Physics movement questions and solutions

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  • 8/12/2019 Physics movement questions and solutions

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    Name: ______________________________________________________________ Total Points: _______

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    Multiple choice questions

    Answer all of the following questions. Read each question carefully. Fill the correctbubble on your scantron sheet. Each question has exactly one correct answer. All

    questions are worth the same amount of points.

    A plane goes 300 km from A to B in a straight line, immediately turns

    around, and returns to A. The time for this round trip is 2 hour. The

    magnitude of the average velocity of the plane for this round trip is

    A.

    0 km/ht

    xvavg

    = and 0=x

    B. 100 km/hC. 150 km/h

    D. 200 km/h

    1.

    E. Cannot be calculated without knowing the acceleration

    A car, initially at rest, travels 20m in 4s along a straight line with

    constant acceleration. The acceleration of the car (in m/s2) is

    A. 0.4B. 1.3C.

    2.52

    2/5.2

    2sm

    t

    xa ==

    D. 4.9

    2.

    E. 9.8

    At a stoplight, a truck traveling at 15m/s passes a car as it starts from

    rest. The truck travels at constant velocity and the car accelerates at

    3m/s2. How many seconds will it take for the car to catch up to the

    truck?

    A. 5

    B.10 txtruck 15= ,

    2321 txcar = , statxx truckcar 10==

    C. 15

    D. 20

    3.

    E. 25

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    An elevator is moving upward with constant acceleration. The

    dashed curve shows the position y of the ceiling of the elevator as a

    function of the time t. At the instant indicated by the dot, a bolt

    breaks loose and drops from the ceiling. Which curve best represents

    the position of the bolt as a function of time?

    4.

    y

    t

    A The trajectory must be

    a parabola. When the

    bolt breaks loose, the

    elevator and the bolt

    have the same velocity

    (the bolt parabola

    should have the same

    tangent as the elevator

    trajectory at the dot)

    C

    DE

    B

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    On a circular track, a car starts from rest at point A and moves in a

    clockwise direction with increasing speed. What is the direction of

    the acceleration vector at A?

    A. Undefined: the acceleration is zero

    B.

    C.

    D.

    5.

    E.

    A

    B

    cAC vvvv rrrr

    == Take C between A and B and bring

    it closer and closer to A. Cvr

    becomes tangent to the

    circle at A ointed u .

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    Referring to the car of the previous question, what would be a valid

    direction of the acceleration at point B?

    A. Undefined: the acceleration is zero

    B.

    C. The car is traveling on a circle: one component of the acceleration

    is toward the center. The car is speeding up: the other component

    of the acceleration is tangent to the circle.

    D.

    6.

    E.

    Four vectors DCBArrrr

    ,,, all have the same magnitude. The angle

    between adjacent vectors is 45as shown. The correct vector

    equation is

    A. 0=+ DCBArrrr

    B.

    02 =+ DCBrrr

    Add Br

    and Dr

    using the parallelogram construction. The sum

    DBrr

    + is equal to Cr

    2

    C. DBBArrrr

    +=+

    D. 0=+++ DCBArrrr

    7.

    E. 02 =++ CBArrr

    Ar

    Br

    Cr

    Dr

    45

    45

    45

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    In the diagram, Ar

    has magnitude 12 m and Br

    has magnitude 8 m.

    The x component of BArr

    + is about

    A. 4.5 mB. 8.5 mC. 12.5 m since Ax+Bx=12cos45 + 8cos60

    D. 14.5 m

    8.

    E. 20 m

    x

    45

    60

    Ar

    Br

    y

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    Two objects, A and B, move with constant speed relative to a straight

    line. The strobe diagram shows the positions of the objects at instant

    1-3, separated by one-second time intervals. (Note that each tick mark

    on the diagram represents 5 meters.)

    At instant 2 what is the direction of the instantaneous velocity of object

    B in the frame of reference of object A?

    A. to the left (B is getting closer and closer to A)B. to the rightC. Undefined: the velocity is zero

    9.

    Still referring to the problem of the previous question, at instant 2, what

    is the magnitude of the instantaneous velocity of object B in the frame

    of reference of object A?

    A. 0 m/sB. 10 m/s

    Between instant 1 and instant 2, the distance between A and B

    changes from 20m to 10mC. 20 m/sD. 30 m/s

    10.

    E. 50 m/s

    B

    AAA

    B B

    1

    1

    2

    2

    3

    35 meters

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    Questions 11 through 18 all refer to the same problem.

    A baseball is thrown from the top of a cliff as shown below. The cliff height

    is marked in each case. Answer the following questions, referring to these

    cases. Take g=10 m/s2.

    In which case will the baseball remain in the air the longest amount of

    time? D

    It is case D that has the largest initial v0y

    11.

    45

    x

    y

    5 m/s

    20 m

    45

    A

    20 m/s

    20 m

    B

    5 m/s

    20 m

    C

    30 m/s

    20 m

    E

    10 m/s

    20 m

    D (Assume that the ball

    doesn't touch the cliff

    on its way down).

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    In which case will the baseball remain in the air the shortest amount

    of time? C

    It is case C that has the smallest initial v0y(since v0y

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    If you wanted to make the ball go farther in the x direction, in case B, you

    could

    A. increase the magnitude of the initial velocity

    OK since the distance is timeflightx tv 0 . A greater initial velocity increases

    v0xB. change the angle between the initial velocity and the x direction to +45

    (i.e. the initial velocity is directed upward at a 45angle).

    If the projectile is fired at an angle at the top of a cliff of height h, the

    flight time is the (positive) solution of

    g

    h

    g

    v

    g

    vttvgth

    2sinsinsin

    2

    12

    22

    00

    0

    2++=+=

    and the distance traveled is

    g

    h

    g

    vv

    g

    vd

    2sincos

    cossin2

    22

    00

    2

    0++=

    Thus

    mg

    h

    g

    vv

    g

    vd

    mg

    hvd

    6.542

    222

    4010

    4020

    2

    2

    2

    00

    2

    045

    00

    =++=

    ===

    =

    =

    C. move to a higher cliffOK since the distance is timeflightx tv 0 . A higher cliff increases tflight time

    D. A and B

    15.

    E. A, B and C

    All questions 16 through 18 refers to case D (take g=10m/s2)

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    Determine how many seconds it takes for the baseball to hit the ground

    A. 1.9B. 2.7

    C. 3.2Solve:

    =+=+= 5110102

    120 2 ttt

    D. 3.8

    16.

    E. 4.8

    What is the magnitude of the velocity when the ball hits the ground?

    A.0 m/sB. 5 m/s

    C. 10.7 m/sD. 15.8 m/s

    17.

    E. 22.4 m/s

    There is only motion along the y axis andsmvtvv y /36.221010 =+==

    What is the direction of the acceleration when the baseball is at its

    highest point?

    A. Undefined: the acceleration is zero.B. xC. xD. y

    18.

    E. y

    The acceleration of the ball is always equal to the acceleration of

    gravity