A 58 kg woman is ice-skating toward the east on a frictionless frozen lake when she collides with a 94 kg man who is ice-skating toward the west. The maximum force exerted on the woman by the man during the collision is 210 N, west. Take east to be the positive direction of the x axis. (a) What is the maximum force on the man from the woman, including direction? (b) What is the maximum acceleration of the woman, including direction? (c) What is the maximum acceleration of the man, including direction?
the women exert the same force on the man but in the opposite direction so if the women's net force exerted by man=-210(west) so the man's net force exerted by woman(a)=210 N (east)
Fnet=ma
210=94a,acceleration of the man (c)a=210/94
acceleration of the woman (b)f=ma
-210=54a
a=-210/54 the acceleration should be negative because the woman is moving west
To answer these questions, we can use the principles of conservation of momentum and Newton's second law.
(a) To determine the maximum force on the man from the woman, we need to find the change in momentum of the woman and divide it by the time of collision. According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
Let's denote the velocity of the woman as Vw, and the velocity of the man as Vm. Since the woman is moving toward the east, and the man is moving toward the west, their velocities will have opposite signs.
Conservation of momentum equation:
(mass of woman x velocity of woman) + (mass of man x velocity of man) = (mass of woman x final velocity of woman) + (mass of man x final velocity of man)
(58 kg x Vw) + (94 kg x (-Vm)) = (58 kg x final velocity of woman) + (94 kg x final velocity of man)
As the collision is elastic and the friction between them is negligible, both the woman and the man stick together and move with the same final velocity.
58 kg x Vw - 94 kg x Vm = (58 kg + 94 kg) x final velocity
From this equation, we can find the magnitude of the final velocity by solving for it:
(58 kg x Vw - 94 kg x Vm) / (58 kg + 94 kg) = final velocity
Now that we have the final velocity, we can calculate the maximum force on the man from the woman using Newton's second law:
Force on the man = mass of man x acceleration of man
So,
Force on the man = 94 kg x acceleration of the man
(b) To find the maximum acceleration of the woman, we can use the equation:
Force on the woman = mass of woman x acceleration of woman
So,
Force on the woman = 58 kg x acceleration of the woman
(c) As the collision is elastic, the maximum acceleration of the man and woman will be the same.
Now, let's calculate these values step by step:
Step 1: Calculate the final velocity of the woman and man using the conservation of momentum equation.
Step 2: Calculate the force on the man using the mass of the man and acceleration of the man.
Step 3: Calculate the force on the woman using the mass of the woman and acceleration of the woman.
Step 4: Determine the maximum acceleration of both the woman and the man, including the direction.
Let's start with step 1:
To solve this problem, we can apply the principles of conservation of momentum and Newton's third law of motion.
(a) According to Newton's third law of motion, the force exerted by the woman on the man will be equal in magnitude but opposite in direction to the force exerted by the man on the woman. Therefore, the maximum force on the man from the woman will also be 210 N, but in the opposite direction, i.e., 210 N, east.
(b) To find the maximum acceleration of the woman, we need to use the equation F = ma, where F is the force and a is the acceleration. Rearranging the equation, we have a = F/m.
Given that the force exerted on the woman by the man is 210 N, west, and the mass of the woman is 58 kg, we can substitute these values into the equation to find the acceleration of the woman.
a = 210 N / 58 kg = -3.62 m/s^2
Since we are taking east as the positive direction, the acceleration of the woman will be negative, indicating that she is decelerating in the positive direction.
Therefore, the maximum acceleration of the woman is 3.62 m/s^2, east.
(c) Similarly, to find the maximum acceleration of the man, we can use the equation a = F/m. Since the maximum force on the man from the woman is equal in magnitude (210 N) but opposite in direction, we will substitute positive 210 N for the force.
a = 210 N / 94 kg = 2.23 m/s^2
The maximum acceleration of the man is 2.23 m/s^2, west.
To summarize:
(a) The maximum force on the man from the woman is 210 N, east.
(b) The maximum acceleration of the woman is 3.62 m/s^2, east.
(c) The maximum acceleration of the man is 2.23 m/s^2, west.