A boy on ice skates is atationary on a frozen lake(no friction).He throws a package of mass 5kg at 4m/s horizontally east.the mass of the boy is 60kg.at the instant the package leaves the boy's hand,the boy starts moving.
Calculate the magnitude of the velocity of the boy immediately after the package leaves his hand.ignore the effect of the friction
conservation of momentum.
initial momtentum=finalmonentum
0=5*4 East+ 60*V
solve for V
V= 1/3 (-E) or .333m/s W
To calculate the magnitude of the velocity of the boy immediately after the package leaves his hand, we can use the principle of conservation of momentum.
The momentum before the package is thrown can be calculated as the product of mass and velocity:
Momentum before = (mass of the boy) × (initial velocity of the boy)
Momentum before = 60 kg × 0 m/s (since the boy is stationary)
The momentum after the package is thrown can be calculated by considering the package's momentum and the resulting momentum of the boy:
Momentum after = (mass of the package) × (velocity of the package) + (mass of the boy) × (velocity of the boy)
Since the package is thrown horizontally east, its horizontal velocity is 4 m/s. Also, as there is no friction, the horizontal component of the boy's velocity will remain constant after the package is released and equal to the horizontal velocity of the package.
Substituting the known values:
Momentum after = (5 kg) × (4 m/s) + (60 kg) × (velocity of the boy)
Since momentum is conserved, we can equate the momenta before and after:
Momentum before = Momentum after
60 kg × 0 m/s = 5 kg × 4 m/s + 60 kg × (velocity of the boy)
0 = 20 kg m/s + 60 kg × (velocity of the boy)
-20 kg m/s = 60 kg × (velocity of the boy)
Dividing both sides of the equation by 60 kg:
velocity of the boy = -20 kg m/s / 60 kg
velocity of the boy = -0.33 m/s
The magnitude of the velocity is always positive, so the magnitude of the velocity of the boy immediately after the package leaves his hand is 0.33 m/s.
To find the magnitude of the velocity of the boy immediately after the package leaves his hand, we can use the principle of conservation of momentum. The total momentum before the package is thrown is equal to the total momentum after the package is thrown.
Before the package is thrown, the boy and the package are at rest, so the total momentum is zero. After the package is thrown, the momentum is conserved and distributed between the boy and the package.
The momentum of an object is calculated by multiplying its mass by its velocity. Let's assign a positive direction to the right (east) and a negative direction to the left (west):
The momentum of the package is calculated as mass times velocity:
Momentum of the package = mass of the package × velocity of the package
= 5 kg × 4 m/s
= 20 kg·m/s to the east (positive)
According to the conservation of momentum, this momentum should be distributed between the package and the boy:
Momentum of the boy = - Momentum of the package
= - 20 kg·m/s to the east (negative)
Since the mass of the boy is 60 kg, we can calculate the velocity of the boy using momentum:
Momentum of the boy = mass of the boy × velocity of the boy
velocity of the boy = Momentum of the boy / mass of the boy
velocity of the boy = (-20 kg·m/s) / 60 kg
= - 0.33 m/s to the east (negative)
The magnitude of the velocity is just the absolute value of the velocity, so the magnitude of the velocity of the boy immediately after the package leaves his hand is 0.33 m/s.