A 60 kg woman and an 85 kg man stand 9.0 m apart on frictionless ice.
(a) How far from the woman is their CM?
(b) If each holds one end of a rope, and the man pulls on the rope so that he moves 1.1 m, how far from the woman will he be now?
(c) How far will the man have moved when he collides with the woman?
To solve these problems, we can use the concept of center of mass (CM) and apply the principle of conservation of momentum.
(a) To find the position of the center of mass, we need to consider the masses and distances of the two individuals. Let's assume the position of the woman is at x = 0 and the man is at x = 9.0 m.
The formula for calculating the center of mass is:
CM = (m1 * x1 + m2 * x2) / (m1 + m2)
Here, m1 represents the mass of the woman, x1 represents her position, m2 represents the mass of the man, and x2 represents his position.
Given:
Mass of the woman (m1) = 60 kg
Position of the woman (x1) = 0 m
Mass of the man (m2) = 85 kg
Position of the man (x2) = 9.0 m
Now we can substitute these values into the formula to find the position of the center of mass:
CM = (60 kg * 0 m + 85 kg * 9.0 m) / (60 kg + 85 kg)
= (765 kg*m) / 145 kg
≈ 5.28 m
Therefore, the center of mass is located approximately 5.28 m from the woman.
(b) When the man moves 1.1 m, we need to calculate how the position of the center of mass changes. Since the system is frictionless, the total momentum of the system remains constant.
We can use the equation:
P_initial = P_final
where P_initial represents the initial momentum and P_final represents the final momentum.
The initial momentum is given by:
P_initial = m1 * v1_initial + m2 * v2_initial
Since the individuals were stationary initially (v1_initial = 0 m/s and v2_initial = 0 m/s), the initial momentum simplifies to:
P_initial = 0 kg * m/s
The final momentum is given by:
P_final = m1 * v1_final + m2 * v2_final
The woman remains stationary, so her velocity remains 0 m/s (v1_final = 0 m/s). The man moves 1.1 m, so his final velocity is given by:
v2_final = (x2_initial - x2_final) / t
= (9.0 m - 1.1 m) / t
= 7.9 m / t
where t represents the time taken.
Now we can set the initial and final momenta equal to each other and solve for t:
0 = 0 kg*m + 85 kg * (7.9 m / t)
0 = 0 kg*m + 85 kg * 7.9 m / t
0 = 664.5 kg*m^2 / t
Since t ≠ 0, we can divide both sides by t to eliminate it:
0 = 664.5 kg*m^2
This equation is always true, so the time taken does not affect the final momentum. Therefore, we can determine that the final momentum is also 0 kg*m/s.
Now we have the final positions of the woman and the man from the center of mass:
x1_final = x1_initial
x2_final = x2_initial + 1.1 m
Substituting the values:
x1_final = 0 m
x2_final = 9.0 m + 1.1 m
= 10.1 m
The new position of the center of mass (CM_final) is then:
CM_final = (m1 * x1_final + m2 * x2_final) / (m1 + m2)
= (60 kg * 0 m + 85 kg * 10.1 m) / (60 kg + 85 kg)
= 853.5 kg*m / 145 kg
≈ 5.88 m
Therefore, the man will be approximately 5.88 m from the woman after moving 1.1 m.
(c) To determine the distance the man will have moved when he collides with the woman, we need to find the distance between their initial positions and the center of mass. Then we can calculate the distance traveled by the man until the center of mass is at the woman's position.
The distance of the woman from the center of mass is:
x1_cm = CM - x1_initial
= 5.28 m - 0 m
= 5.28 m
Similarly, the distance of the man from the center of mass is:
x2_cm = x2_initial - CM
= 9.0 m - 5.28 m
= 3.72 m
When the man collides with the woman, the center of mass will move to the woman's position. Therefore, the distance traveled by the man until the collision occurs is equal to x2_cm:
Distance traveled by the man = x2_cm
= 3.72 m
Therefore, the man will have moved approximately 3.72 m when he collides with the woman.