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

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.

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