A 30-kg child stands at one end of a floating 20-kg canoe that is 5.0-m long and initially at rest in the water. The child then slowly walks to the other end of the canoe. How far does the canoe move in the water, assuming water friction is negligible?
To solve this problem, we can apply the conservation of momentum principle. The total momentum before and after the child walks to the other end of the canoe should be equal.
Before the child walks, the canoe and the child are at rest. Thus, the total momentum is zero.
After the child walks to the other end, they will both be in motion. Let's assume that the child moves with velocity v relative to the canoe.
The canoe's velocity relative to the child can be found using the conservation of momentum equation:
(Mass of child × Velocity of child) + (Mass of canoe × Velocity of canoe) = 0
(30 kg × v) + (20 kg × Vc) = 0
Rearranging the equation, we get:
20 kg × Vc = -30 kg × v
Vc = (-30 kg × v) / 20 kg
Vc = -3/2 v
Since the child walked from one end of the canoe to the other, the relative velocity of the canoe to the water will be twice the velocity of the child:
Vw = 2v
The distance the canoe moved in the water can be found using the formula:
Distance = Velocity × Time
Since the canoe and the water initially are at rest, the time it takes for the child to walk from one end to the other will be equal to the time it takes for the canoe to move a distance d in the water:
d = Vw × t
Now to find the relation between velocity, time, and distance, we need to use the kinematic equation:
d = (1/2)at^2 + v0t
Since the canoe is initially at rest, v0 will be equal to 0:
d = (1/2)at^2 + 0
d = (1/2)at^2
Rearranging the equation, we get:
2d = at^2
d = (a/2)t^2
Now let's find the acceleration a:
Using Newton's second law:
F = ma
The only external force acting on the system is the force exerted by the child on the canoe. According to Newton's third law, the reaction force exerted by the canoe on the child will be equal in magnitude but opposite in direction. Thus, their net force is zero, and the acceleration of the system is zero.
Therefore, a = 0.
Plugging this value into the equation:
d = (0/2)t^2
d = 0
Therefore, the canoe does not move in the water when the child walks from one end to the other, assuming water friction is negligible.
To determine how far the canoe moves in the water, we need to apply the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum of a system remains constant if no external forces act on it. In this case, the system consists of the child and the canoe.
Initially, both the child and the canoe are at rest, so the momentum of the system is zero. When the child begins to walk to the other end of the canoe, an internal force is applied within the system. As a result, the system's center of mass remains in the same position, but the canoe and child move in opposite directions to conserve momentum.
According to the principle of conservation of momentum, the momentum of the child and the canoe is equal in magnitude and opposite in direction. Therefore, we can write:
(mass of child) × (velocity of child) = (mass of canoe) × (velocity of canoe)
Given:
Mass of child (m1) = 30 kg
Mass of canoe (m2) = 20 kg
Length of the canoe (L) = 5.0 m
Initially, the child stands at one end of the canoe, so the initial center of mass coincides with the child's position.
When the child walks to the other end of the canoe, the center of mass must remain in the same position. Since the child has a smaller mass compared to the canoe, their velocities will be different but scaled accordingly to conserve momentum.
Let x be the distance the canoe moves in the water when the child reaches the other end.
Using the principle of conservation of momentum, we can set up the following equation:
(m1 × vf1) = (m2 × vf2)
Here, vf1 is the final velocity of the child and vf2 is the final velocity of the canoe.
To determine the velocities, we need to use the concept of conservation of center of mass.
The initial center of mass position of the system is located at the child's initial position. When the child moves to the other end of the canoe, the center of mass position must remain the same.
The center of mass (cm) of the system can be determined using the following formula:
cm = (m1 × x1 + m2 × x2) / (m1 + m2)
Here, x1 represents the initial distance of the child from the center of mass, and x2 represents the initial distance of the canoe from the center of mass.
Since the child starts at one end of the canoe, x1 = L/2 (half the length of the canoe) and x2 = -L/2 (negative half the length of the canoe).
The final center of mass position should be at the child's new location at the other end of the canoe, so x1 = -L/2 and x2 = L/2.
Using the formula for the center of mass, we can write:
0 = (m1 × L/2 + m2 × (-L/2)) / (m1 + m2)
Simplifying this equation gives:
0 = (30 kg × L/2 - 20 kg × L/2) / (30 kg + 20 kg)
0 = (10 kg × L) / 50 kg
Since there are no external forces or friction acting on the system, the center of mass remains in the same position, and the distance it moves must also be zero. Therefore:
0 = L
Thus, the canoe does not move in the water.
Therefore, the distance the canoe moves in the water is 0 meters.
the center of mass remains at the same place.
As I do this in my head, cm at start measureing from the center of the canoe, (20*0+30*2.5)/50=7/5 m from the center of the boat. So it shifts then to that far on the other side, so the boat moved 14/5 meters relative to the water.
Thanks,.following your work(20*0+30*2.5)/50
I got 1.5 then I doubled and got 3m.
Answer choices we're:
1m 2m 3m 4m 5m