Work, Energy and Power


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









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    Let the initial velocity of the shell be v, then by the conservation of momentum mv = Mv' ​
    where v' = velocity of gun.

    ∴ v' =
    m
    v
    M

    Now, total K.E. =
    1
    mv2 +
    1
    Mv'2
    22

    =
    1
    mv2 +
    1
    M
    m
    2v2
    22M

    =
    1
    mv2 1 +
    m
    2M

    =
    1
    × 0.21 +
    0.2
    v2 = (0.1 × 1.05)v2
    24

    But total K.E. = 1.05 kJ = 1.05 × 103 J ​
    ∴ 1.05 × 103 = 0.1 × 1.05 × v2
    v2 =
    1.05 × 103
    = 104
    0.1 × 1.05

    ∴ v = 102 = 100 ms–1.

    Correct Option: A

    Let the initial velocity of the shell be v, then by the conservation of momentum mv = Mv' ​
    where v' = velocity of gun.

    ∴ v' =
    m
    v
    M

    Now, total K.E. =
    1
    mv2 +
    1
    Mv'2
    22

    =
    1
    mv2 +
    1
    M
    m
    2v2
    22M

    =
    1
    mv2 1 +
    m
    2M

    =
    1
    × 0.21 +
    0.2
    v2 = (0.1 × 1.05)v2
    24

    But total K.E. = 1.05 kJ = 1.05 × 103 J ​
    ∴ 1.05 × 103 = 0.1 × 1.05 × v2
    v2 =
    1.05 × 103
    = 104
    0.1 × 1.05

    ∴ v = 102 = 100 ms–1.