Which one of the following is the quantity of transfer of linear momentum to the floor, when a dumbbell of mass 500 g falls from a height of 5 m and stops after hitting the floor (take $g=10 \text{ m/s}^2$)?
0·5 kg-m/s
5·0 kg-m/s
10·0 kg-m/s
1·0 kg-m/s
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC Geoscientist – 2023
First, calculate the velocity of the dumbbell just before hitting the floor. Using the equation $v^2 = u^2 + 2gh$, where $u=0$ (starts from rest), $g=10 \text{ m/s}^2$, and $h=5 \text{ m}$.
$v^2 = 0^2 + 2 \times 10 \times 5 = 100 \text{ m}^2/\text{s}^2$.
$v = \sqrt{100} = 10 \text{ m/s}$. The direction is downwards.
The mass of the dumbbell is $m = 500 \text{ g} = 0.5 \text{ kg}$.
The momentum of the dumbbell just before impact is $p_{initial} = m \times v = 0.5 \text{ kg} \times 10 \text{ m/s} = 5 \text{ kg-m/s}$ (downwards).
After hitting the floor, the dumbbell stops, so its final velocity is 0.
The momentum of the dumbbell after impact is $p_{final} = m \times 0 = 0$.
The change in momentum of the dumbbell is $\Delta p_{dumbbell} = p_{final} – p_{initial} = 0 – (5 \text{ kg-m/s downwards}) = -5 \text{ kg-m/s downwards} = 5 \text{ kg-m/s upwards}$.
By the impulse-momentum theorem and Newton’s third law, the momentum transferred to the floor is equal in magnitude and opposite in direction to the change in momentum of the dumbbell.
Momentum transfer to floor = $-\Delta p_{dumbbell} = -(5 \text{ kg-m/s upwards}) = 5 \text{ kg-m/s downwards}$.
The question asks for the *quantity* of transfer, which is the magnitude. The magnitude is 5 kg-m/s.