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A man weighing 700 NN and a woman weighing 440 NN have the same momentum. What is the ratio of the man's kinetic energy KmKmK_m to that of the woman K

User Ruwen
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1 Answer

2 votes

Answer:

The ratio of the man's kinetic energy to that of the woman's kinetic energy is 0.629.

Step-by-step explanation:

Given;

weight of the man, W = 700 N

Weight of the woman, W = 440 N

momentum is given by;


P = mv\\\\v = (P)/(m)

Kinetic energy of the man;


K_m = (1)/(2)m_m((P_m)/(m_m))^2 \\\\K_m = (P_m^2)/(2m_m)

Momentum of the man is calculated as;


P_m^2 = 2m_mK_m

The kinetic energy of the woman is given by;


K_w = (P_w^2)/(2m_w)

The momentum of the woman is given;


P_w^2 = 2m_wK_w

Since, momentum of the man = momentum of the woman


P_m^2 = P_w^2


2m_mK_m = 2m_wK_w\\\\(K_m)/(K_w) = (2m_w)/(2m_m)\\\\(K_m)/(K_w) = (m_w)/(m_m)

mass of the mas = 700 / 9.8 = 71.429

mass of the woman is = 440 / 9.8 = 44.898


(K_m)/(K_w) = (44.898)/(71.429)\\\\(K_m)/(K_w) =0.629

Therefore, the ratio of the man's kinetic energy to that of the woman's kinetic energy is 0.629.

User Robin Whittleton
by
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