Respuesta :
Answer:
the final speed of object A changed by a factor of [tex]\frac{1}{\sqrt{3} }[/tex] = 0.58
the final speed of object B changed by a factor of [tex]\sqrt{\frac{5}{3} }[/tex] = 1.29
Explanation:
Given;
kinetic energy of object A, = 27 J
let the mass of object A = [tex]m_A[/tex]
then, the mass of object B = [tex]m_B = \frac{m_A}{4}[/tex]
work done on object A = -18 J
work done on object B = -18 J
let [tex]v_i[/tex] be the initial speed
let [tex]v_f[/tex] be the final speed
For object A;
[tex]K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2 = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ - v_i^2 )\ =- 18\\\\v_f^2 \ - v_i^2 = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ - v_i^2 = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\[/tex]
[tex]v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\[/tex]
Thus, the final speed of object A changed by a factor of [tex]\frac{1}{\sqrt{3} }[/tex] = 0.58
To obtain the change in the final speed of object B, apply the following equations.
[tex]K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B = \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\[/tex]
[tex]\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i[/tex]
Thus, the final speed of object B changed by a factor of [tex]\sqrt{\frac{5}{3} }[/tex] = 1.29
Velocity is the rate of change of position with respect to time. The velocity of objects A and B is 0.577 and 1.29 times the initial velocities.
What is Velocity?
Velocity is the rate of change of position with respect to time.
Given to us
The kinetic energy of object A before work is applied, [tex]KE_A = 27\rm\ J[/tex]
Mass of object A = [tex]m_a[/tex]
Mass of object B = [tex]m_b = \dfrac{m_a}{4}[/tex]
Work done on both the objects, w = -18 J
We know that the kinetic energy of object A is 27 j, therefore,
[tex]KE_A = 27 J\\\\\dfrac{1}{2}m_av_i^2 = 27\\\\m_av_i^2 = 54\\\\m_a = \dfrac{54}{v_i^2}[/tex]
We know that the work done on object A is -18 J, therefore, the difference in the Kinetic energy of the object will be work,
[tex]KE_f - KE_i = -18\\\\\dfrac{1}{2}m_Av_f^2 -\dfrac{1}{2}m_Av_i^2 = 18\\\\v_f^2 -v_i^2 = \dfrac{-18 \times 2}{m_A}\\\\v_f^2 -v_i^2 = \dfrac{-36}{m_A}[/tex]
Substitute the value of [tex]m_a[/tex],
[tex]v_f^2 -v_i^2 = \dfrac{-36}{m_A}\\\\v_f^2 -v_i^2 = \dfrac{-36}{\dfrac{54}{v_i^2}}\\\\v_f^2 -v_i^2 = \dfrac{-36 v_i^2}{54}\\\\v_f^2 = \dfrac{1v_i^2}{3}+v_i^2\\\\v_f^2 = \dfrac{-36 v_i^2}{54}\\\\v_f = v_i\sqrt{\dfrac{1}{3}}\\\\v_f = 0.577v_i[/tex]
Hence, the velocity of object A is 0.577 times the initial velocity.
Kinetic Energy of the object B,
[tex]KE_B = \dfrac{1}{2}m_B v_i^2\\\\KE_B = \dfrac{1}{2}\dfrac{m_A}{4} v_i^2\\\\KE_B =\dfrac{m_A}{8} v_i^2\\\\KE_B =\dfrac{m_A}{8} v_i^2\\\\KE_B =\dfrac{54}{8} \\[/tex]
We know that the work done on object B is -18 J, therefore, the difference in the Kinetic energy of the object will be work,
[tex]KE_f - KE_i = -18\\\\\dfrac{1}{2}m_Bv_f^2 -\dfrac{1}{2}m_Bv_i^2 = 18\\\\v_f^2 -v_i^2 = \dfrac{-18 \times 2}{m_B}\\\\v_f^2 -v_i^2 = \dfrac{-36}{m_B}\\\\v_f^2 -v_i^2 = \dfrac{-36}{\dfrac{m_A}{4}}\\\\v_f^2 -v_i^2 = \dfrac{-144}{m_A}\\\\}}\\\\v_f^2 -v_i^2 = \dfrac{-144}{\dfrac{54}{v_i^2}}[/tex]
[tex]v_f^2 -v_i^2 = \dfrac{-144}{\dfrac{54}{v_i^2}}\\\\v_f^2 -v_i^2 = \dfrac{-144v_i^2}{54}\\\\v_f^2 = \dfrac{-144v_i^2}{54} + v_i^2\\\\v_f^2 = \dfrac{-144v_i^2}{54} + v_i^2\\\\v_f^2 = \dfrac{-5v_i^2}{3}\\\\ | v_f |= \sqrt{\dfrac{5v_i^2}{3}}\\\\v_f = \sqrt{\dfrac{5v_i^2}{3}}\\v_f = v_i\sqrt{\dfrac{5}{3}}\\\\v_f = 1.29 v_i[/tex]
hence, the velocity of object B is 1.29 times the initial velocity.
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