Question

How much momentum does 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, and take $g=10m{s}^{-2}$)?

• A. 40 kg m/s
• B. 50 kg m/s
• C. 30 kg m/s
• D. 35 kg m/s

Answer 1

The correct option is A

40 kg m/s

Given: 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, $u=0$

Let the final velocity of the dumbbell(when it is about to hit the floor) be $v$.
According to the third equation of motion:
$v^2-u^2=2as\Rightarrow v=\sqrt{u^2+(2\times a\times s)}$
$\Rightarrow v=\sqrt{0+(2\times 10\times 0.8)}=4m{s}^{-1}$

$\text{Momentum}=\text{mass}\times \text{velocity}$
$p=10kg\times 4m{s}^{-1}=40kgm{s}^{-1}$

Hence, the momentum with which the dumbbell hits the floor is
$40kgm{s}^{-1}$