Laws of Motion

  • Question 521
    CBSEENPH11020539

    A block A of mass m1 rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of the table and from its other end, another block B of mass m2 is suspended. The coefficient of kinetic friction between the block and the table is straight mu subscript straight k. When the block A is sliding on the table, the tension in the string is

    • fraction numerator left parenthesis straight m subscript 2 plus straight mu subscript straight k straight m subscript 1 right parenthesis straight g over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction
    • fraction numerator left parenthesis straight m subscript 2 minus straight mu subscript straight k straight m subscript 1 right parenthesis straight g over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction
    • fraction numerator straight m subscript 1 straight m subscript 2 left parenthesis 1 plus straight mu subscript straight k right parenthesis straight g over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction
    • fraction numerator straight m subscript 1 straight m subscript 2 left parenthesis 1 minus straight mu subscript straight k right parenthesis straight g over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction

    Solution

    C.

    fraction numerator straight m subscript 1 straight m subscript 2 left parenthesis 1 plus straight mu subscript straight k right parenthesis straight g over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction

    FBD of block A,

    T -m1afk ..... (i)
    FBD  of block B

    m2g -T = m2a ... (ii)
    Adding Eqs. (i) and (ii), we get
    m2g-m1a = m2a +fkrightwards double arrow space straight m subscript 2 straight g minus straight m subscript 1 straight a space equals space straight m subscript 2 straight a space plus straight f subscript straight k
rightwards double arrow space straight a equals space fraction numerator left parenthesis straight m subscript 2 minus straight u subscript straight k straight m subscript 1 right parenthesis straight g over denominator straight m subscript 1 plus straight m subscript 2 end fraction
from space equation space left parenthesis ii right parenthesis
straight T equals space straight m subscript 2 minus left parenthesis straight g minus straight a right parenthesis
equals space straight m subscript 2 open square brackets 1 minus fraction numerator left parenthesis straight m subscript 2 minus straight mu subscript straight k straight m subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight g
straight T equals space fraction numerator straight m subscript 1 straight m subscript 2 space left parenthesis 1 plus straight mu subscript straight k right parenthesis over denominator straight m subscript 1 plus straight m subscript 2 end fraction straight g
    Question 522
    CBSEENPH11020548

    A stone is dropped from a height h. It hits the ground with a certain momentum p. If the same stone is dropped from a height 100% more than the previous height, the momentum when it hits the ground will change by

    • 68%

    • 41%

    • 200%

    • 100%

    Solution

    B.

    41%

    velocity space straight v space equals space square root of 2 gh end root space space space space space space space.... left parenthesis straight i right parenthesis
and space momentum space straight p space equals space mv space space space space... space left parenthesis ii right parenthesis
From space left parenthesis Eqs. right parenthesis space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space have
space straight p proportional to square root of straight h
Here comma
straight p subscript 2 over straight p subscript 1 space equals space square root of straight h subscript 2 over straight h subscript 1 end root
therefore space fraction numerator straight p subscript 2 over denominator straight p subscript 1 space end fraction space equals square root of fraction numerator 2 straight h over denominator straight h end fraction space end root space equals square root of 2

straight p subscript 2 space equals 1.414 straight p subscript 1
percent sign change space equals space fraction numerator straight p subscript 2 minus straight p subscript 1 over denominator straight p subscript 1 end fraction space straight x space 100 space equals space 41 percent sign
    Question 523
    CBSEENPH11020550

    A car of mass m is moving on a level circular track of radius R. If straight mu subscript straight s represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is given by

    • square root of straight mu subscript straight s mRg end root
    • square root of Rg divided by straight mu subscript straight s end root
    • square root of mRg divided by straight mu subscript straight s end root
    • square root of straight mu subscript straight s Rg end root

    Solution

    D.

    square root of straight mu subscript straight s Rg end root

    In this condition, centripetal force is equal to static frictional force between road and tyres, 
    so
    straight mu subscript straight s space mg space equals mv squared over straight R
straight v subscript max space equals space square root of straight mu subscript straight s Rg end root

    Question 524
    CBSEENPH11020553

    Three masses are placed on the x- axis: 300g at origin, 500g at x = 40 cm and 400 g at x = 70 cm. The distance of the centre of mass from the origin is 

    • 40 cm

    • 45 cm

    • 50 cm

    • 30 cm

    Solution

    A.

    40 cm

    straight x subscript cm space equals fraction numerator straight m subscript 1 straight x subscript 1 space plus straight m subscript 2 straight x subscript 2 space plus straight m subscript 3 straight x subscript 3 over denominator straight m subscript 1 plus straight m subscript 2 plus straight m subscript 3 end fraction
space equals space fraction numerator 300 space left parenthesis 0 right parenthesis space plus 500 space left parenthesis 40 right parenthesis space plus 400 space left parenthesis 70 right parenthesis over denominator 300 plus 500 plus 400 end fraction
equals space fraction numerator 20000 plus 28000 over denominator 1200 end fraction
space equals 48000 over 1200 space equals 40 space cm

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