A boy of mass 60 kg stands on a boat of mass 250 kg. This system is stationary. From the boat, the boy throws a rock of mass 1 kg into the water with a velocity of 10 m/s. What is the magnitude and direction of the boat's motion as a result of this?
A)
The boat moves in the direction of the rock with a velocity of 31 m/s.
B)
The boat moves in the direction of the rock with a velocity of 0.02 m/s.
C)
The boat moves in a direction opposite to the rock with a velocity of 0.03 m/s
D)
The boat moves in a direction opposite to the rock with a velocity of 10.1 m/s.
clearly not A or B
since the rock is so much smaller than the boat system, the boat will move much more slowly, right?
To find the magnitude and direction of the boat's motion, we can apply the principle of conservation of momentum:
The initial momentum of the system (boat + boy + rock) is zero since it is stationary.
The final momentum of the system must also be zero, as there are no external forces acting on it.
The total momentum of the system is given by the equation:
Total Momentum = (Mass of the boy * Velocity of the boy) + (Mass of the boat * Velocity of the boat) + (Mass of the rock * Velocity of the rock)
For the initial system, the momentum is zero:
0 = (Mass of the boy * 0) + (Mass of the boat * 0) + (Mass of the rock * 0)
Now, when the boy throws the rock with a velocity of 10 m/s, the momentum changes:
0 = (Mass of the boy * 0) + (Mass of the boat * Velocity of the boat) + (Mass of the rock * 10 m/s)
Since the boy does not contribute to the system's momentum change, we can ignore the term with the mass of the boy.
Now we can solve for the velocity of the boat:
0 = (Mass of the boat * Velocity of the boat) + (Mass of the rock * 10 m/s)
⇒ Mass of the boat * Velocity of the boat = - (Mass of the rock * 10 m/s)
Dividing through by the mass of the boat:
Velocity of the boat = - (Mass of the rock * 10 m/s) / Mass of the boat
Substituting the given values:
Velocity of the boat = - (1 kg * 10 m/s) / 250 kg
Calculating this expression:
Velocity of the boat = - 0.04 m/s
The negative sign indicates that the boat moves in the opposite direction to the rock.
Therefore, the correct answer is:
C) The boat moves in a direction opposite to the rock with a velocity of 0.03 m/s.
To solve this problem, we can apply the principle of conservation of momentum. The total momentum before the rock is thrown is equal to the total momentum after the rock is thrown.
The momentum of an object is given by the product of its mass and its velocity.
Before the rock is thrown, the total momentum of the system (boy + boat) is given by:
Total momentum before = (Mass of the boy * Velocity of the boy) + (Mass of the boat * Velocity of the boat)
Since the system is stationary, the velocity of the boat is 0. So the total momentum before the rock is thrown is:
Total momentum before = (Mass of the boy * Velocity of the boy) + (Mass of the boat * 0)
= Mass of the boy * Velocity of the boy
After the rock is thrown, the total momentum of the system will still be conserved. The rock moves in the direction opposite to the motion of the boat.
Let's calculate the momentum of the rock:
Momentum of the rock = Mass of the rock * Velocity of the rock
= 1 kg * 10 m/s
= 10 kg*m/s
Since the momentum is conserved, the total momentum of the system after the rock is thrown is equal to the total momentum before the rock is thrown:
Total momentum before = Total momentum after
Mass of the boy * Velocity of the boy = Total momentum after
Now, let's solve for the velocity of the boat:
Velocity of the boat = Total momentum after / Mass of the boat
Velocity of the boat = (Mass of the boy * Velocity of the boy) / Mass of the boat
Plugging in the given values, we get:
Velocity of the boat = (60 kg * 10 m/s) / 250 kg
= 6 m/s
Therefore, the magnitude and direction of the boat's motion as a result of the rock being thrown is 6 m/s in the direction opposite to the rock's motion.
So, the correct answer is:
C) The boat moves in a direction opposite to the rock with a velocity of 0.03 m/s.