NEET-XI-Physics

exam-2 year:2016

with Solutions - page 3
  • #117
    A bullet of mass 10g moving horizontally with a
    velocity of 400 ``ms^{-1}`` strikes a wooden block of mass
    2 kg which is suspended by a light inextensible string
    of length 5 m. As a result, the centre of gravity of
    the block is found to rise a vertical distance of
    10 cm. The speed of the bullet after it emerges out
    horizontally from the block will be :-
    (1) 120 ``ms^{-1}``
    (2) 160 ``ms^{-1}``
    (3) 100 ``ms^{-1}``
    (4) 80 ``ms^{-1}``
    digAnsr:   1
    Ans : (1)
    Sol.
    2kg
    2kg v1
    v210gm
    400 m/s
    0.1 m
    5m
    Applying momentum conservation
    1 2
    10 10
    400 0 2 v v
    1000 1000
    ï‚´ + = ï‚´ + ï‚´
    &implies; 4 = 2v1 + 0.01v2 ......(1)
    Applying work energy theorem for block
    W = â–µKE
    &implies; 2 × 10 × 0.1 =
    1
    2
    × 2 × v12
    &implies; v1 = 2 = 1.4 m/s
    Putting the value of v1 in equation (1)
    4 = 2 × 1.4 + 0.01 v2 &implies;v2 = 120 m/s