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With a mass of 109 kg, Baby Bird is the smallest monoplane ever
flown. Suppose the Baby Bird and pilot are coasting along the
runway when the pilot jumps horizontally to the runway behind
the plane. The pilot's velocity upon leaving the plane is 5.11 m/s
backward. After the pilot jumps from the plane, the plane coasts
forward with a speed of 3.27 m/s. If the pilot's mass equals 61.4
kg, what is the velocity of the plane and pilot before the pilot
jumps? Round to the nearest hundredth.

Respuesta :

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]

onyebuchinnaji

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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