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Questions: 1 2 3 4 5 6

Physics 1120: Math Review Solutions

  1. Convert the following:
    (a) 3.5 km2 to m2


    (b) 56 g/cm3 to kg/m3


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  2. Write vectors equations for each diagram below.
    (a) (b)
    (c) (d)
    (e) (f)

    (a) (b)
    (c) (d)
    (e) (f)

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  3. Momentum is a vector quantity given by p = mv. A system consists of 2 particles: one of mass 150 g moving at 125 cm/s at 15.0° west of north, and the other of mass 200 g moving at 90 cm/s at 30.0° south of east. What is the net momentum of the system?

    First we get the magnitude of the momenta: and . The net momentum is the vector sum of the given momenta, Pnet = P1 + P2. Next we neatly sketch the problem and its solution.

    Vector problems are solved by breaking the given vectors into their i and j components.

    P1 = −i0.1875sin(15°) + j0.1875cos(15°) = −i0.04853 + j0.18111 kg·m/s.

    P2 = i0.1800cos(30°) + −j0.1800sin(30°) = i0.15588 −j0.09000 kg·m/s

    The resultant Pnet = P1 + P2 is

    Pnet = i0.10735 + j0.09111 kg·m/s.

    Using Pythagoras' Theorem, . We use trigonometry to find the direction. The angle is . Note that the components are in the N-E quadrant. The total or net momentum is thus 0.141 kg m/s at 40.3° north of east.

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  4. Two forces act on a body. The first is F1 = 135 N at 25.0° west of south, and the second is F2 = 190 N acting at 40.0° south of east. Find the net force acting on the body.

    The net force is the vector sum of the given forces, Fnet = F1 + F2.

    First we neatly sketch the problem and its solution.


    Vector problems are solved by breaking the given vectors into their i and j components.

    F1 = −i135sin(25°) + −j135cos(25°) = −i57.05 − j122.35 N.

    F2 = i190cos(40°) + −j190sin(40°) = i145.55 − j122.13 N

    The resultant Fnet = F1 + F2 is

    F2 = i85.50 − j244.48 N.

    Using Pythagoras' Theorem, . We use trigonometry to find the direction. The angle is . Since the components are in the S-E quadrant, the total or net force is thus 260 N at 70.1° south of east. It would be more common to state this as 260 N at 19.9° east of south.

  5. A person on the shore at English Bay observes that the net velocity of a person in a boat is 2.75 m/s at 17.0° north of east. The boat is moving relative to the water at 1.90 m/s heading 10.0° south of east. Determine the velocity of the water current relative to the person on shore.

    First, we must realize that the velocity that the observer sees is the sum of the velocity of the boat and the velocity of the current, i.e. . Since we are looking for , we are dealing with the subtraction of vectors vcurrent = vobserved − vboat. To solve this, or any other vector problem, we sketch the solution first.

    Vector problems are solved by breaking the given vectors into their i and j components.

    vobserved = i2.75cos(17°) + j2.75sin(17°) = i2.630 + j0.804 m/s.

    vboat = i1.90cos(10°) + −j1.90sin(10°) = i1.871 − j0.330 m/s

    The resultant vcurrent = vobserved − vboat is

    vcurrent = i0.759 + j1.134 m/s.

    Using Pythagoras' Theorem, . We use trigonometry to find the direction. The angle is . The components are in the N-E quadrant, thus the velocity of the current is thus 1.36 m/s at 56.2° north of east.

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  6. The net force of two forces acting on a body is 3500 N at 14.5° west of south. One force is 2100 N at 22.0° east of south. What is the other force?

    Since we are told the total force and one of the forces, the other force is given by F2 = Fnet − F1. To solve this, or any other vector problem, we sketch the solution first.

    Vector problems are solved by breaking the given vectors into their i and j components.

    Fnet = −i3500sin(14.5°) + −j3500cos(14.5°) = −i876.33 − j3388.52 N.

    F1 = i2100sin(22°) + −j2200cos(22°) = i786.67 − j1947.07 N

    The resultant F2 = Fnet − F1 is

    F2 = −i1663.00 − j1441.43 N.

    Using Pythagoras' Theorem, . We use trigonometry to find the direction. The angle is . Since the components are in the W-S quadrant, the second force is thus 2200 N at 40.9° south of west.

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