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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 particles of masses m_(1) and m_(2) in projectile motion have velocities vecv_(1) and vecv_(2) respectively at time t= 0. They collide at time t_(0) . Their velocities become vecv_(1) and vecv_(1) at time 2t_(0) while still moving in air. The value of |(m_(1)vecv_(1)^(1)+m_(2)vecv_(2)^(1))-(m_(1)vecv_(1)+m_(2)vecv_(2))| is

    A
    zero
    B
    `(m_(1)+m_(2))g t_(0)`
    C
    `2(m_(1)+m_(2))g t_(0)`
    D
    `(1)/(2)(m_(1)+m_(2))g t_(0)`

  • Two particles of masses m_1 and m_2 in projectile motion have velocities vecv_1 and vecv_2 respectively at time t=0. They collide at time t_0. Their velocities become vecv_1' and vecv_2' at time 2t_0 while still moving in air. The value of |(m_1vecv_1' + m_2vecv_2') - (m_1vecv_1 + m_2vecv_2)| is

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    zero
    B
    `(m_1+m_2)gt_0`
    C
    `1/2(m_1+m_2)gt_0`
    D
    `2(m_1+m_2)gt_0`

  • If two bodies are in motion with velocity vec(v)_(1) and vec(v)_(2) :

    A
    `|v_("rel")| = sqrt(v_(1)^(2) + v_(2)^(2))`
    B
    `|vec_("rel")| = v_(1) +- v_(2)|`
    C
    `v_("re") = 0`
    D
    `v_("rel") gt c` (speed of light)

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