NEET-XII-Physics

pp year:2018

with Solutions -
  • #16
    A moving block having mass m, collides with
    another stationary block having mass 4m. The
    lighter block comes to rest after collision.
    When the initial velocity of the lighter block
    is v, then the value of coefficient of
    resistitution (e) will be :-
    (1) 0.5
    (2) 0.25
    (3) 0.8
    (4) 0.4
    digAnsr:   2
    Ans : (2)
    Sol. By conservation of linear momentum
    mv = 4mv' &implies; =
    v
    v '
    4
    coefficient of restitution (e) =
    Velocity of separation
    Velocity of approach
    
    =
    
    v
    0
    4
    v 0
    = =
    1
    0.25
    4