6 problems found
A car of mass \(m\) makes a journey of distance \(2d\) in a straight line. It experiences air resistance and rolling resistance so that the total resistance to motion when it is moving with speed \(v\) is \(Av^2 +R\), where \(A\) and \(R\) are constants. The car starts from rest and moves with constant acceleration \(a\) for a distance \(d\). Show that the work done by the engine for this half of the journey is \[ \int_0^d (ma+R+Av^2) \, \d x \] and that it can be written in the form \[ \int_0^w \frac {(ma+R+Av^2)v}a\; \d v \,, \] where \(w =\sqrt {2ad\,}\,\). For the second half of the journey, the acceleration of the car is \(-a\).
Solution: The force delivered by the engine must be \(ma + R + Av^2\), (so the net force is \(ma\)). Therefore the work done is \(\displaystyle \int_0^d F \d x = \int_0^d (ma + R + Av^2) \d x\) Notice that \(a = v \frac{\d v}{\d x} \Rightarrow \frac{a}{v} = \frac{\d v}{\d x}\) and so \begin{align*} && WD &= \int_0^d (ma + R + Av^2) \d x \\ &&&= \int_{x=0}^{x=d} (ma + R + Av^2) \frac{v}{a} \frac{\d v}{\d x} \d x \\ &&&= \int_{x=0}^{x=d} \frac{ (ma + R + Av^2)v}{a} \d v \\ \end{align*} Also notice that if we move with constant acceleration from rest for a distance \(d\) the final speed is \(v^2 = 2ad \Rightarrow v = \sqrt{2ad}\)
A particle of unit mass is projected vertically upwards with speed \(u\). At height \(x\), while the particle is moving upwards, it is found to experience a total force \(F\), due to gravity and air resistance, given by \(F=\alpha \e^{-\beta x}\), where \(\alpha\) and \(\beta\) are positive constants. Calculate the energy expended in reaching this height. Show that \[ F= {\textstyle \frac12} \beta v^2+ \alpha - {\textstyle \frac12} \beta u^2 \;, \] where \(v\) is the speed of the particle, and explain why \( \alpha = \frac12 \beta u^2 +g\), where \(g\) is the acceleration due to gravity. Determine an expression, in terms of \(y\), \(g\) and \(\beta\), for the air resistance experienced by the particle on its downward journey when it is at a distance \(y\) below its highest point.
Solution: Considering the energy of the particle, we have initial kinetic energy of \(\frac12 u^2\) and final energy is \(\frac12 v^2\), the change in energy is the work done by the force, \begin{align*} &&\text{Work done against resistance} &= \text{loss in kinetic energy} \\ &&\int F \, \d x &= \int \alpha e^{-\beta x} \, \d x \\ &&&= \frac{\alpha}{\beta} \l 1 - e^{-\beta x} \r \\ &&&= \frac{1}{\beta} \l \alpha - F\r \\ &&&= \frac12 u^2 - \frac12 v^2 \\ \Rightarrow && F &= \frac12 \beta v^2 + \alpha - \frac12 \beta u^2 \end{align*} When \(v = 0\) there is no air resistance, ie \(F = g\), but \(g = 0 + \alpha - \frac12 \beta u^2 \Rightarrow \alpha = g + \frac12 \beta u^2\) \(F = \frac12 \beta v^2 + g\), ie air resistance is \(\frac12 \beta v^2\) Looking at forces acting on the particle when it's descending, \begin{align*} && v \frac{dv}{dx} &= g - \frac12 \beta v^2 \\ \Rightarrow && \frac{v}{g - \frac12 \beta v^2} \frac{dv}{dx} &= 1 \\ \Rightarrow && \int \frac{v}{g - \frac12 \beta v^2} \, dv &= \int dx \\ \Rightarrow && \frac1{\beta}\l\ln(g - \frac12\beta v^2) - \ln(g)\r &= y\\ \Rightarrow && \ln \l 1 - \frac12 \frac{\beta}{g}v^2 \r &= \beta y \\ \Rightarrow && \frac{g}{\beta} \l 1-e^{-\beta y} \r = \frac12 v^2 \end{align*} Since force is the rate of change of work, we can say that the force is \(ge^{-\beta y}\) and the air resistance is \(g \l 1-e^{-\beta y} \r\)
In an aerobatics display, Jane and Karen jump from a great height and go through a period of free fall before opening their parachutes. While in free fall at speed \(v\), Jane experiences air resistance \(kv\) per unit mass but Karen, who spread-eagles, experiences air resistance \mbox{\(kv + (2k^2/g)v^2\)} per unit mass. Show that Jane's speed can never reach \(g/k\). Obtain the corresponding result for Karen. Jane opens her parachute when her speed is \(g/(3{k})\). Show that she has then been in free fall for time \(k^{-1}\ln (3/2)\). Karen also opens her parachute when her speed is \(g/(3{k})\). Find the time she has then been in free fall.
