Respuesta :
Answer:
the maximum deformation undergone by the spring = 47.46 cm
Explanation:
Using conservation of momentum:
[tex]m_Av_A + m_Bv_B = (m_A+m_B )v[/tex]
where:
[tex]m_A = \frac{250\ kN}{g}[/tex]
[tex]v_A = 40[/tex]
[tex]m_B = \frac{550\ kN}{g}[/tex]
[tex]v_B =10[/tex]
Then;
[tex]m_Av_A + m_Bv_B = (m_A+m_B )v[/tex]
[tex]\frac{250*10^3}{9.81}*40 + \frac{550*10^3}{9.81}*10 = (\frac{800*10^3}{9.81} )v[/tex]
[tex]1580020.387 = 81549.43935 \ v[/tex]
[tex]v = \frac{1580020.387}{81549.43935}[/tex]
v = 19.375 m/s
However ; using conservation of energy to determine the maximum deformation undergone by the spring ; we have:
[tex]\frac{1}{2} [m_Av_A^2 +m_Bv_B^2] =\frac{1}{2}[(m_A+m_B)v^2 + kx^2][/tex]
[tex][m_Av_A^2 +m_Bv_B^2] =[(m_A+m_B)v^2 + kx^2][/tex]
[tex][\frac{250*10^3}{9.81}*40^2 + \frac{550*10^3}{9.81}*10^2] =[ (\frac{800*10^3}{9.81} )*19.375^2 + 70 *10^6 \ * x^2][/tex]
x = 0.4746 m
x = 47.46 cm
Thus, the maximum deformation undergone by the spring = 47.46 cm
Answer:
The maximum deformation undergone by the spring is 0.475 m
Explanation:
The conservation of momentum is equal:
[tex]m_{A} v_{A} +m_{B} v_{B} =v(m_{A} +m_{B} )[/tex]
Replacing:
[tex]\frac{250x10^{3}*40 }{9.8} +\frac{550x10^{3}*10 }{9.8} =v(\frac{(250+550)x10^{3} }{9.8}) \\v=19.375m/s[/tex]
The conservation of energy is:
[tex]\frac{1}{2} (m_{A}v_{A}^{2}+m_{B}v_{B}^{2} )=\frac{1}{2} (v^{2}(m_{A} +m_{B} )+kx^{2} )\\\frac{250x10^{3}*40^{2} }{9.8} +\frac{550x10^{3}*10^{2} }{9.8} =19.375^{2} (\frac{800x10^{3} }{9.8} )+70x10^{6} x^{2} \\ x=0.475m[/tex]
