physics 140 — midterm exam 2 version a — 2014 nov 7physics 140 — midterm exam 2 version a —...

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Physics 140 — Midterm Exam 2 Version A — 2014 Nov 7 There are 10 multiple choice questions. Select the correct answer for each one and mark it on the bubble form on the cover sheet. Each question has only one correct answer. (2 marks each) 1. A satellite is in orbit around a planet. Which of the following quantities would change the speed the satellite needs to orbit at? (a) the mass of the satellite (b) the mass of the planet (c) the radius of the orbit (d) Both A and C (e) Both B and C correct Let v be the speed of the satellite, M the mass of the planet, m the satellite’s mass and R the radius of its orbit. Then v 2 R = G M R 2 v = G M R v depends on both the mass of the planet and the radius of the orbit, but the mass of the satel- lite is not in the relation. , 2. A box sits on the horizontal bed of a moving truck. Static friction between the box and the truck keeps the box from sliding as the truck ac- celerates with acceleration a. The mass of the truck is M, the mass of the box is m and the co- ecient of static friction is μ s . What is the mag- nitude of the force of static friction between the box and the truck floor? (a) F f = ma (b) F f = Ma (c) F f = μ s mg (d) F f = μ s Mg (e) None of the above The friction F f causes the box of mass m to ac- celerate with acceleration a. It’s just Newton’s second law. 3. The box is sitting on the bed of a moving truck in the previous question while the truck slows to a stop. The work done by static friction on the box while stopping is (a) zero. (b) positive. (c) negative. correct (d) will depend on the directions that are de- fined to be positive and negative. If the box slows to a stop, the force on the box is opposite to the direction of motion. Hence the work done by the force is negative. 4. What is the direction of the dot product of two vectors A and B? (a) Half way between the direction of each of the vectors. (b) arccos A· B | A|| B| (c) arccos | A|| B| A· B (d) Perpendicular to both vectors. (e) None of the above. Correct The dot product of two vectors is a scalar and has no direction. 5. Work is done on an object to triple its kinetic energy. By what factor does the object’s speed change? (a) 9 (b) 1/9 (c) 3 correct (d) 1/ 3 (e) 3 1 2 mv 2 2 1 2 mv 2 1 = 3 v 2 v 1 = 3 6. Two balls of masses 2m an m each experience a net force of the same magnitude. The 2m-ball experiences a net force of magnitude F straight downward and the m ball has a net force of mag- nitude F on it straight upward. What happens to the centre of mass while the forces are applied? 2m m F F 1

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  • Physics 140 — Midterm Exam 2 Version A — 2014 Nov 7

    There are 10 multiple choice questions. Select the correct answer for each one and mark it on thebubble form on the cover sheet. Each question has only one correct answer. (2 marks each)

    1. A satellite is in orbit around a planet. Which of

    the following quantities would change the speed

    the satellite needs to orbit at?

    (a) the mass of the satellite

    (b) the mass of the planet

    (c) the radius of the orbit

    (d) Both A and C

    (e) Both B and C correct

    Let v be the speed of the satellite, M the massof the planet, m the satellite’s mass and R theradius of its orbit. Then

    v2

    R= G

    MR2

    v =

    �G

    MR

    v depends on both the mass of the planet andthe radius of the orbit, but the mass of the satel-lite is not in the relation. ,

    2. A box sits on the horizontal bed of a moving

    truck. Static friction between the box and the

    truck keeps the box from sliding as the truck ac-

    celerates with acceleration a. The mass of thetruck is M, the mass of the box is m and the co-efficient of static friction is µs. What is the mag-nitude of the force of static friction between the

    box and the truck floor?

    (a) F f = ma

    (b) F f = Ma

    (c) F f = µsmg

    (d) F f = µsMg

    (e) None of the above

    The friction F f causes the box of mass m to ac-celerate with acceleration a. It’s just Newton’ssecond law.

