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The figure represents two bodies of mass...

The figure represents two bodies of masses 10 kg and 20 kg, moving with an initial velocity of `10 ms^(-1)` and `5 ms^(-1)` respectively. They collide with each other. After collision, they move with velocities `12 ms^(-1)` and `4 ms^(-1)` respectively. The time of collision is 2s. Now calculate `F_(1)` and `F_(2)`.

Text Solution

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`m_(1) = 10 kg " " m_(2) = 20 kg`
`u_(1) = 10 ms^(-1) " " u_(2) = 5 ms^(-1)`
`v_(1) = 12ms^(-1)" " v_(2) = 4ms^(-1)`
Time of collision, t = 2s
`therefore` Force acting on 20 kg object
`F_(1) = m_(2)((v_(2)-u_(2))/(t)`
= `20((4-5)/(2))`
`F_(1) = -10 N`
Force acting on 10 kg object
`F_(2) = m_(1)((v_(1)-u_(1))/(t))`
` = 10((12-10)/(2))`
`F_(2) = 10N`
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