The man speeds up by 1.60 m/s and then has the same kinetic energy as the boy. What is the speed (in meters/second) of the boy?
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Let m be the mass of the boy. then 2m is the mass of the man
man k energy= 0.5*2m*V^2
boy k energy=0.5*m*v^2
therefor 0.5*m*v^2=2*0.5*2m*V^2
v^2=4V^2
V=v/2
now man speed is V+1.60 and
0.5mv^2=0.5*2m*(v/2 +1.60)^2
v^2=2*((v+3.20)/2)^2
2v^2=v^2+6.40v+10.24
v^2-6.40v-10.24=0
v=7.725 m/s
man k energy= 0.5*2m*V^2
boy k energy=0.5*m*v^2
therefor 0.5*m*v^2=2*0.5*2m*V^2
v^2=4V^2
V=v/2
now man speed is V+1.60 and
0.5mv^2=0.5*2m*(v/2 +1.60)^2
v^2=2*((v+3.20)/2)^2
2v^2=v^2+6.40v+10.24
v^2-6.40v-10.24=0
v=7.725 m/s