    3. The box is sitting on the bed of a moving truck in

    the previous question while the truck slows to a

    stop. The work done by static friction on the box

    while stopping is

    (a) zero.

    (b) positive.

    (c) negative. correct(d) will depend on the directions that are de-

    fined to be positive and negative.

    If the box slows to a stop, the force on the boxis opposite to the direction of motion. Hencethe work done by the force is negative.

    4. What is the direction of the dot product of two

    vectors �A and �B?

    (a) Half way between the direction of each of

    the vectors.

    (b) arccos

    ��A·�B|�A||�B|

    (c) arccos

    �|�A||�B|�A·�B

    (d) Perpendicular to both vectors.

    (e) None of the above. Correct

    The dot product of two vectors is a scalar andhas no direction.

    5. Work is done on an object to triple its kinetic

    energy. By what factor does the object’s speed

    change?

    (a)

    √9

    (b) 1/9

    (c)

    √3 correct

    (d) 1/√

    3

    (e) 3

    1

    2mv2

    2

    1

    2mv2

    1

    = 3

    v2v1=√

    3

    6. Two balls of masses 2m an m each experience anet force of the same magnitude. The 2m-ballexperiences a net force of magnitude F straightdownward and the m ball has a net force of mag-nitude F on it straight upward. What happens tothe centre of mass while the forces are applied?

    2mm

    F

    F

    1

  • Physics 140 — Midterm Exam 2 Version A — 2014 Nov 7

    (a) The centre of mass accelerates straight

    down

    (b) The centre of mass accelerates straight up

    (c) The centre of mass accelerates downward

    at an angle

    (d) The centre of mass accelerates upward at an

    angle

    (e) The centre of mass does not accelerate.correct

    The net external force is zero because bothforces are of equal magnitude but opposite di-rections.

    7. In Case 1, a mass M hangs from a vertical springhaving spring constant k and is at rest in its equi-librium position. In Case 2 the mass has been

    lifted a distance D vertically upward. If we definethe potential energy in Case 1 to be zero, what is

    the potential energy of Case 2?

    (a)1

    2kD2 correct

    (b) MgD

    (c)1

    2kD2 + MgD

    When measured from the new equilibriumpoint, the gravitational potential energy is in-cluded in 1

    2ky2

    2

  • Physics 140 — Midterm Exam 2 Version A — 2014 Nov 7

    8. Here is a graph of potential energy as a function

    of position of an object moving on the x axis. At

    which point is the force greatest in the negative

    direction?

    U(x)

    x

    a b c d e

    0

    (e) is correct. The first is the negative of thederivative of the potetial. Where the slope ofU(x) is positive, F is negative.

    F(x) = −dU(x)dx

    9. Two boys, Al and Bob, are on the balcony of

    an apartment and each throws an object at the

    same time. The objects land on the flat parking

    lot below at the same time. Al throws his object

    upward at 30 degrees to the horizontal and Bob

    throws his object downward at an angle of 30 de-

    grees to the horizontal. Ignore air resistance.

    (a) Al throws his object with initial speed that

    is faster than Bob’s.

    (b) Al throws his object with initial speed that

    is slower than Bob’s.

    (c) They both throw their objects at thesame speed. correct

    (d) The relative speed of the throws depends on

    the relative masses of the objects

    This activity is not recommended.

    Only vA = vB = 0 will allow the ball to begin andend at the same time.

    10. A circus performer of weight W is walking alonga high wire as shown and the wire deviates from

    horizontal by an angle θ.

    The tension in the wire FT is

    (a) FT = (W/2) sin θ

    (b) FT = (W/2) cos θ

    (c) FT = W/(2 sin θ) correct

    (d) FT = W/(2 cos θ)

    (e) None of the above

    The weight of the performer is in equilibriumwith the vertical components of the tension ofthe rope on the right side and left side: mg =2Ty. When the angle of the rope is small, themagnitude of the tension must be large in or-der for the vertical component to be sufficientlylarge to balance the weight: FT = mg/(2 sin θ).

    3