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A child in a boat throws a 5.80-kg package out horizontally with a speed of 10.0 m/s. The mass of the child is 24.6kg and the mass of the boat is 39.0kg . (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 force is in the direction of the velocity of the box and negative value if the direction of the force is in the direction opposite to the velocity of the box.

Answer :

MechEngineer

Answer:

-0.912 m/s

Explanation:

When the package is thrown out, momentum is conserved. The total momentum after is the same as the total momentum before, which is 0, since the boat was initially at rest.

[tex] (m_c + m_b)v_b + m_pv_p = 0[/tex]

where [tex]m_c = 24.6 kg, m_b = 39 kg, m_p = 5.8 kg[/tex] are the mass of the child, the boat and the package, respectively. [tex], v_p = 10m/s, v_b[/tex] are the velocity of the package and the boat after throwing.

[tex] (24.6 + 39)v_b + 5.8*10 = 0[/tex]

[tex]63.6v_b + 58 = 0[/tex]

[tex]v_b = -58/63.6 = -0.912 m/s[/tex]

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