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Physics 101 Exam 2
12 November 2004 Prof L. Weinstein

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

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
Sun's mass $M_S = 2*10^{30}$ kg	Sun's radius $R_S = 7*10^8$ m
Earth-Moon distance $d_{E-m} = 3.8\cdot 10^8$ m	Earth-Sun distance $d_{E-S} = 1.5\cdot10^{11}$ m
G = $6.67\cdot10^{-11}$ N$\cdot$m$^2$/kg$^2$

1. You throw a 2-kg block upward at an angle of 45$^o$ and a speed of 6 m/s toward your friend. What is its kinetic energy as it leaves your hand?

   KE = (1/2)mv^2 = 0.5*(2 kg)*(6 m/s)^2 = 36 kg-m^2/s^2 = 36 J
   

2. You throw a 3-kg block upward at an angle of 37$^o$ with a kinetic energy of 24 J toward your friend. What is its kinetic energy immediately (ie: a split second) before your friend catches it? (Assume that your friend catches it at the same height that you threw it from. Ignore air resistance.)

   = 0.5in

   KE -> PE and KE -> KE. You start with 24 J of kinetic energy and end with 24 J of kinetic energy. Energy is conserved.
   

3. Aircraft carriers use catapults to rapidly accelerate airplanes to high enough speeds that they can remain airborne. (Airplanes crash if they fly too slowly.) Assume the catapult can do a fixed amount of work on each airplane. If the catapult can give a plane with mass $m$ an initial speed $v$, how much speed can the catapult give a lighter plane with mass ${1\over2}m$?
   1. ${1\over2}v$
   2. between ${1\over2}v$ and $v$
   3. $v$
   4. between $v$ and $2v$
   5. $2v$
   6. need more information
   
   
   Work = change in KE = change in (1/2)mv^2. If mass is halved, then v must increase. v cannot double, because v^2 would then quadruple and that would be too much. The answer is between v and 2v.
   
   

4. You compete in the pumpkin dropping competition and build a 1-m tall pumpkin catcher that successfully catches (without breaking) a 4-kg pumpkin dropped from a 10-story building. If a new competition is held where the pumpkin is dropped from a 20-story building, how tall should your new pumpkin catcher be in order to successfully catch the pumpkin? (Assume that the new pumpkin catcher is built just like the old one, except for its size. Ignore air resistance.)
   1. 1/4 m
   2. 1/2 m
   3. 1 m
   4. 2 m
   5. 4 m
   6. need more information
   
   You double the initial PE of pumpkin so that you are also doubling the KE of the pumpkin immediately before it hits the pumpkin catcher. The catcher does work to stop the pumpkin. W = Fd = change in KE. The work done must double, therefore the stopping distance must also double.
   

5. When you drive your car one mile at a speed of 30 mph, the tires of your car make a certain number of rotations. Next you test drive a new vehicle with tires twice the diameter of your car's tires. How many rotations do these tires make in one mile?
   1. 1/4 as many
   2. half as many
   3. the same
   4. double
   5. quadruple
   6. need more information
   
   
   Speed is irrelevant. A tire covers a distance = 2*pi*r with each rotation. Therefore the larger tire requires half as many rotations to cover the same distance.
   

6. Norfolk is at a latitude of about 40 degrees N. Our linear speed due to the rotation of the Earth about its axis is about 350 m/s. If you move to the equator your linear speed due to the rotation of the Earth about its axis will
   1. decrease
   2. stay the same
   3. increase
   4. need more information
   
   Look at a globe. The Earth rotates on its axis. The axis passes through the North and South poles. Therefore, any point on the equator is as far as possible from the axis as you can possibly get on the Earth. By increasing your distance form the axis of rotation, you increase your linear speed.
   
   

7. A favorite playground item is a large horizontal tire held up by chains. The tire is free to swing and also to rotate. Let's just consider the rotational motion. Ignore friction. The pictures show the swing as viewed from above. Two children of about equal mass get on the swing and sit up straight (the left-hand picture). Their parents start the swing rotating at some rotational speed. If the children now lean way out (the right-hand picture), the rotational speed of the swing will

   = 1.5in

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

   Angular momentum is conserved. By leaning out, the children increase their distance to the axis of rotation, therefore increasing their rotational inertia. Therefore, their rotational speed must decrease. This is identical to the rotating ice skater and to the demo with the person spinning on the stool with the weights in her hand.
   

