Question

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

Answer 1

Answer:

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

Explanation:

Given;

weight of the man, W = 700 N

Weight of the woman, W = 440 N

momentum is given by;

$P = mv\\\\v = \frac{P}{m}$

Kinetic energy of the man;

$K_m = \frac{1}{2}m_m(\frac{P_m}{m_m})^2 \\\\K_m = \frac{P_m^2}{2m_m}$

Momentum of the man is calculated as;

$P_m^2 = 2m_mK_m$

The kinetic energy of the woman is given by;

$K_w = \frac{P_w^2}{2m_w}$

The momentum of the woman is given;

$P_w^2 = 2m_wK_w$

Since, momentum of the man = momentum of the woman

$P_m^2 = P_w^2$

$2m_mK_m = 2m_wK_w\\\\\frac{K_m}{K_w} = \frac{2m_w}{2m_m}\\\\\frac{K_m}{K_w} = \frac{m_w}{m_m}$

mass of the mas = 700 / 9.8 = 71.429

mass of the woman is = 440 / 9.8 = 44.898

$\frac{K_m}{K_w} = \frac{44.898}{71.429}\\\\\frac{K_m}{K_w} =0.629$

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