A child in a boat throws a 5.90-kg package out horizontally with a speed of 10.0 m/s. Calculate the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is 28.0 kg and that of the boat is 35.0 kg. (Take the package's direction of motion as positive.)

0.94 m/s opposite the direction of the package

Explanation:

From the law of conservation of momentum,

Total momentum before the throw = Total momentum after the throw.

Note: Assuming the child, the boat and the package, where initially at the same velocity

U (m+m'+m'') = V (m+m') + v'' (m'') ... Equation 1

Note: since the boat was initially at rest, U = 0 m/s

0 = V (m+m') + v' (m'')

-V (m+m') = v'' (m'') ... Equation 2

Where,

V = Velocity of the boat immediately after the throw, m = mass of the child, m' = mass of the boat, m'' = mass of the package, v'' = velocity of the package after the throw.

make V the subject of the equation

V = - v'' (m'') / (m+m') ... Equation 3

Given: v'' = 10 m/s, m'' = 5.9 kg, m = 28 kg, m' = 35 kg

Substitute into equation 3

V = - 10 (5.9) / (28+35)

V = - 59/63

V = - 0.94 m/s.

The Negative sign shows that the velocity of the both immediately after the throw is in opposite direction to the velocity of the package