Question

A child in a boat throws a 5.90-kg package out horizontally with a speed of 10.0 m/s. The mass of the child is 25.0 kg and the mass of the boat is 38.4 kg. (Figure 1)
Calculate the velocity of the boat immediately after, assuming it was initially at rest.
Express your answer to three significant figures and include the appropriate units. Enter positive value if the direction of the velocity is in the direction of the velocity of the box and negative value if the direction of the velocity is in the direction opposite to the velocity of the box.

Answer 1

-0.930 m/s is the velocity of the boat.

Given:

Mass of child and boat , $m_1$ = (25.0 + 38.4 )kg

                                               = 63.4 kg

Mass of the package, $m_2$ = 5.90 kg

Velocity of package thrown from boat , $v_2$ = 10.0m/s

$v_1 =?$

Initial velocity v = 0 m/s

As the boat is at rest, $(m_1 + m_2) v=0$

According, to the law of conversation of momentum;

∴ Momentum before = Momentum after

   $( m_1 + m_2 ) v = m_1v_1 + m_2v_2\\0 = m_1v_1 + m_2v_2\\0 = 63.4 v_1 + 5.90*10.0\\63.4 v_1 = - 5.9\\v_1 = - 0.930 m/s$

Negative direction shows the velocity in the direction opposite to the motion of the package.

Therefore, -0.930 m/s is the velocity of the boat.

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