How much momentum will a dumb-bell of mass `10kg` transfer to the floor if it falls a height of `80 cm`? Take its downward acceleration to be `10 m//s^(2)`.

Explanation

We will first calculate the final velocity of falling dumb-bell (just before touching the floor)
Here, Initial velocity,u = 0 (Falls from rest)
Final velocity, v = ? (To be calculated)
Acceleration, a=10 m `s^(-2)`
And, Distance,s=80 cm
(or Height)=`80/100m`
`=0.8m`
Now, `v^(2)=u^(2)+2as`
So, `v^(2)=(0)^(2)+2xx10xx0.8`
`v^(2)=16`
`v=sqrt(16)`
`v=4ms^(-1)`
Momentum=mass`xx`velocity
`=10kgxx4ms^(-1)`
`=40kgms^(-1)`
Now, the momentum of falling dumbell just beforetouchng thr floor is 40 kg `ms^(-1)`. So, the dumbell will transfer an equal amount of momentum (40 kg m`s^(-1)`) to the floor.