# Achild in a boat throws a 5.80kg package out horizontally with a speed of 10.0 m/s, fig. 7-31.calculate the velocity of

###### Question:

with a speed of 10.0 m/s, fig. 7-31.calculate the velocity of the

boat immediately after, assuming itwas initially at rest. the mass

of the child is 29.0 kg and that of the boat is 50.0 kg. ignore

water resistance. find itsmagnitude.

## Answers

0.734 m/s

Explanation:

Since there's no external force outside the system, by the law of momentum conservation, momentum before the throw (which is 0) must be the same as after the throw:

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

where m_p = 5.8 kg is the mass of the package

v_p = 10m/s is the velocity of the package

m_c = 29 kg is the mass of the child

m_b = 50 kg is the mass of the boat

As the package and the boat has opposite directions of motion, they should have opposite velocity:

[tex]5.8*10 + (29 + 50)v_b = 0[/tex]

[tex]b_b = \frac{-58}{79} = -0.734 m/s[/tex]

So the magnitude of the boat velocity is 0.734 m/s

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explanation: