Laws of Motion

  • Question 529
    CBSEENPH11020605

    The upper half of an inclined plane of the inclination straight theta is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by,

    • straight mu space equals space fraction numerator 1 over denominator tan space straight theta end fraction
    • straight mu space equals space fraction numerator 2 over denominator tan space straight theta end fraction
    • straight mu space equals space 2 space tan space straight theta
    • straight mu space equals space tan space straight theta

    Solution

    C.

    straight mu space equals space 2 space tan space straight theta
    The block will be stationary when,

    mg sin straight theta . L = straight alpha mg cos straight theta space. space straight L over 2
    straight mu space equals space fraction numerator mg space sin space straight theta space straight L over denominator mg space cos space straight theta space begin display style bevelled straight L over 2 end style space end fraction

space space equals space 2 space fraction numerator sin space straight theta over denominator cos space straight theta end fraction

space space equals space 2 space tan space straight theta

straight mu equals space 2 space tan space straight theta
    Question 530
    CBSEENPH11020607

    An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 m/s and the second part of mass 2 kg moves with second part of mass 2 kg moves with 8 m/s speed. If the third part flies off with 4 m/s speed, then its mass is,

    • 3 kg

    • 5 kg

    • 7 kg

    • 17 kg

    Solution

    B.

    5 kg

    We have,
    p1 + p2 + p3 = 0 [ because p = mv]
    Therefore,
    1 x 12 i  + 2 x 8 j + p3 = 0
    rightwards double arrow space 12 space straight i space plus space 16 space straight j space plus space straight p subscript 3 space equals space 0

rightwards double arrow space straight p subscript 3 space equals space minus left parenthesis 12 space straight i space space plus space 16 space straight j right parenthesis

therefore space straight p subscript 3 space equals space square root of left parenthesis 12 right parenthesis squared plus left parenthesis 16 right parenthesis squared end root

space space space space space space space space space space equals space square root of 144 space plus space 256 space end root

space space space space space space space space space space equals space 20 space kg minus straight m divided by straight s
    Now. p3 = m3 v3
    This implies,


    straight m subscript 3 space equals space straight p subscript 3 over straight v subscript 3 space
space space space space space equals space 20 over 4 space
space space space space space equals space 5 space kg
    Question 531
    CBSEENPH11020625

    A person of mass 62 kg is inside of a lift of mass 940 kg and presses the button on control panel. The lift starts moving upwards with an acceleration 1.0 m/s2. If g = 10 m/s2 ,the tension in the supporting cable is

    • 9680 N

    • 11000 N

    • 1200 N 

    • 8600 N

    Solution

    B.

    11000 N


    Total mass = Mass of lift + Mass of person
    = 940 +60 = 1000 kg
    T - mg = ma
    Hence, T - 1000 x 10 = 1000 x 1
     T = 11000 N
    Question 532
    CBSEENPH11020631

    A planet moving along an elliptical orbit is closer to the sun at a distance r1 and farthest away at a distance of r2. if v1 and v2 are the linear velocities at these points respectively, then the ratio v1/v2 is

    • r2/r1

    • (r2/r1)2

    • r1 /r2

    • (r1/r2)2

    Solution

    A.

    r2/r1

    From the law of conservation of angular momentum
    mr1v1 = mr2v2
    r1v1 = r2v2
    v1/v2 = r2/r1

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