Question

How much momentum will a dumb-bell of mass $10\,{\text{kg}}$ transfer to the floor if it falls from a height of $80\,{\text{cm}}$ and does not rebound? Take its downward acceleration to be $10\,{\text{m}}{{\text{s}}^{ - 2}}$
A. $40\,{\text{kgm}}{{\text{s}}^{ - 1}}$
B. $20\,{\text{kgm}}{{\text{s}}^{ - 1}}$
C. $80\,{\text{kgm}}{{\text{s}}^{ - 1}}$
D. $120\,{\text{kgm}}{{\text{s}}^{ - 1}}$

Answer 1

First of all, we will find the final velocity using the laws of motion. Then we will find the value of momentum using the expression for momentum. We will substitute the values and manipulate it accordingly.

Stepwise solution:

In the given problem,
Mass of the dumb-bell is $10\,{\text{kg}}$ .
The distance covered by the dumb-bell is $80\,{\text{cm}}$ .
Acceleration due to gravity is $10\,{\text{m}}{{\text{s}}^{ - 2}}$ .

Since, the dumb-bell is dropped from a height of $80\,{\text{cm}}$ , so the initial velocity is $0\,{\text{m}}{{\text{s}}^{ - 1}}$ . The object moves from the position of rest.
We know,
$1\,{\text{cm}} = {10^{ - 2}}\,{\text{m}}$
So,
80\,{\text{cm}} = 80 \times {10^{ - 2}}\,{\text{m}} \\\
80\,{\text{cm}} = 0.80\,{\text{m}} \\\
We need to calculate the final velocity with which it hits the ground.
To calculate the final velocity, we will use the third law of motion:
${v^2} = {u^2} + 2as$ …… (1)
Where,
$v$ indicates final velocity.
$u$ indicates initial velocity.
$a$ indicates acceleration due to gravity.
$s$ indicates the distance covered by the dumb-bell.

Substituting the required values in the equation (1), we get:
{v^2} = {u^2} + 2as \\\
${v^2} = {0^2} + 2 \times 10 \times 0.80 \\\$
{v^2} = 16 \\\
v = 4\,{\text{m}}{{\text{s}}^{ - 1}} \\\
The final velocity with which the dumb-bell hits the ground is $4\,{\text{m}}{{\text{s}}^{ - 1}}$ .

We know, momentum is the product of the mass of the body and the velocity.
To calculate the momentum, we have a formula:
$p = mv$ …… (2)
Where,
$p$ indicates the momentum.
$m$ indicates mass of the dumb-bell.
$v$ indicates velocity of the dumb-bell with which it is falling.

Substituting, the required values in the equation (2), we get:
p = 10\,{\text{kg}} \times 4\,{\text{m}}{{\text{s}}^{ - 1}} \\\
p = 40\,{\text{kgm}}{{\text{s}}^{ - 1}} \\\
Hence, the dumb-bell transfer of momentum of $40\,{\text{kgm}}{{\text{s}}^{ - 1}}$ to the ground.
The correct option is A.

Note: In this problem, we are required the momentum at the final stage, when it hits the ground. For this, take the initial velocity as zero, as it is dropped from a height. Any object when it is dropped from a height, falls under the action of gravity.