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A 92 kg man standing on a surface of negligible friction kicks forward a 66 g stone lying at his feet, so that it acquires a velocity of +4.0 m/s. What velocity does the man acquire as a result? m/s

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Answer #1

let,


mass of the person, m1=92 kg and velocity is v1


mass of the stone, m2=66 kg and velocity is v2=4 m/sec

by using law of conservation of momentum,


P_initial =P_final

0=m1*V1+m2*v2


===>


92*v1+(66*10^-3)*4=0


===> v2=2.87*10^-3 m/sec

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