8. In order to be in Low Earth Orbit (ie: about 200 km from the surface of the Earth or about 6600 km from the center of the Earth), you need to have a tangential speed of 8 km/s. If the Earth expands so that it has the same mass but twice the radius, what speed would you need to be in Low Earth Orbit 200 km above the (expanded) surface?
   1. less than 8 km/s
   2. 8 km/s
   3. more than 8 km/s
   4. need more information
   
   There are two effects here. Remember that we derived the 8 km/s from a) teh fact that the curvature of the Earth is 5 m in 8 km and b) that you fall 5 m in 1 s. If the Earth has twice the radius, then the curvature will only be 2.5 m in 8 km (half as much). However, the gravitational force is 1/4 as large (since the distance to the center of the Earth is doubled and F ~ 1/r^2). Therefore you will only fall 5/4 m in 1 s. The net effect is that the orbital speed is less than 8 km/s. I gave almost full credit to people who argued that the radius of curvature was increased.
   
   

9. Calculate the gravitational force between the Earth and the Sun (see the numbers at top of exam).

   F = G M_sun M_earth / (d_earth_sun)^2
   F = (6.7*10^-11 Nm^2/kg^2)*(2*10^30 kg)*(6*10^24 kg) / (1.5*10^11 m)^2
   F = 3.56*10^22 N
   
   

10. You drive your car on the curved highway exit ramp at 20 mph. In order for your car to follow the curve and stay on the road, you need 2,000 N of frictional force.

    a) draw an arrow showing the direction of the frictional force when the car is at point $A$.

    b) How much force will you need to follow the curve and stay on the road if you drive on the same curve at 60 mph?

    = 1.5in

    The force points from point A to the center of the circle.
    F = mv^2/r. If you double v, then F must quadruple. Here we tripled v so that F must be 9 times larger. Thus F = 18,000 N.
    
    

11. If you moved to another planet with twice the mass of the Earth but the same radius, the acceleration due to gravity at the surface would be
    1. four times smaller than Earth's (ie: 2.5 m/s$^2$)
    2. two times smaller than Earth's (ie: 5 m/s$^2$)
    3. the same as Earth's (ie: 10 m/s$^2$)
    4. two times larger than Earth's (ie: 20 m/s$^2$)
    5. four times larger than Earth's (ie: 40 m/s$^2$)
    6. need more information
    
    F = G*M_planet*M_you/r^2
    If M_planet doubles and everything else is unchanged, then the force will double.
    a = F/M_you. If F doubles and your mass is unchanged, then your gravitational acceleration will double.
    
    
    

12. (no explanation needed) When you accelerate your standard (ie: non-hybrid) gasoline-powered car from 0 to 60 mph, you are converting chemical energy in the gasoline to kinetic energy (and other forms of energy). When you use your brakes to stop, what form of energy is the kinetic energy converted to?
    1. chemical
    2. potential
    3. thermal (ie: heat)
    4. other
    5. need more information

13. You throw a 7-kg rock with an initial speed of $v=25$ m/s such that its horizontal speed is 15 m/s and its vertical speed is 20 m/s. How much time is the rock in the air? Ignore air resistance. Assume the rock lands at the same height that you threw it from.

    = 1.5in

    Only the vertical speed matters. The gravitational acceleration is 10 m/s^2 downward. Therefore, after 1 second, the upward velocity is 10 m/s, after 2 s it is zero (the maximum height). Then it takes 2 more seconds to fall back down. The answer is 4 seconds.
    
    

14. Olympics divers need to rotate their bodies a certain number of times for specific dives. At the beginning of the dive, a diver's body is typically fully extended and she is rotating at a certain rate. When the diver 'tucks' (ie: draws her legs up to her chest and brings her arms in), will she rotate faster or slower than when her body is extended?

    = 1.1in

    1. faster (more rotations per second)
    2. same rotational speed
    3. slower (fewer rotations per second)
    4. need more information

    There are no external torques, therefore angular momentum is conserved. When they are tucked, the average distance to the center of rotation is decreased, decreasing rotational inertia and therefore increasing rotational velocity.
    

15. You jump out of an airplane and are in free fall. While you are falling, you take out a cup and a bottle of water. You tilt the bottle in order to pour the water from the bottle into the cup. Ignore air resistance. What happens?
    1. the water pours normally
    2. the water pours faster than normal
    3. the water pours slower than normal
    4. the water does not pour (it stays in the bottle)
    5. need more information
    
    
    Everything is falling as fast as it can. The water can't fall faster by falling out of the bottle. The water does not pour. This is the same as the styrofoam cup of water with a hole in it. When I dropped the cup, the water stopped flowing out of the cup.
    
    

16. In one of the Star Trek movies, Captain Kirk falls off of a high cliff. If he hit the ground, he would die. Fortunately, Mr. Spock swoops in horizontally with his aircar and saves Kirk by catching him about 6 inches from the ground. Would this work in real life?

    = 1.1in

    1. yes
    2. no
    3. need more information
    
    Kirk is falling at a very high speed. It does not matter whether Kirk hits the ground or the air car. He will hit it too hard. The only way for this to work would be if the aircar caught him higher up and slowed his descent gradually. With only 6 inches of space, there is no way for the air car to do that.