Solutions to Exam #1

SBCC 2017Summer2 P 101 Solutions to Exam 01 2017Jul11A Page 1 of 9 Solutions to Exam #1 1. Which of the following natural sciences most directly involves and applies physics? a) Botany (plant biology) b) Physical chemistry c) Synthetic chemistry d) Biochemistry 2. What is the value, with proper units if any, of the product 3 kg/m x 3 m/s? a) 9 kg m/s b) 9 kg m c) 9 kg/s d) Undefined 3. What is the sum 2 kg + 3 s? a) 5 kg s b) 5 kg/s c) 5 d) Undefined. 4. The mass of an object is measured as 12.0 kg but with an absolute uncertainty of 1.2 kg. What is the relative uncertainty? a) 0.6 kg b) 12.6 kg c) 1/20 = 5% d) 1/10 = 10% 5. Which of the following is the proper way to write 0.0213 m in scientific notation? a) 0.0213 x 10 0 m

SBCC 2017Summer2 P 101 Solutions to Exam 01 2017Jul11A Page 2 of 9 b) 0.213 x 10-1 m c) 2.13 x 10-2 m d) 2130 x 10-5 m 6. Suppose you measure the length of a board and find its length to be 2.00 m + 0.01 m, then cut the end of the board off so that its new shorter length is measured as 1.50 m + 0.03 m. From this, you report that the length that was cut off is 2.00 m 1.50 m = 0.50 m. What is the uncertainty in the length that was cut off? a) + 0.01 m b) + 0.02 m c) + 0.03 m d) + 0.04 m 7. Tell which of the following quantities is a vector. a) The number of questions on this exam. b) A volume of air. c) 2 N m East. d) Speed 80 mph. 8. What is the sum of a vector pointed to the right and a vector pointed downward? The vectors are drawn below. (Yes, I didn't tell the magnitudes.) a) A vector pointed to the right. b) A vector pointing on a slant downward and to the right. c) A vector pointing on a slant upward and to the right. d) Zero or the null vector.

SBCC 2017Summer2 P 101 Solutions to Exam 01 2017Jul11A Page 3 of 9 9. Suppose a body has mass, 1 kg and velocity 20 m/s East, but the net force on the body is 4 N North. What is the acceleration of this body? (1 N = 1 kg m/s 2.) a) 80 m/s 2 East. b) 24 m/s 2 Northeast. c) 4 m/s 2 North. d) 4 m/s 2 South. Newton's Second Law of Motion 10. Which of the following is a measure of inertia of a body? a) The volume of the body. b) The length of the body. c) The temperature of the body. d) The mass of the body. 11.Suppose a rock is dropped at the top of a well so the rock freely falls and hits the bottom 4 seconds after being dropped. How deep is the well? (g = 9.8 m/s 2.) a) 39.2 meters b) 78.4 meters c) 100 meters d) Infinitely deep,. y = ½ g t 2

SBCC 2017Summer2 P 101 Solutions to Exam 01 2017Jul11A Page 4 of 9 12. The total force on a closed system is 20 N to the right. What is the rate of change of the total linear momentum of the system? (1 N = 1 kg m/s 2.) a) 10 kg m/s 2 to the left. b) 10 kg m/s 2 to the right c) 20 kg m/s 2 to the right d) Cannot be determined without knowing the total mass of the system. 13. A man pushes a cart with a force 100 newtons toward the East. What force does the cart exert on the man? a) 100 newtons East b) 100 newtons West c) 100 newtons Northeast d) Zero, the man is pushing but the cart is not. Newton's Third Law of Motion 14. Suppose the total external force on a closed system is zero and continues to be zero. If two bodies in the system push on each other so each gains a linear momentum with magnitude 200 kg m/s (but the total external force is still zero) then what is the change in total linear momentum of the system? a) 200 kg m/s b) 400 kg m/s c) Zero d) Cannot be determined without knowing the directions of travel of the two bodies after they push on each other. 15. Suppose a body with mass 10.0 kg is moving at a velocity of 20.0 m/s East. What is the linear momentum of the body? a) 2.0 kg m/s East b) 20.0 kg m/s West c) 30.0 kg m/s West d) 200 kg m/s East

SBCC 2017Summer2 P 101 Solutions to Exam 01 2017Jul11A Page 5 of 9 10 kg 20 m/s East = 200 kg m/s East 16. Suppose the linear momentum p of a body is toward the left but is changing at a constant rate toward the right. What is the direction of the net force on this body? a) To the left b) To the right c) Zero because the change counters the linear momentum. d) Zero because there is no net force on the body. Newton's Second Law of Motion 17. Suppose the net force on a body is 20.0 newtons downward for 10.0 seconds. What is the change in linear momentum in those 10.0 seconds? (1 newton = 1 kg m/s 2.) a) 2 kg m/s upward b) 2 kg m/s downward c) 200 kg m/s upward d) 200 kg m/s downward Impulse F Δt = Δp (K=1 for metric units). 18. If I push on a car with a force of 20 N East and the car moves 20 m North while I am pushing it, then how much work did I do on the car? a) 400 N m = 400 J b) 1 N / m = 1 J c) Zero d) Cannot be determined because I didn't say how much I would get paid for pusing the car. ΔW = F Δx = F Δx cos θ = 0 since cos θ = 0. Force perpendicular to displacement does zero work. 19. Suppose two tiny bodies a distance r from each other have a gravitational attraction toward each other. What happens to the magnitude of the gravitational force on one body if the distance between the two bodies changes to 5r? a) The gravitational force becomes 25 times what it originally was. b) The gravitational force becomes 5 times what it originally was.

