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A body of mass 2 kg moving with a velocity `3 m//sec` collides with a body of mass 1 kg moving with a velocity of `4 m//sec` in opposite direction. If the collision is head-on perfect inelastic, thenA. both particles will move together with velocity `(2//3) m//sec` after collisionB. the momentum of the system is `2 kg m//sec` throughoutC. the momentum of the system is `10 kg m//sec` throughoutD. The loss in KE of the system is `(49//3)` Joule |
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Answer» Correct Answer - A::B::D Applying momentum conservation. `m_(1)u_(1) - m_(2) u_(2) = (m_(1) + m_(2))v` `2 (3) - 1 (4) = (2 + 1) v` `rArr v = (2)/(3)` m/sec Net mometum `= m_(1) u_(1) - m_(2) u_(2)` `2 (3) - 1 (4) = 2 kg-m//sec` Loss in `KE = (1)/(2) m_(1) u_(1)^(2) + (1)/(2) m_(2) u_(2)^(2) - (1)/(2) (m_(1) + m_(2)) v^(2)` `= (1)/(2) xx 2 xx (3)^(2) + (1)/(2) xx (1) xx (4)^(2) - (1)/(2) (2 + 1) ((2)/(3))^(2)` `= 9 + 8 - (2)/(3) = (49)/(3)` Joule |
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