Answer:
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
