#### Question

A hammer of mass 500 g, moving at 50 m s^{−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?

#### Solution

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=(m(v-u))/t=(0.5(0-50))/0.01=-2500N`

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.

Is there an error in this question or solution?

Solution A hammer of mass 500 g, moving at 50 m s^(−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? Concept: Mathematical Formulation of Second Law of Motion.