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)`