Question

How much momentum will a dùmbbell of 10 kg transfer to the floor, if it falls from a height of 80 cm? Take its downward accleration to be 10 ms$^{-2}$​

Answer 1

Answer:

Given:-

Mass of the body (m) = 10 kg

Height (H/S) = 80 cm = 0.8 m

Acceleration (a) = 10 ms$^{-2}$

Initial velocity (u) = 0

Final velocity (v) = ?

To find:-

Momentum of the dùmbbell before hitting ground.

Solution:-

First we need to find the final velocity in order to find the momentum of the dùmbbell.

By third law of motion

v$^{2}$ - u$^{2}$ = 2 as

$\Longrightarrow$ $v^{2}$ - (0)$^{2}$ = 2 × 10 × 0.8

$\Longrightarrow$ v$^{2}$ = 16

$\Longrightarrow \: v = \sqrt{16} \\ \\ = 4 \: ms {}^{ - 1}.$

Hence, momentum of dùmbbell before hitting ground = mv

(p) = mv

$= 10 \: kg \: \times 4 \: ms {}^{ - 1} \\ \\ = 40 \: kg \: ms {}^{ - 1}.$

Answer 2

Given :

• Mass of ďumbbell (m) = 10 kg
• Height (h) = 80 cm = 0.8 m
• Acceleration (a) = 10 m/s²
• Initial velocity (u) = 0 m/s

To find :

• Momentum (p)

According to the question,

In this question we will use Newton's third equation of motion,

↦ v² = u² + 2as

Where,

• v stands for Final velocity
• u stand for Initial velocity
• s stands for Distance
• a stands for Acceleration

↦ Substituting the values,

↦ v² = (0)² + 2 × 10 × 0.8

↦ v² = 0 + 16

↦ v² = 16

↦ v = √16

↦ v = 4

So,the velocity is 4 m/s..

Now,

↦ Momentum = Mass × Velocity

Or,

↦ p = mv

↦ p = 10 kg × 4 m/s

↦ p = 40 kg m/s

So,the momentum is 40 kg m/s...

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