brought her arms (and the weights) in and rotated slower when she held her arms (and the weights) out.

    
    

13. When you walk, you keep your legs straight. When you run, you bend your legs at the knee. In both cases, your leg rotates around your hip (obviously, not ALL the way around). Which has a larger rotational inertia when rotated around the hip, a straight leg, or a bent leg?
    1. straight leg
    2. bent leg
    3. both the same
    4. need more information

    The farther the mass is from the center of rotation, the harder it is to make it rotate and greater its rotational inertia. In equation terms, $I = mr^2$. The foot of the straight leg is farther from the axis of rotation (the hip) so the straight leg has larger rotational inertia.

    When you run, you bend your leg to decrease its rotational inertia so you can bring it forward faster to take the next step.

    
    

14. There are two types of bicycle brakes, brakes that apply their force to the outside (rim) of the wheel and brakes that apply their force to the inside (near the axle) of the wheel. If both types of brakes can apply the same force to the wheel, which type applies more torque (and is thus more effective at stopping the bicycle)?
    1. rim brakes
    2. hub brakes
    3. both the same
    4. need more information

    Torque = force times lever arm. Both brakes apply the same force. The one that has the larger lever arm will be more effective at stopping the bike.

    The rim brakes apply their force far from the axle. This means that there is a large lever arm. The hub brakes apply their force very close to the axle. This means that there is a short lever arm. This is exactly the same physical situation as pushing on a door. You want to push on the door as far from the axis of rotation (the hinges) as possible. This is why the door knob is not next to the hinges.

    
    

15. You use a slingshot to throw a 7 kg rock from the top of a building horizontally at a speed of 40 m/s. The rock hits the ground 2 seconds later. How tall is the building?

    The horizontal speed is irrelevant. This is identical to dropping a rock and asking how far it falls in 2 s. The distance it falls is $d = 1/2 gt^2$ where $g = 10 m/s^2$. Thus
    

    $$d = 1/2 (10 m/s^2) (2 s)^2 = 20 m$$
    
    

    
    

16. Your car gets 25 miles per gallon (mpg) when driving at 65 mph on the highway with the windows closed. If you continue driving at the same speed and turn on the air conditioner in the car, you car will get
    1. more than 25 mpg
    2. 25 mpg
    3. less than 25 mpg
    4. need more information

    Energy is conserved. The energy to run the car comes from the gasoline. The air conditioner uses energy that comes from burning gasoline. This means that you will not be able to travel as far on 1 gallon of gas. Note that this probably changes your mileage by about 1 mpg.

    
    

17. Two billion years ago, the moon was much closer to the Earth than it is now. At that time, the difference in water level between high tide and low tide was
    1. greater than today
    2. the same as today
    3. smaller than today
    4. need more information

    When the moon was closer, the gravitational force on the Earth was larger since $F\sim 1/d^2$. But this is not enough, since the tides depend on the difference between the gravitational force of the moon on the Earth at the near side of the Earth and at the center of the Earth. This difference (and hence the tidal forces and hence the height of the tides) also increases. There are two ways to justify this: a: use the ugly equation: Tidal force = $\Delta F = \frac{GMm (2R_{Earth})}{d^3}$ or b: The moon exerts more tidal force on the Earth than the Sun does because the moon is much closer to the Earth than the Sun. Therefore, if the moon was even closer, the tides would be even bigger.