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Two rowboats move in opposite directions towards each other. When the boats
are next to each other, a bag of mass m = 25 kg is carefully transferred from boat 1 to boat 2. As a result, boat 2 stops moving, and boat 1 continues moving with speed v = 8 m/s. What are the initial speeds of the boats v1 and v2 if the mass of boat 2 is m2 = 1 tonne.

Answer :

ONYEBUCHINNAJIGO

The initial velocity of the first boat is [tex]\frac{7456 -8m_1}{m_1 + 907}[/tex]

The  initial velocity of the second boat is [tex]8 - ( \frac{7456 -8m_1}{m_1 + 907})[/tex]

Conservation of linear momentum

The principle of conservation of linear momentum states that the total momentum of an isolated system is always conserved.

The initial velocity of each boat is calculated by applying the principle of conservation of linear momentum as follows;

[tex]m_1 u_1 +m_2u_2 =m_ 1v_1 + m_2 v_2[/tex]

let the mass of the first boats = m1

mass of the second boat = 907 kg

[tex]m_1 (-u_1 )+ 907 (u_2 )= (m_1-25)(8) + (907 + 25) (0)\\\\-m_1u_1 + 907 u_2 = 8m_1 - 200\\\\200 + 907u_2 = 8m_1 + m_1u_1\\\\200 + 907u_2 = m_1(8 + u_1) \ \ ---(1)[/tex]

Apply one direction velocity equation

[tex]u_1 + v_1 = u_2 + v_2\\\\-u_1 + 8 = u_2 + 0\\\\u_2 = 8 - u_1 \ \ \ ---(2)[/tex]

solve one and two together

[tex]200 + 907(8 - u_1) = m_1(8 + u_1)\\\\200 + 7256-907u_1 = 8m_1 + m_1 u_1\\\\7456 -907u_1 = 8m_1 + m_1 u_1\\\\7456 -8m_1 = m_1 u_1 + 907u_1\\\\7456 - 8m_1 = u_1 (m_1 + 907)\\\\u_1 = \frac{7456 -8m_1}{m_1 + 907}[/tex]

m1 is the mass of the boat 1 (not given)

Initial velocity of the boat 2

[tex]u_2 = 8 - (\frac{7456 -8m_1}{m_1 + 907} )[/tex]

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

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