Question

A man weighing 750 n and a woman weighing 500 n have the same momentum. what is the ratio of the man's kinetic energy km to that of the woman kw?

Answer 1

The ratio of the man's kinetic energy to the woman's when they have the same momentum is 2/3. This is calculated using the fact that kinetic energy is inversely proportional to mass for objects with the same momentum.

Step-by-step explanation:

The question asks for the ratio of the kinetic energies of a man and a woman who have the same momentum. To solve this, we can use the relationship between kinetic energy (K) and momentum (p). The kinetic energy can be expressed as K = p²/(2m), where m is mass and p is momentum.

Since both individuals have the same momentum, their momenta are equal, and we denote this as p. The man's mass (m1) can be found by dividing his weight by gravitational acceleration (g), that is, m1 = 750 N/g. Similarly, for the woman, m2 = 500 N/g. Consequently, the kinetic energy ratio (Km/Kw) can be found by plugging these masses into the kinetic energy formula and simplifying.

Thus, Km/Kw will be:

Km = p²/(2×750N/g)

Kw = p²/(2×500N/g)

By dividing Km by Kw, we find the kinetic energy ratio Km/Kw = (500/750) = 2/3. This means the man's kinetic energy is two-thirds that of the woman's when they have equal momentum.

Answer 2

Because weight W = M g, the ratio of weights equals the ratio of masses.

(M_m g)/ (M_w g) = [ (p^2 Man )/ (2 K_man)] / [ (p^2 Woman )/ (2 K_woman)

but p's are equal, so

K_m/K_m = (M_w g)/(M_m g) = W_woman / W_man = 450/680 = 0.662