Laws of Motion

  • Question 541
    CBSEENPH11020753

    A shell of mass 200 g is ejected from a gun of mass 4 g by an explosion that generates 1.05 kJ of energy. The initial velocity of the shell is

    • 100 ms-1

    • 80 ms-1

    • 40 ms-1

    • 120 ms-1

    Solution

    A.

    100 ms-1

    In the given problem conservation of linear momentum and energy hold good.
    Conservation of momentum yields.
    m1v1 + m2v2 = 0
    4v1 + 0.2 v2  = 0
    Conservation of energy yields

    1 half straight m subscript 1 straight v subscript 1 squared space plus 1 half space straight m subscript 2 straight v subscript 2 squared space equals space 1050 space space space... left parenthesis straight i right parenthesis
or
1 half space straight x 4 straight v subscript 1 squared space plus space 1 half 0.1 straight v subscript 2 squared space equals space 1050 space space space.... space left parenthesis ii right parenthesis

solving space eqs. space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis space we space have
straight v subscript 1 space equals space 100 space straight m divided by straight s

    Question 542
    CBSEENPH11020759

    Sand is being dropped on a conveyor belt at the rate of M kg/s. The force necessary to keep the belt moving with a constant velocity of v m/s will be

    • Mv newton

    • 2 Mv newton

    • Mv/2 Newton

    • zero

    Solution

    A.

    Mv newton

    Force required,
    straight F space equals space fraction numerator straight d left parenthesis mv right parenthesis over denominator dt end fraction
equals space straight v open parentheses dm over dt close parentheses space equals space vM

    as velocity v is constant, hence
    F= Mv newton

    Question 543
    CBSEENPH11020772

    A block B is pushed momentarily along a horizontal surface with an initial velocity v. if μ is the coefficient of sliding friction between B  and the surface, block B will come to rest after a time:

    • v/gμ

    • gμ/v

    • g/v

    • v/g

    Solution

    A.

    v/gμ

    Block B will come to rest, if force applied to it will vanish due to frictional force acting between block B and surface, ie,
    force applied = frictional force
    i.e., μmg = ma
    μmg = m (v/t)
    t =v/μg

    Question 544
    CBSEENPH11020795

    300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking =10 m/s2, work done against friction is

    • 200 J

    • 100 J

    • Zero

    • 1000 J

    Solution

    B.

    100 J

    Net work done in sliding a body up to a height h on inclined plane
        = Work done against gravitational force + Work done against frictional force
    rightwards double arrow space space space space space straight W space equals space straight W subscript straight g plus straight W subscript straight f space space space space space space space space space space space... left parenthesis straight i right parenthesis
But space space space space space straight W space equals space 300 space straight J
space space space space space space straight W subscript straight g space equals space mgh space equals space 2 space cross times 10 cross times 10 space equals space 200 space straight J
putting space in space Eq. space left parenthesis straight i right parenthesis comma space we space get
space space space space space space space space space 300 space equals space 200 plus straight W subscript straight f
rightwards double arrow space space space space space space space space space space space space space straight W subscript straight f space equals space 300 space minus space 200 space equals space 100 space straight J
     

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