Laws of Motion

  • Question 525
    CBSEENPH11020565

    A system consists of three masses m1, m2 and m3 connected by a string passing over a pulley P. The mass m1 hangs freely and m2 and m3 are on a rough horizontal table (the coefficient of friction= straight mu). the pulley is frictionless and of negligible mass. The downward acceleration of mass m1 is,

    • fraction numerator straight g space left parenthesis 1 minus gμ right parenthesis over denominator straight g end fraction
    • fraction numerator 2 gμ over denominator 3 end fraction
    • fraction numerator straight g space left parenthesis 1 minus 2 straight mu right parenthesis over denominator 3 end fraction
    • fraction numerator straight g left parenthesis 1 minus 2 straight mu right parenthesis over denominator 2 end fraction

    Solution

    C.

    fraction numerator straight g space left parenthesis 1 minus 2 straight mu right parenthesis over denominator 3 end fraction

    First of all consider the forces on the blocks,

    For the 1st block,
    mg - T1 = m x a        ... (i)
    Let us consider second and third block as a system,
    Then,
    T12 μmg = 2m x a
    mg (1 space minus 2 straight mu) = 3m x a
    straight a space equals space 2 over 3 left parenthesis 1 minus 2 straight mu right parenthesis 

    Question 526
    CBSEENPH11020573

    The ratio of the accelerations for a solid sphere (mass m and radius R) rolling down an incline of angle straight theta without slipping and slipping down the incline without rolling is,

    • 5:7

    • 2:3

    • 2:5

    • 7:5

    Solution

    A.

    5:7

    A solid sphere rolling without slipping down an inclined plane is,

    We have in this case,
    a subscript 1 space equals space fraction numerator straight g space sin space straight theta over denominator 1 plus space begin display style straight k squared over straight R squared end style end fraction space equals space fraction numerator g space sin space theta over denominator 1 space plus open parentheses space begin display style bevelled 2 over 5 end style close parentheses end fraction
F o r space s o l i d space s p h e r e comma space K squared space equals space 2 over 5 R squared
space space space space space space equals space fraction numerator g space sin space theta over denominator begin display style bevelled 7 over 5 end style end fraction
rightwards double arrow space a subscript 1 space equals space 5 over 7 space g space sin space theta
    For a sphere slipping down an inclined plane,
    straight a subscript 2 space equals space straight g space sin space straight theta

That space is comma

space space space space space space straight a subscript 1 over straight a subscript 2 space equals space fraction numerator bevelled 5 over 7 space straight g space sin space straight theta over denominator straight g space sin space straight theta end fraction
rightwards double arrow space straight a subscript 1 over straight a subscript 2 space equals space 5 over 7
    Question 527
    CBSEENPH11020580

    A car of mass 1000 kg negotiates a banked curve of radius 90m on a frictionless road. If the banking angle is 45o, the speed of the car is 

    • 20 ms-1

    • 30ms-1

    • 5 ms-1

    • 10 ms-1

    Solution

    B.

    30ms-1

    The angle of banking
    tan space straight theta space equals space straight v squared over rg
Given comma space straight theta space equals space 45 to the power of straight o comma space straight r equals space 90 space straight m
space and space straight g space equals space 10 space straight m divided by straight s squared
tan space 45 to the power of straight o space equals space fraction numerator straight v squared over denominator 90 space straight x space 10 end fraction
straight v space equals space square root of 90 space straight x 10 space straight x space tan space 45 to the power of straight o end root
speed space of space car space straight v space equals space 30 space straight m divided by straight s

    Question 528
    CBSEENPH11020597

    Two sphere A and B of masses m1 and m2 respectively collide. A is at rest initially and B is moving with velocity v along the x -axis. After the collision, B has a velocity v/2 in a direction perpendicular to the original direction. The mass A moves after collision in the direction

    • same as that of B

    • opposite to that of B

    • straight theta space equals space tan to the power of negative 1 end exponent space open parentheses 1 half close parentheses space to space the space straight x space minus axis
    • straight theta space equals space tan to the power of negative 1 end exponent space open parentheses fraction numerator negative 1 over denominator 2 end fraction close parentheses space to space the space straight x space minus axis

    Solution

    C.

    straight theta space equals space tan to the power of negative 1 end exponent space open parentheses 1 half close parentheses space to space the space straight x space minus axis

    Here, Pi = m2vi +m2 x 0

    straight P subscript straight f space equals space straight m subscript 2 straight v over 2 straight j space plus space straight m subscript 1 space straight x space straight v subscript 2
Law space of space conservation space of space momentum
straight P subscript straight i space equals space straight P subscript straight f
straight m subscript 2 straight v subscript straight i space equals straight m subscript 2 space straight v over 2 straight j space plus straight m subscript 1 straight x space straight v subscript 1
straight v subscript 1 space equals space fraction numerator straight m subscript 2 over denominator straight m subscript 1 space end fraction vi space plus straight m subscript 2 over straight m subscript 1 straight v over 2 straight j
From space this space equation space we space can space find
tan space straight theta space equals space 1 half
straight theta space equals space tan to the power of negative 1 end exponent space open parentheses 1 half close parentheses space to space the space straight x minus space axis

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