10 people,each of mass 70 kg, stand on a boat with a mass of 1000 kg. they jump off one end of the boat with velocity 1 m/s relative to the boat. the boat rolls in the opposite direction without friction.
a what is the final velocity of the boat if all the people jump at the same time?
b. what is the final velocity of the boat if the people jump off one at a time?
I don't get it. Why should the boat roll? In what direction are the people jumping?
To solve this problem, we can apply the law of conservation of momentum.
a) When all the people jump at the same time, the total initial momentum of the system is zero (since the initial velocity of the boat is zero), and according to the law of conservation of momentum, the total final momentum of the system will also be zero.
The momentum of an object is given by the product of its mass and velocity. Initially, the momentum of the 10 people is given by:
momentum_initial = (mass of one person) * (velocity of one person) * (number of people)
We can calculate this as:
momentum_initial = (70 kg) * (1 m/s) * (10) = 700 kg m/s
Since the total final momentum of the system is zero, the final momentum of the boat (opposite in direction to the momentum of the people) should also be 700 kg m/s.
The final velocity of the boat can be calculated using the equation of momentum:
momentum_final = (mass of the boat) * (final velocity of the boat)
Solving for the final velocity of the boat:
momentum_final = (1000 kg) * (final velocity of the boat)
Since momentum_final = 700 kg m/s and the mass of the boat is 1000 kg, we have:
700 kg m/s = (1000 kg) * (final velocity of the boat)
Simplifying:
final velocity of the boat = 0.7 m/s
Therefore, the final velocity of the boat, when all the people jump at the same time, is 0.7 m/s.
b) When the people jump off one at a time, the momentum of the boat will change after each person jumps.
Assuming all the people jump with the same velocity of 1 m/s relative to the boat, the momentum they transfer to the boat will be equal to their own momenta.
The momentum transferred to the boat by each person will be:
momentum_transferred = (mass of one person) * (velocity of one person)
Using the value given, this simplifies to:
momentum_transferred = (70 kg) * (1 m/s) = 70 kg m/s
Since there are 10 people, the total momentum transferred to the boat will be:
total_momentum_transferred = (momentum_transferred) * (number of people) = (70 kg m/s) * (10) = 700 kg m/s
Thus, the final momentum of the boat will be the sum of the transferred momenta from each of the people, which is 700 kg m/s.
Using the equation of momentum:
momentum_final = (mass of the boat) * (final velocity of the boat)
Substituting the known values:
700 kg m/s = (1000 kg) * (final velocity of the boat)
Simplifying:
final velocity of the boat = 0.7 m/s
Therefore, the final velocity of the boat, when the people jump off one at a time, is also 0.7 m/s.
To answer these questions, we need to apply the principle of conservation of momentum. According to this principle, the total momentum of an isolated system remains constant before and after an event.
a. If all the people jump off the boat at the same time, the initial momentum is the sum of the individual momenta of the people and the boat. The boat has a mass of 1000 kg and initially has zero velocity. The momentum of the boat before the people jump is therefore 0 kg⋅m/s.
The momentum of an object is given by the product of its mass and velocity. Each person has a mass of 70 kg and jumps off the boat with a velocity of 1 m/s relative to the boat. Since there are 10 people, the total mass of the people is 10 * 70 kg = 700 kg.
The momentum of the people before they jump off the boat is given by the product of their total mass and their initial velocity:
momentum = mass * velocity = 700 kg * 1 m/s = 700 kg⋅m/s
According to the conservation of momentum, the total momentum before the people jump off the boat must equal the total momentum after they jump off. Therefore, the momentum of the boat and people combined after they jump off must be 700 kg⋅m/s.
The mass of the boat remains the same, so we can calculate its final velocity by dividing the total momentum by the mass of the boat:
final velocity = total momentum / mass of the boat = 700 kg⋅m/s / 1000 kg = 0.7 m/s
Therefore, the final velocity of the boat, when all people jump at the same time, is 0.7 m/s in the opposite direction.
b. If the people jump off one at a time, the situation is slightly different. After each person jumps, the boat will experience a change in momentum and therefore a change in velocity.
Let's consider the scenario where the first person jumps off the boat. The momentum of the boat and the first person before the jump is 0 kg⋅m/s. The momentum of the first person before the jump is 70 kg * 1 m/s = 70 kg⋅m/s.
According to the conservation of momentum, the total momentum before and after the first person jumps must be the same.
After the first person jumps, the boat gains a momentum of 70 kg⋅m/s in the opposite direction. Since the mass of the boat is 1000 kg, the change in velocity can be calculated using the equation:
change in velocity = change in momentum / mass of the boat = 70 kg⋅m/s / 1000 kg
Repeat this process for each person jumping off one at a time, and add up the changes in velocity to find the final velocity of the boat.
For example, if each person jumps off with the same velocity of 1 m/s, the final velocity of the boat after all 10 people have jumped off can be calculated as follows:
final velocity = sum of changes in velocity
final velocity = (70 kg⋅m/s / 1000 kg) + (70 kg⋅m/s / 1000 kg) + ... + (70 kg⋅m/s / 1000 kg)
= 10 * (70 kg⋅m/s / 1000 kg)
= 7 m/s
Therefore, the final velocity of the boat, when the people jump off one at a time, is 7 m/s in the opposite direction.