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An object of mass `1kg` travelling in a straight line with a velocity of `10m//s` collides with, and sticks to, a stationary wooden block of mass `5kg`. Then, they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

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Here, Mass of object,`m_(1)`= 1 kg
And, Velocity of object, `v_(1)=10ms^(-1)`
So, Momentum of object=`m_(1)xxv_(1)`
`1xx10 kg ms^(-1)`…(1)
`=10kgms^(-1)`
Now, Mass of wooden block, `m_(2)` =5 kg
And, Velocity of wooden block, `v_(2) = 0` (It is stationary)
So, Momentum of wooden block = `m_(2) xx v_(2)`
=`5xx0`
`=0kgms^(-1)`....(2)
Total momentum=10+0
before impact=10 kgm`s^(-1)`......(3)
Now, according to the law of conservation of momentum, the total momentum just after the impact will be the same as the tota lmomentum just before the impact. This means that the total momentum just after the impact will also be 10 kg m `s^(-1)`.
Total momentum=10 kg m `s^(-1)`
Total mass of object = 1 kg + 5 kg
and wooden block= 6 kg
And Velocity of object = v m `s^(1)` (Supposed)
and wooden block
So, `10=6xxv`
And, `v=10/6`
Thus, the velocity of the object and wooden block together is 1.67 metres per second.

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