Laws of Motion

  • Question 533
    CBSEENPH11020686

    A ball moving with velocity 2 ms-1 collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in ms-1) after collision will be

    • 0,1

    • 1,1

    • 1,0.5

    • 0,2

    Solution

    A.

    0,1

    If two bodies collide head on with coefficient of restitution
    straight e space equals space fraction numerator straight v subscript 2 minus straight v subscript 1 over denominator straight u subscript 1 minus straight u subscript 2 end fraction
From space the space law space of space conservation space of space linear space momentum

straight m subscript 1 straight u subscript 1 plus straight m subscript 2 straight u subscript 2 space equals space straight m subscript 1 straight v subscript 1 space plus straight m subscript 2 straight v subscript 2
rightwards double arrow space straight v subscript 1 space equals space open square brackets fraction numerator straight m subscript 1 minus em subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight u subscript 1 plus open square brackets fraction numerator left parenthesis 1 plus straight e right parenthesis straight m subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight u subscript 2
Substituting space straight u subscript 1 space equals space 2 space ms to the power of negative 1 end exponent comma space straight u subscript 2 space equals 0 comma space straight m subscript 1 equals straight m subscript 2 space and space straight m subscript 2 space equals space 2 straight m comma space straight e space equals space 0.5
we space get space space straight v subscript 1 space equals space open square brackets fraction numerator straight m minus straight m over denominator straight m plus 2 straight m end fraction close square brackets space straight x 2
similarly comma
straight v subscript 2 space equals open square brackets fraction numerator left parenthesis 1 plus straight e right parenthesis straight m subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight u subscript 1 space plus open square brackets fraction numerator straight m subscript 2 minus em subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight u subscript 2
equals space open square brackets fraction numerator 1.5 space straight x space straight m over denominator 3 straight m end fraction close square brackets space straight x space 2
space equals space 1 space ms to the power of negative 1 end exponent

    Question 534
    CBSEENPH11020699

    A man of 50 kg mass standing in a gravity free space at height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed 2 ms-1. When the stone reaches the floor, the distance of the man above the floor will be

    • 9.9 m

    • 10.1 

    • 10 m

    • 20 m

    Solution

    B.

    10.1 

    m r = constant
    straight m subscript 1 straight r subscript 1 space equals straight m subscript 2 straight r subscript 2

straight r subscript 2 space fraction numerator straight m subscript 1 straight r subscript 1 over denominator straight m subscript 2 end fraction

equals space fraction numerator 0.5 space straight x space 10 over denominator 50 end fraction space equals space 0.1
    The distance of the man above the floor (total height) = 10+0.1 = 10.1

    Question 535
    CBSEENPH11020714

    A student measures the distance traversed in free fall of a body, initially at rest in a given time. He uses this data to estimate g, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance  and the time are e1 and e2 respectively, the percentage error in the estimation of g is

    • e1 - e2

    • e1 + 2e2

    • e1+ e2

    • e1 - 2e2

    Solution

    B.

    e1 + 2e2

    straight h space equals 1 half gt squared

straight g space equals space straight h over straight t squared
therefore space log space equals space log space straight h space minus 2 space log space straight t
open parentheses fraction numerator increment straight g over denominator straight g end fraction straight x 100 close parentheses subscript max space equals space open parentheses fraction numerator increment straight h over denominator straight h end fraction straight x 100 close parentheses plus 2 open parentheses fraction numerator increment straight t over denominator straight t end fraction straight x 100 close parentheses
straight e subscript 1 plus 2 straight e subscript 2
    Question 536
    CBSEENPH11020716

    An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are 1 kg first part moving with a velocity of 12 ms-1 and 2 kg second part moving with a velocity of 8 ms-1.If the third part flies off with a velocity of 4 ms-,its mass would be

    • 5 kg 

    • 7 kg

    • 17 kg

    • 3 kg

    Solution

    A.

    5 kg 

    apply the law of conservation of linear momentum. 
    momentum of first part = 1 x 12 = 12 kg ms-1
    Momentum of the second part  = 2 x 8 = 16 kg ms-1 '
    Resultant monmentum
    = [(12)2 +(16)2]1/2 = 20 kg ms-1
    The third part should also have the same momentum.


    Let the mass of third part be M, then 
    4 x M = 20
    M = 5 kg

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