Question

A $10\text{kg}$ iron dumbbell is dropped from a height of $80\text{cm}$ onto a stone floor. If the dumbbell comes to rest after hitting the floor, how much momentum (in ${\text{kg m s}}^{-1}$ ) will the floor acquire? Take the acceleration caused in the dumbbell due to gravity, to be $10{\text{m s}}^{-2}$

• A. 40

Answer 1

The correct option is A 40
Given:
The mass of the dumbbell, $\text{m}=10\text{kg}$
Height, $\text{h}=80\text{cm}=80\times {10}^{-2}\text{m}$
Acceleration due to gravity, $\text{g}=10{\text{m s}}^{-2}$
Initial velocity of the dumbbell, $\text{u}=0$
According to the third equation of motion,
${\text{v}}^2-{\text{u}}^2=2\text{gh}$
Final velocity of the dumbbell,

$\text{v}=\sqrt{{\text{u}}^2+\text{2gh}}$

$⟹\text{v}=\sqrt{0+2\times 10\times 80\times {10}^{-2}}$

$⟹\text{v}=\sqrt{16}=4{\text{m s}}^{-1}$

According to the law of conservation of momentum, the momentum acquired by the floor = the final momentum of the dumbbell
$⟹\text{mv}=10\times 4=40{\text{kg m s}}^{-1}$