Question

How much momentum will a dumbbell of mass $10kg$ transfer to the floor if it falls from a height of $80cm$ (assume that the dumbbell stops after hitting the ground)? Take $g=10m{s}^{-2}$

• A. $40{kgms}^{-1}$​
• B. $50{kgms}^{-1}$​
• C. $30{kgms}^{-1}$​
• D. $35{kgms}^{-1}$​

Answer 1

The correct option is A

$40{kgms}^{-1}$​

Mass of the dumbbell, $m$= $10kg$
Distance covered by the dumbbell, $s=80cm=0.8m$
Acceleration in the downward direction,$a=10m{s}^{-2}$
Initial velocity of the dumbbell, $u=0$
Final velocity of the dumbbell(when it was about to hit the floor) = $v$
According to the third equation of motion
$v^2-u^2=2as\Rightarrow v=\sqrt{u^2+(2\times a\times s)}$
$\Rightarrow v=\sqrt{0+(2\times 10\times 0.8)}$ = $4{ms}^{-1}$​
Hence, the momentum(p) with which the dumbbell hits the floor is
$p=m\times v=10\times 4=40{kgms}^{-1}$​
Therefore, dumbbell will exert a momentum of $40{kgms}^{-1}$​ to the floor