Asked by nan
A father racing his son has 1/3 the kinetic energy of the son, who has 1/3 the mass of the father. The father speeds up by 1.4 m/s and then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son?
Answers
Answered by
drwls
M = father's mass
m = son's mass = M/3
V = father's initial speed
v = son's initial speed
(1/2)MV^2 = (1/3)*(1/2)*m v^2
M*V^2 = (1/3)(M/3)v^2
V^2/v^2 = 1/9
V = v/3
Second equation:
(1/2)M*(V + 1.4)^2 = (1/2)m*v^2
= (1/2)*(M/3)*(3V)^2
cancel out the M's and (1/2)'s
(V + 1.4)^2 = 3V^2
V^2 + 2.8V + 1.96 = 3V^2
V^2 -1.4V -0.98 = 0
Solve for V; take the positive root.
m = son's mass = M/3
V = father's initial speed
v = son's initial speed
(1/2)MV^2 = (1/3)*(1/2)*m v^2
M*V^2 = (1/3)(M/3)v^2
V^2/v^2 = 1/9
V = v/3
Second equation:
(1/2)M*(V + 1.4)^2 = (1/2)m*v^2
= (1/2)*(M/3)*(3V)^2
cancel out the M's and (1/2)'s
(V + 1.4)^2 = 3V^2
V^2 + 2.8V + 1.96 = 3V^2
V^2 -1.4V -0.98 = 0
Solve for V; take the positive root.
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