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A 76 kg boater, initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock. If the
boater moves out of the boat with a velocity of 2.5 m/s to the right, what is the final velocity of the boat?

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onyebuchinnaji

The final velocity of the boat is 4.22 m/s in opposite direction to the boater.

Final velocity of the boat

The final velocity of the boat is determined by applying the principle of conservation of linear momentum as follows;

Let the initial velocity of the boater and the boat = v

[tex]m_1 u_1 + m_2 u_2 = v(m_1 + m_2)\\\\76(2.5) + 45u_2 = 0(76 + 45)\\\\190 + 45u_2=0\\\\45u_2 = -190\\\\u_2 = \frac{-190}{45} \\\\u_2 = -4.22 \ m/s[/tex]

Thus, the final velocity of the boat is 4.22 m/s in opposite direction to the boater.

Learn more about conservation of linear momentum here: https://brainly.com/question/7538238

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