Answer to Question #156198 in Algebra for joel

Question

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)

Answer 1

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.

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

• Then separate variables & proceed forward

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

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

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

• Then the general equation is

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

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

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

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

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