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# 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? - CBSE Class 9 - Science

ConceptMathematical Formulation of Second Law of Motion

#### 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.

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#### APPEARS IN

NCERT Solution for Science Textbook for Class 9 (2018 to Current)
Chapter 9: Force and Laws of Motion
Q: 3 | Page no. 130
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.
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