An explosion blows a stationary rock into three parts. Two of masses 1kg and 2kg moves at right angles to one another with velocities `12ms^(-1)` and `8 ms^(-1)`, respectively. If the velocity of third part is `4 ms^(-1)`, the mass of the rock is

Options

A. 8 kg
B. 5 kg
C. 17 kg
D. 3 kg

Answer

A

Explanation

Given, `m_(1) = 1 kg, m_(2) = 2 kg`,
`v_(1) = 12 ms^(-1), v_(2) = 8 ms^(-1)` and `v_(3) = 4 ms^(-1)`
Since, in the explosion of stationary rock, the momentum is converted,
So `P_(i) = p_(f)`
`0 = p_(f) = p_(1) + p_(2) + p_(3)`
where, `p_(1) = m_(1)v_(1), p_(2) = m_(2)v_(2)` and `p_(3) = m_(3)v_(3)`
`p_(3) = -(p_(1) + p_(2))`
`p_(3) = sqrt((p_(1)^(2) + p_(2)^(2) + 2p_(1).p_(2) cos theta)`
`p_(3) = sqrt(12^(2) + 16^(2) + 2 xx 12 xx 16 cos 90^(@))`
`p_(3) = 20`
`m_(3)v_(3) = m_(3) xx 4 = 20 rArr m_(3) = 5` kg
Hence, the mass of the rock is,
`m = m_(1) + m_(2) + m_(3)`
`m = 1 + 2 + 5 = 8` kg
So, the correct option is (a).