Question

A hammer weighing 100N falls freely on a nail from a height of 1.25m. Calculate the force of the blow of the impact last for 10-2 sec.

Answer 1

Given:

A hammer weighing 100N falls freely on a nail from a height of 1.25m.

To find:

Force of impact ?

Calculation:

First of all, we need to calculate the change in momentum of the hammer upon striking the nail:

Mass of hammer = 100/g = 10 kg.

Velocity of hammer just before striking the nail:

$\therefore \: {v}^{2} = {u}^{2} + 2gh$

$\implies \: {v}^{2} = {0}^{2} + (2 \times 10 \times 1.25)$

$\implies \: {v}^{2} = 25$

$\implies \: v = 5 \: m {s}^{ - 1}$

Now, let's assume that the nail stopped after striking :

So, change in momentum :

$\therefore \: \Delta P = 10(v - u)$

$\implies \: \Delta P = 10(0 - 5)$

$\implies \: \Delta P = - 50 \: kgm {s}^{ - 1}$

So, momentum transferred to nail :

$\implies \: \Delta P_{nail} = 50 \: kgm {s}^{ - 1}$

So, force experienced by nail:

$\implies \: force = (\Delta P_{nail} ) \times t$

$\implies \: force =50 \times {10}^{ - 2}$

$\implies \: force =0.5 \: N$

So, force experienced by nail is 0.5 N.

Answer 2

answer in pic...

Hope it will helps you.