Question

A hammer of mass 500 g moving at 50 m/s,strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is Force of nail on the hammer. ​

Answer 1

Given :

• Mass(m) = 500 g = $\sf\dfrac{500}{1000}$ kg = 0.5 kg

• Initial Velocity (u) = 50 m/s

• Final Velocity (v) = 0 m/s

• Time (t) = 0.01 seconds

Let's calculate the acceleration first :

$\dag \sf \: From \: 1st \: equation \: of \: motion$

★ v = u + at

→ 0 = 50 + a × 0.01

→ 0 = 50 + 0.01a

→ 0 - 50 = 0.01 a

→ -50 = 0.01 a

→ a = $\sf\dfrac{-50}{0.01}$

→ a -5000 m/s²

Now let's calculate the force acquired

$\dag \sf \: From \: 2nd \: law \: of \: motion$

★ Force = Mass × Acceleration

→ F = 0.5 × ( -5000) N

→ F = -2500 N

$\therefore$ Force of the nail on the hanmer is 2500 Newtons. The negative sign indicates that this force is in opposing motion.

Answer 2

Answer:

$\boxed{\huge{Force \ of \ nail \ on \ the \ hammer=2500 \ N}}$

Explanation:

Mass of the hammer:

m = 500 g = 0.5 kg

Initial velocity of the hammer:

u = 50 m/s

Time taken by the nail to the stop the hammer:

t = 0.01 s

Velocity of the hammer:

v = 0 (since the hammer finally comes to rest)

From Newton’s second law of motion:

Force;

$f =\frac{Change \ in \ momentum}{time} \\\\ = \frac{m(v-u)}{ t}\\\\=\frac{ 0.5(0-50)}{0.01} \\\\=\frac{-25}{0.01} \\\\=-2500 N$

The hammer strikes the nail with a force of -2500 N.

Hence, from Newton’s third law of motion, the force of the nail on the hammer is equal and opposite, i.e., +2500 N.