Answer to Question #156198 in Algebra for joel

Question #156198
A particle P, of mass m, is projected vertically upwards from a point O with speed 2g/k m/s in a medium whose resistance to motion is of magnitude kv per unit mass, where v is the speed of P and k a positive contant.
(i) Show that v(dv/dx) = -g - kv
The particle P attains the highest point H of its path.
(ii) Show further that OH = g/k^2 (2 - In3)metres
Given that P reaches another point M with speed g/2k m/s as it traces its path back to O, the point of projection,
(iii) Prove that HM = g/k^2 ( In2 - 1/2)
1
Expert's answer
2021-01-26T19:12:35-0500

Solution


  • When it goes upwards resistive forces that act on it are its own weight (mg) & the medium resistance (mkv).
  • Then it travels under a deceleration upwards & F=ma in the form F=m(dv/dt) can be written as the deceleration is not constant as F varies with the velocity.
  • Then a differential equation generates & by solving it with respect to relevant variables, answers for the rest can be found.

"\\qquad\\qquad\n\\begin{aligned}\n\\uparrow F&=m\\frac{dv}{dt}=m\\frac{dv}{dx}\\times\\frac{dx}{dt}=mv\\frac{dv}{dx}\\\\\n-mg-mkv&=mv\\frac{dv}{dx}\\\\\n-g-kv&=v\\frac{dv}{dx}\\\\\n\\\\\n\n\\end{aligned}"

  • Then separate variables & proceed forward

"\\qquad\\qquad\n\\begin{aligned}\n\\frac{v}{(g+kv)}.dv&=-dx\\\\\n\\frac{1}{k}\\bigg[\\frac{(kv+g)-g}{(g+kv)}\\bigg].dv&=-dx\\\\\n\\frac{1}{k}\\bigg[1-\\frac{g}{(g+kv)}\\bigg].dv&=-dx\\\\\n\\frac{1}{k}\\bigg[dv-\\frac{g}{(g+kv)}.dv\\bigg]&=-dx\\\\\n\\frac{1}{k}\\bigg[\\int dv-\\frac{g}{k}\\int \\frac{k}{(g+kv)}dv\\bigg]&=-\\int dx\\\\\n\\frac{1}{k}\\bigg[v-\\frac{g}{k}ln(g+kv)\\bigg]&=-x+c\n\\end{aligned}"

  • Solving for the arbitrary constant c, a general equation for the velocity & distance can be obtained.
  • Substitute the values: @ x=0m v=2g/k

"\\qquad\\qquad\n\\begin{aligned}\n\\frac{1}{k}\\bigg[\\frac{2g}{k}-\\frac{g}{k}ln(g+2g)\\bigg]&=-0+c\\\\\nc&= \\frac{g}{k^2}\\big[2-ln(3g)\\big]\n\\end{aligned}"

  • Then the general equation is

"\\qquad\\qquad\n\\begin{aligned}\n\\frac{1}{k}\\bigg[v-\\frac{g}{k}ln(g+kv)\\bigg]&=-x+\\frac{g}{k^2}\\big[2-ln(3g)\\big]\n\\end{aligned}"

  • Then at the highest point: x=OH & v=0

"\\qquad\\qquad\n\\begin{aligned}\n\\frac{1}{k}\\bigg[0-\\frac{g}{k}ln(g)\\bigg]&=-OH+\\frac{g}{k^2}\\big[2-ln(3g)\\big]\\\\\nOH&= \\frac{g}{k^2}\\big[2-ln(3g)\\big]+\\frac{g}{k^2}ln(g)\\\\\n&=\\frac{g}{k^2}\\big[2-(ln(3g)-ln(g))\\big]\\\\\n&=\\frac{g}{k^2}\\big[2-ln3\\big] \n\\end{aligned}"

  • Then at the return @ M, x=OM=(OH-HM) & v=-g/2k. Mind the direction of the velocity.

"\\qquad\\qquad\n\\begin{aligned}\n\\frac{1}{k}\\bigg[-\\frac{g}{2k}-\\frac{g}{k}ln(g+(-\\frac{g}{2}))\\bigg]&=-(OH-HM)+\\frac{g}{k^2}\\big[2-ln(3g)\\big]\\\\\n-\\frac{g}{k^2}\\bigg[\\frac{1}{2}+ln(\\frac{g}{2})\\bigg]&=(HM-OH)+\\frac{g}{k^2}\\big[2-ln(3g)\\big]\\\\\nHM&= OH-\\frac{g}{k^2}\\big[2-ln(3g)\\big]-\\frac{g}{k^2}\\bigg[\\frac{1}{2}+ln(\\frac{g}{2})\\bigg]\\\\\n&=\\frac{g}{k^2}\\big[2-ln3\\big]-\\frac{g}{k^2}\\big[2-ln(3g)\\big]-\\frac{g}{k^2}\\bigg[\\frac{1}{2}+ln(\\frac{g}{2})\\bigg]\\\\\n&=\\frac{g}{k^2}\\bigg[2-ln3-2+ln(3g)-\\frac{1}{2}-ln(\\frac{g}{2})\\bigg]\\\\\n&=\\frac{g}{k^2}\\bigg[ln(\\frac{3g}{3\\times\\frac{g}{2}})-\\frac{1}{2}\\bigg]\\\\\n&=\\frac{g}{k^2}\\bigg[ln2-\\frac{1}{2}\\bigg]\n\\end{aligned}"


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