Question

a man weighing 740 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 that of the woman's is approximately 0.676, which is calculated by taking the inverse proportion of their masses since they both have the same momentum.

Step-by-step explanation:

The question asks about the ratio of kinetic energies of a man and a woman who have the same momentum. Since kinetic energy is given by the formula K = 1/2m·v² and momentum is p = m·v, we can express the kinetic energy in terms of momentum and mass as K = p²/2m. If the man and woman have the same momentum (p), the ratio of their kinetic energies will be inversely proportional to their masses.

Given the weights (which can be used to find mass since weight = mass × gravity), we can say the man's mass is 740 N/g and the woman's mass is 500 N/g, where g is the acceleration due to gravity (9.81 m/s²). Therefore, the ratio of the man's kinetic energy (Km) to the woman's kinetic energy (Kw) is:

Km/Kw = (woman's mass / man's mass) = 500/740 = 5/7.4 or approximately 0.676.

Answer 2

The ratio of the man's kinetic energy
`(\(K_m\))` to that of the woman's
`(\(K_w\))` is approximately 0.676. This means the man's kinetic energy is about 67.6% of the woman's kinetic energy.

To find the ratio of the man's kinetic energy
`(\(K_m\))` to that of the woman's
`(\(K_w\)),` we need to understand the relationship between momentum, kinetic energy, and mass. Momentum
`(\(p\))` and kinetic energy
`(\(K\))` are given by:

`\[ p = mv \]`

`\[ K = (1)/(2)mv^2 \]`

where
`\( m \)` is mass and
`\( v \)` is velocity.

Given that the man and woman have the same momentum, their momenta can be equated:

`\[ p_m = p_w \]`

`\[ m_m v_m = m_w v_w \]`

From the given weights (which are equivalent to gravitational force), we can find their masses. Weight \( W \) is given by \( W = mg \), where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, m/s^2 \)).

1. Calculate Masses :

`\[ m_m = (W_m)/(g) = (740 \, N)/(9.8 \, m/s^2) \]`

`\[ m_w = (W_w)/(g) = (500 \, N)/(9.8 \, m/s^2) \]`

2. Substitute Masses in the Momentum Equation :

`\[ (740)/(9.8) v_m = (500)/(9.8) v_w \]`

3. Solve for Ratio of Velocities:

`\[ (v_m)/(v_w) = (500)/(740) \]`

4. Calculate Ratio of Kinetic Energies :

`\[ (K_m)/(K_w) = ((1)/(2) m_m v_m^2)/((1)/(2) m_w v_w^2) \]`

`\[ = (m_m)/(m_w) * \left( (v_m)/(v_w) \right)^2 \]`

`\[ = (740/9.8)/(500/9.8) * \left( (500)/(740) \right)^2 \]`

`\[ = (740)/(500) * \left( (500)/(740) \right)^2 \]`

5. Calculate the Final Ratio :

`\[ = (740)/(500) * \left( (500^2)/(740^2) \right) \]`

`\[ = (740 * 500^2)/(500 * 740^2) \]`

Let's compute the final ratio.

The ratio of the man's kinetic energy
`(\(K_m\))` to that of the woman's
`(\(K_w\))` is approximately 0.676. This means the man's kinetic energy is about 67.6% of the woman's kinetic energy.