A child in a boat throws a 5.30-kg package out horizontally with a speed of 10.0 m/s Fig. 7–31. Calculate the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is 24.0 kg and the mass of the boat is 35.0 kg.

FIGURE 7-31

Problem 7.

The mass of the boat is \({m_1} = 35.0\;{\rm{kg}}\).

The mass of the child is \({m_2} = 24.0\;{\rm{kg}}\).

The mass of the package is \({m_3} = 5.30\;{\rm{kg}}\).

After the throw, the speed of the package is \({v_3} = 10.0\;{\rm{m/s}}\).

Let \(v\) be the velocity of the boat immediately after throwing the package. Here, the child stays in the boat; therefore, the velocity of the child is the same as the velocity of the boat.

You can assume the velocity of the package is in the positive direction.

The total system of boat, child and package, is at rest initially; therefore, the initial momentum of the system is zero.

After throwing the package, the final momentum will also equal zero from the momentum conservation principle.

The total momentum after the collision is \(\left( {{m_1} + {m_2}} \right)v + {m_3}{v_3}\).

Now from the concept of momentum conservation,

\(\begin{array}{c}\left( {{m_1} + {m_2}} \right)v + {m_3}{v_3} = 0\\\left( {{m_1} + {m_2}} \right)v = - {m_3}{v_3}\\v = - \frac{{{m_3}{v_3}}}{{{m_1} + {m_2}}}\end{array}\)

Substituting the values in the above equation,

\(\begin{array}{c}v = - \frac{{5.30\;{\rm{kg}} \times \left( {10.0\;{\rm{m/s}}} \right)}}{{\left( {35.0\;{\rm{kg}}} \right) + \left( {24.0\;{\rm{kg}}} \right)}}\\ = - 0.898\;{\rm{m/s}}\end{array}\)

The negative sign of velocity suggests that the velocity is in the opposite direction of the velocity of the package.

Hence, the velocity of the boat is \(0.898\;{\rm{m/s}}\) in the opposite direction of the package.