A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 25 m/s in 0.05 s. (a) What acceleration is imparted to the paper? (b) If the mass of the paper is 0.003 kg, what force does the boxer exert on it? (c) How much force does the paper exert on the boxer?
Answer : Given: Initial velocity u= 0 Final velocity v= 25 m/s Mass of the paper m= .003 kg, Time = 0.05 s. Calculate : Force exerted by the boxer on the paper(F )= b Acceleration of the paper( a )= p What is the force exerted by the paper on the boxer hand= 1) Calculating the force exerted by the boxer on the paper can be given by formula Force = mass x acceleration , So We know mass of the paper(m) is Given . We have to calculate acceleration Acceleration of the paper= final velocity - Initial velocity / time = 25 m/s - 0 m/s / 0.05 s ap = 500 m/s 2 2) Calculating the force exerted by the boxer, we can write as, Force exerted by the boxer (F ) = mab of the paper x acceleration of the paper = .003 kg, x 500 m/s 2 = 1.5 N 3) Calculating the force exerted by the paper on the boxer hand: According to newton’s third law: Every action has an equal and opposite reaction Force exerted by the paper on the boxer hand = Force exerted by the boxer on the paper Fp = Fb Therefore the force exerted by the paper on the boxer hand ( F ) = 1.5 N p