Solution: Looking at the forces on Jane, \(kv < g \Rightarrow v < \frac{g}{k}\). For Karen we have \begin{align*} kv + (2k^2/g)v^2 &< g\\ -g^2 + gkv + (2k^2)v^2 &< 0 \\ (2kv-g)(kv+g) &< 0\\ \Rightarrow v &< \frac{g}{2k} \end{align*} \begin{align*} && \dot{v} &= g - kv \\ \Rightarrow && \frac{\dot{v}}{g - kv} &= 1 \\ \Rightarrow && T &= \int_0^{g/(3k)} \frac{1}{g - kv} dv \\ && &= \int_0^{g/(3k)} \frac{1}{g - kv} dv\\ && &= \int_0^{g/(3k)} \frac{1}{g - kv} dv \\ && &= \left [-\frac{1}{k} \ln \l g - kv \r \right ]_0^{g/(3k)} \\ && &= \frac{1}{k} \ln \l g \r - \frac{1}{k} \ln \l \frac{2}{3}g \r\\ &&&= \frac{1}{k} \ln \l \frac{3}{2} \r \end{align*} \begin{align*} && \dot{v} &= g - kv - (2k^2/g)v^2 \\ \Rightarrow && \frac{\dot{v}}{g - kv - (2k^2/g)v^2} &= 1 \\ \Rightarrow && T &= \int_0^{g/(3k)} \frac{1}{g - kv - (2k^2/g)v^2} dv \\ && &= \int_0^{g/(3k)} \frac{g}{(g-2kv)(kv+g)} dv\\ && &= \int_0^{g/(3k)} \l \frac{2}{3(g-2kv)} + \frac{1}{3(kv+g)} \r dv\\ && &= \left [ \l -\frac{1}{3k} \ln (g-2kv) + \frac{1}{3k}\ln(kv+g) \r \right ]_0^{g/(3k)} \\ && &= \left [ \l -\frac{1}{3k}\ln \l \frac{g}{3} \r + \frac{1}{3k}\ln \l \frac{4g}{3} \r \r \right ] - \left [- \frac1{3k} \ln(g) + \frac{1}{3k} \ln (g) \right ] \\ && &= \frac{1}{3k} \ln \l 4 \r \end{align*} NB: \(\sqrt[3]{4} \approx 1.58 > \frac{3}{2}\) so Karen has been in free-fall for longer, but not \emph{much} longer than Jane.
A particle of unit mass is projected vertically upwards in a medium whose resistance is \(k\) times the square of the velocity of the particle. If the initial velocity is \(u\), prove that the velocity \(v\) after rising through a distance \(s\) satisfies \begin{equation*} v^{2}=u^{2}\e^{-2ks}+\frac{g}{k}(\e^{-2ks}-1). \tag{*} \end{equation*} Find an expression for the maximum height of the particle above he point of projection. Does equation \((*)\) still hold on the downward path? Justify your answer.
Solution: \begin{align*} \text{N2}(\uparrow): && 1 \cdot v\frac{\d v}{\d s} &= -g - kv^2 \\ \Rightarrow && \int \frac{v}{g+kv^2} \d v &= \int -1 \d s \\ \Rightarrow && \frac{1}{2k}\ln(g+kv^2) &= -s + C \\ s =0, v = u: && \frac{1}{2k} \ln(g+ku^2) &= C \\ \Rightarrow && s &= \frac{1}{2k} \ln \frac{g+ku^2}{g+kv^2} \\ \Rightarrow && e^{-2ks} &= \frac{g+kv^2}{g+ku^2} \\ \Rightarrow && v^2 &= u^2e^{-2ks} + \frac{g}{k}(e^{-2ks}-1) \end{align*} The maximum height will be when \(v = 0\), ie \(\displaystyle s = \frac{1}{2k}\ln\left(1 + \frac{k}{g}u^2 \right)\). On the downward path the resistance will be going upwards, ie \begin{align*} \text{N2}(\uparrow): && 1 \cdot v\frac{\d v}{\d s} &= -g + kv^2 \end{align*} but our solution is solving a different differential equation, therefore unless \(k=0\) the equation will be different.
