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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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Correct Answer - 10 kg m `s^(-1)`; 10 kg m `s^(-1)`; 5/3 m `s^(-1)`
Mass of the object, `m_(1) = 1 kg`
Velocity of the object before collision, `v_(1) = 10 m//s`
Mass of the stationary wooden block, `m_(2) = 5 kg`
Velocity of the wooden block before collision, `v_(2) = 0 m//s`
`:.` Total momentum before collision `= m_(1) v_(1) + m_(2) v_(2)`
` = 1 (10) + 5 (0) = 10 kg ms^(-1) ` ltbr gt It is given that after collision, the object and the wooden block stick together.
Total mass of the combined system `= m_(1) + m_(2) `
Velocity of the combined object = v
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
` m_(1) v_(1)+ m_(2) v_(2) = (m_(1)+ m_(2)) v`
`implies1 (10) + 5 (0) = (1 + 5) v`
`impliesv=(10)//(6)`
`=(5)//(3) m//s`
The total momentum after collision is also 10 kg m/s.
Total momentum just aftet the impact = 10 kg `ms^(-1)`
Total momentum just after the impact `=(m_(1)+m_(2))v=6xx5//3=10 kg ms^(-1)`
Hence, velocity of the combined object after collision `=5//3 ms^(-1)`

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