SBCC 2017Summer2 P 101 Solutions to Exam 01 2017Jul11A Page 6 of 9 c) The gravitational force becomes 1/25 of what it originally was. Newton s Law of Gravity gives the magnitude as proportional to the product of the masses and inversely proportional to the square of the distance between the bodies. 20. A ball is sitting on a hill at rest but then is released and rolls down the hill. The gravitational potential energy of the ball and Earth decreases, but what happens to the total energy of the ball and Earth as a result of this decrease? a) Part of the total energy is destroyed. b) The total energy is constant. c) Part of the total energy became linear momentum. d) Part of the total energy was transformed in to fat. The ball needs to lose weight. 21. Consider a ball raised to a height above the ground and then dropped so it bounces several times but then comes to rest on the ground. What happened to the potential energy of the ball and Earth in the end when the ball is at rest on the ground? a) The gravitational potential energy was converted to elastic potential energy. b) The energy was destroyed by bringing the ball to rest. c) The potential and kinetic energies were eventually converted to sound and heat energy. d) The gravitational potential energy of the ball and Earth did not change. Energy is conserved but it can be converted from one form to another. In this case, the mechanical energy (potential and kinetic energy) was converted to sound and heat. 22. What is the kinetic energy of a body of mass 2.0 kg moving at 2.0 m/s (relative to the state of being at rest)? a) 2 J b) 4 J c) 8 J d) 16 J Kinetic energy = ½mv 2. 23. Suppose we choose either a line as a reference and place a tiny mass m = 1.0 kg at a location r = 3.0 m from the reference line. What is the moment of inertia I about this reference line?

SBCC 2017Summer2 P 101 Solutions to Exam 01 2017Jul11A Page 7 of 9 a) 3.0 kg m 2 b) 9.0 kg m 2 c) 12 kg m 2 d) 18 kg m 2 For a tiny mass, the moment of inertia is I=mr 2 where m is the mass and r is the separation from the reference line. 24. Suppose we look down the axis of a circular wheel and see it spinning counterclockwise. What is the direction of the angular momentum? (Hint: Use the right-hand rule.) a) Toward us b) Away from us c) To the right d) To the left 25. Suppose a astronaut spins in outer space with arms and legs curled so the astronaut looks fairly similar to a ball. There is zero net torque on the astronaut so we know the angular momentum of the astronaut does not change. Thus, the astronaut cannot stop spinning. How can the astronaut slow his or her angular velocity ω when the net torque on the astronaut is zero? a) The astronaut converts his or her own linear velocity to angular velocity. b) The astronaut can change his or her own moment of inertia by extending his or her arms and legs. c) The astronaut can exert forces to change his or her own angular momentum even with zero torque. d) The astronaut cannot change his or her own angular velocity in outer space. We know the angular momentum is constant because the torque is zero, but angular momentum is the product of moment of inertia and angular velocity. Thus, changing the moment of inertia is accompanied by a change in angular velocity. 26. In class, a torque was applied to a spinning wheel with its angular momentum pointing North, and the change in angular momentum at that moment was toward the West. What was the direction of the torque when the angular momentum was changing West. a) North

SBCC 2017Summer2 P 101 Solutions to Exam 01 2017Jul11A Page 8 of 9 b) Northwest c) West d) Upward 27. Suppose a car travels in a circle with a centripetal acceleration a R of 20 m/s 2 toward the center of the circle. The passengers inside the car feel an inertial force which is the centrifugal force. What is this centrifugal force for a 100 kg mass? a) 2000 N toward the center of the circle b) 2000 N outward away from the center of the circle c) 20 N in the direction the car is traveling d) Zero The magnitude of the centripetal force on the body with mass m = 100 kg is the product of the centripetal acceleration a R = 20 m/s 2 and the mass m. The centrifugal force has the same magnitude as the centripetal force, but has the opposite direction. 28. What is the weight of a body (the gravitational force on the body) if the mass of the body is 10 kg? (Hint: The weight is the gravitational force downward on the body. Using Newton's Second Law of Motion may be the quickest way to determine this since I did not give the mass or radius of the Earth. g = 9.8 m/s 2.) a) 9.8 N b) 10 N c) 98 N d) 100 N 29. Suppose two astronauts are at rest relative to each other, but then the first astronaut, one with a mass of 150 kg, pushes the other astronaut so the two separate. If the second astronaut travels at half the speed of the first astronaut, then what is the mass of the second astronaut? (Hint: The initial linear momentum is zero. The total external force on this system of two astronauts is still zero. Use the Law of Conservation of Linear Momentum and our definition of linear momentum to determine the mass of the second astronaut.) a) 75 kg b) 150 kg c) 300 kg

SBCC 2017Summer2 P 101 Solutions to Exam 01 2017Jul11A Page 9 of 9 d) Zero Total linear momentum is conserved for this closed and isolated system of two astronauts. Since the total linear momentum was zero before they separated, the total linear momentum must still be zero. Linear momentum is the product of mass and velocity. After separation, one astronaut (with given mass) had half the magnitude of velocity as the second astronaut, the second astronaut must have half the mass of the first astronaut.