Explanation:
It is given that,
Weight of the man, [tex]F_m=750\ N[/tex]
Weight of the woman, [tex]F_w=440\ N[/tex]
Both man and woman have same momentum as :
[tex]p_m=p_w[/tex]
[tex]m_mv_m=m_wv_w[/tex]
[tex]\dfrac{v_m}{v_w}=\dfrac{F_w/g}{F_m/g}=\dfrac{F_w}{F_m}[/tex]...........(1)
Ratio of man's kinetic energy to that of woman is :
[tex]R=\dfrac{E_m}{E_w}[/tex]
[tex]R=\dfrac{(1/2)m_mv_m^2}{(1/2)m_wv_w^2}[/tex](from equation (1))
[tex]R=\dfrac{F_m}{F_w}\times (\dfrac{v_m}{v_w})^2[/tex]
[tex]R=\dfrac{750}{440}\times (\dfrac{440}{750})^2[/tex]
R = 0.58
So, the ratio of the man's kinetic energy to that of the woman is 0.58. Hence, this is the required solution.
The ratio of the man's kinetic energy Km to that of the woman Kw is 0.59.
The mass of the man can be determined by using Newton's second law of motion.
W = mg
m = W/g
Mass of the man = 750/9.8 = 76.53 kg
Mass of the woman = 440/9.8 = 44.9 kg
Momentum, P = mv
76.53v₁ = 44.9v₂
v₂ = 76.53v₁ /44.9
v₂ = 1.7v₁
[tex]K_m = \frac{1}{2} mv_1^2= \frac{1}{2} \times 76.53 \times v_1^2 = 38.27v_1^2\\\\K_w = \frac{1}{2} mv_2^2= \frac{1}{2} \times 44.9 \times(1.7 v_1)^2 = 64.88v_1^2\\\\K_m : K_w = 38.27v_1^2 : 64.88v_1^2 = 0.59[/tex]
Thus, the ratio of the man's kinetic energy Km to that of the woman Kw is 0.59.
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