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Two particles of masses m(1) and m(2) in...

Two particles of masses `m_(1)` and `m_(2)` in projectile motion have velocities `vec(v)_(1)` and `vec(v)_(2)` , respectively , at time `t = 0`. They collide at time `t_(0)`. Their velocities become `vec(v')_(1)` and `vec(v')_(2)` at time ` 2 t_(0)` while still moving in air. The value of `|(m_(1) vec(v')_(1) + m_(2) vec(v')_(2)) - (m_(1) vec(v)_(1) + m_(2) vec(v)_(2))|`

Text Solution

Verified by Experts

The correct Answer is:
C
By applying impulse-momentum theorem
`=|(m_(1)vec(v)_(1) + m_(2)vec(v)_(2)) - (m_(1)vec(v)_(1) + m_(2)vec(v)_(2))|`
`= |(m_(1) + m_(2)) vec(g) (2L_(0))| - 2(m_(1) + m_(2)) g t_(0)`
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Knowledge Check

  • Two particle of masses m_(1).m_(2) moves with initial velocities u_(1) and u_(2) . On collision one of the particles get excited to higher level after absorbing energy epsilon . If final velocities of particles be v_(1) and v_(2) then we must have :

    A
    `1/2m_(1)u_(1)^(2) +1/2 m_(2)u_(2)^(2) =1/2m_(1)v_(1)^(2)+1/2m_(2)v_(2)^(2)-epsilon`
    B
    `1/2 m_(1)u_(1)^(2) +1/2m_(2)u_(2)^(2) - epsilon =1/2 m_(1)v_(1)^(2) +1/2m_(2)v_(2)^(2)`
    C
    `1/2m_(1)^(2)u_(1)^(2)+1/2m_(2)^(2)u_(2)^(2)+epsilon =1/2 m_(1)^(2)v_(1)^(2)+1/2m_(2)^(2)v_(2)^(2)`
    D
    `m_(1)^(2)u_(1)+m_(2)^(2)u_(2)-epilon=m_(1)^(2)v_(1)+m_(2)^(2)v_(2)`

  • If vec(a) is a unit vector and vec(b)=(2,1,-1) and vec( c )=(1,0,3) . Then the maximum value of [vec(a)vec(b)vec( c )] is…………..

    A
    `-1`
    B
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    C
    `sqrt(6)+sqrt(10)+1`
    D
    `sqrt(60)`

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