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Sagot :
Total resultant velocity=5.11-3.27=1.84m/s
- m_1=61.4kg
- m_2=109kg
- v_1=1.84m/s
- v_2=?
[tex]\\ \sf\longmapsto ∆P=P[/tex]
[tex]\\ \sf\longmapsto m_1v_1=m_2v_2[/tex]
[tex]\\ \sf\longmapsto v_2=\dfrac{m_1v_1}{m_2}[/tex]
[tex]\\ \sf\longmapsto v_2=\dfrac{61.4(1.84)}{109}[/tex]
[tex]\\ \sf\longmapsto v_2=112.976/109[/tex]
[tex]\\ \sf\longmapsto v_2\approx 1.3m/s[/tex]
The velocity of the plane and the pilot before the pilot jumps is 0.25 m/s.
The given parameters;
- mass of the pilot, m₁ = 61.4 kg
- velocity of the pilot, u₁ = 5.11 m/s backwards
- velocity of the plane, u₂ = 3.27 m/s forward
- mass of the plane, m₂ = 109 kg
The velocity of the plane and the pilot before the pilot jumps is calculated by applying the principle of conservation of linear momentum for inelastic collision as follows;
[tex]m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\61.4(-5.11) \ + \ 109(3.27) = v(61.4 + 109)\\\\42.68 = v(170.4)\\\\v = \frac{42.68}{170.4} \\\\v = 0.25 \ m/s[/tex]
Thus, the velocity of the plane and the pilot before the pilot jumps is 0.25 m/s.
Learn more about inelastic collision here: https://brainly.com/question/7694106
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