Class 9
Chapter 9 Class 9 - Force and Laws Of Motion




Additional Question 3 A hammer of mass 500 g, moving at 50 m sβˆ’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? Given Mass of hammer = m = 500 g = 500/1000 kg = 0.5 kg Initial velocity of the hammer = u = 50 m/s Time taken as 0.03 instead of 0.01 Since the hammer comes to rest, Final velocity of hammer = v = 0 m/s We know that Force = (π‘š (𝑣 βˆ’ 𝑒))/𝑑 = (0.5 (0 βˆ’ 50))/0.01 = (0.5 Γ— βˆ’50)/0.03 = (5/10 Γ— βˆ’50)/(3/100) = (5 Γ— βˆ’50 Γ— 100)/(10 Γ— 3) = βˆ’2500 N This is the Force exerted by Hammer on the nail ∴ Hammer strikes the nail with a force of βˆ’2500 N. By Newton’s third law of motion, nail exerts an equal force in opposite direction on the hammer ∴ Force exerted by nail on hammer = 2500 N

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Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 14 years and a teacher from the past 18 years. He teaches Science, Economics, Accounting and English at Teachoo