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We will assume that when the dumb-bell falls to the floor, it does not bounce and remains on the floor i.e. all of its momentum gets transferred to the floor at the same moment. Now we will use the third equation of motion by Newton to find what will be the velocity of the dumb-bell just before it strikes the ground.

$2as={{v}^{2}}-{{u}^{2}}$

Here, a is the acceleration on the body (dumb-bell here) which here will be the acceleration due to gravity. Its magnitude has been given as 10 $m{{s}^{-2}}$ in the question.

s is the distance traveled by the body under that acceleration which here is given as 80 cm i.e. 0.8 m.

v is the final velocity of the dumb-bell and u will be its initial velocity.

We will take u to be zero as the body starts from rest and we will take the downward direction as positive. So,

\[\begin{align}

& 2\times 10\times 0.8={{v}^{2}}-{{0}^{2}} \\

& {{v}^{2}}=16 \\

& \therefore v=4m{{s}^{-1}} \\

\end{align}\]

The momentum of the body at that moment will be the product of its mass and velocity.

$P=mv=10\times 4=40kgm{{s}^{-1}}$

As all the momentum gets transferred, $40kgm{{s}^{-1}}$ will be transferred.