Question

4. A hammer of mass 500 g, moving at 50 m 5-1, strikes a nail. The nail stops the hammer in a very short time of
0.01 s. What is the force of the nail on the hammer?
(Ans. 2500 N)​

Answer 1

Answer:

✔️ f = -2500 N

Explanation:

How Answer is 2500N⤵️

➪Mass of the hammer (m) = 500 g = 0.5 kg

➪The initial velocity of the hammer (u) = 50 m/s

➪Time is taken by the nail to the stop the hammer (t) = 0.01 s

➪The velocity of the hammer (v) = 0 (since the hammer finally comes to rest)

From Newton’s second law of motion

f = m (v – u) / t

Where f = force

f = 0.5 × (0 – 50) / 0.01

✔️f = -2500 N

-ve sign indicates the force of nail on the hammer moving in the opposite direction to that of hammer movement.

Answer 2

Given :-

Mass of the hammer = 500 g

Initial velocity of the hammer = 50 m/s

Time is taken by the nail = 0.01 sec

To Find :-

The force of the nail on the hammer.

Analysis :-

Here we are given with the mass, initial velocity and time taken by the nail.

Using the Newton’s second law of motion, find the force of the nail on the hammer by substituting the values accordingly.

Solution :-

We know that,

• f = Force
• u = Initial velocity
• m = Mass
• v = Final velocity
• t = Time

Using the formula,

$\underline{\boxed{\sf Second \ law \ of \ motion=f=m\dfrac{(v-u)}{t} }}$

Given that,

Mass (m) = 500 g = 0.5 kg

Final velocity (v) = 0 m/s

Initial velocity (u) = 50 m/s

Time (t) = 0.01 sec

Substituting their values,

⇒ f = 0.5 × (0 - 50) / 0.01

⇒ f = 0.5 × (-50) / 0.01

⇒ f = 0.5 × -5000

⇒ f = -2500 N

Therefore, the force of the nail on the hammer is -2500 N.