Final answer:
Using conservation of momentum principle, the speed of the boat and child system after the package is thrown is found to be 0.544 m/s in the direction opposite to the package's motion.
Step-by-step explanation:
Calculating the Speed of the Boat After Throwing an Object
The problem describes a scenario where conservation of momentum applies because no external forces are acting on the system of the child, boat, and package.
The total momentum before and after the package is thrown must remain constant as long as the system is closed and isolated.
Therefore, the momentum of the child and boat combined must be equal and opposite to the momentum of the thrown package.
Initially, the momentum of the system is zero because both the boat and the child are at rest.
When the child throws the package, the momentum of the package is given by:
momentum of package = mass of package × velocity of package
Substituting the given values:
momentum of package = 5.1 kg × 8 m/s
= 40.8 kg·m/s
Since momentum is conserved, the boat and child system must have momentum equal to 40.8 kg·m/s in the opposite direction after the throw.
Considering the total mass of the boat and child is 75 kg (55 kg + 20 kg),
we can use the following equation to find the speed v of the boat and child:
momentum of boat and child = total mass × speed of boat
40.8 kg·m/s = 75 kg × v
Solving for v gives us:
v = − 40.8 kg·m/s ÷ 75 kg
v = − 0.544 m/s
The negative sign indicates that the boat moves in the opposite direction to the throw.
So, the speed of the boat immediately after the child throws the package is 0.544 m/s in the direction opposite to the package's motion.