In a certain race, runners run 5\(\,\)km in a straight line to a fixed point and then turn and run back to the starting point. A steady wind of 3\(\,\text{ms}^{-1}\) is blowing from the start to the turning point. At steady racing pace, a certain runner expends energy at a constant rate of 300\(\,\)W. Two resistive forces act. One is of constant magnitude \(50\,\text{N}\). The other, arising from air resistance, is of magnitude \(2w\,\mathrm{N}\), where \(w\,\text{ms}^{-1}\) is the runner's speed relative to the air. Give a careful argument to derive formulae from which the runner's steady speed in each half of the race may be found. Calculate, to the nearest second, the time the runner will take for the whole race. \textit{Effects due to acceleration and deceleration at the start and turn may be ignored.} The runner may use alternative tactics, expending the same total energy during the race as a whole, but applying different constant powers, \(x_{1}\,\)W in the outward trip, and \(x_{2}\,\)W on the return trip. Prove that, with the wind as above, if the outward and return speeds are \(v_{1}\,\)ms\(^{-1}\) and \(v_{2}\,\)ms\(^{-1}\) respectively, then \(v_{1}+v_{2}\) is independent of the choices for \(x_{1}\) and \(x_{2}\). Hence show that these alternative tactics allow the runner to run the whole race approximately 15 seconds faster.
Solution: Note that \(P = Fv\). Since he is running at a steady pace, we can say that \(F\) must be equal to the resistive forces (as net force is \(0\)). Therefore \(F = 50 + 2(v+3)\) on the way out. ie, \(300 = (2v + 56)v \Rightarrow 150 = v^2 + 28v \Rightarrow v = \sqrt{346}-14\) On the way back, \(F = 50 + 2(v-3)\), ie \(300 = (2v+44)v \Rightarrow 150 = v^2 +22v \Rightarrow v = \sqrt{271}-11\) Therefore the total time will be \(\frac{5000}{\sqrt{346}-150} + \frac{5000}{\sqrt{271}-11} \approx 2002\), or 33 minutes, 22 seconds. Very respectable! The total energy in this first run is \(E = Pt = 2002 \cdot 300\). Now suppose we apply two different powers as in the question, then we must have: \begin{align*} && x_1 &= 2v_1^2 + 56v_1 \\ && x_2 &= 2v_2^2 + 44v_2 \\ && E &= x_1 \frac{5000}{v_1} + x_2 \frac{5000}{v_2} \\ &&&= 5000 \left ( \frac{x_1}{v_1} + \frac{x_2}{v_2} \right) \\ \Rightarrow && \frac{x_1}{v_1} &= 2v_1 + 56 \\ && \frac{x_2}{v_2} &= 2v_2 + 44 \\ \Rightarrow && \frac{E}{5000} &= 2(v_1+v_2) + 100 \\ \Rightarrow && v_1+v_2 &\text{ is independent of the choices for }x_i \end{align*} We wish to minimize \begin{align*} && \frac{5000}{v_1} + \frac{5000}{v_2} &\underbrace{\geq}_{AM-HM} 10\,000 \cdot \frac{2}{v_1+v_2} \\ &&&= 10\,000 \cdot \frac{2}{\sqrt{346}-14+\sqrt{271}-11} \\ &&&\approx 1987 \end{align*} ie they can go 15 seconds quicker with better strategy.
A goalkeeper stands on the goal-line and kicks the football directly into the wind, at an angle \(\alpha\) to the horizontal. The ball has mass \(m\) and is kicked with velocity \(\mathbf{v}_{0}.\) The wind blows horizontally with constant velocity \(\mathbf{w}\) and the air resistance on the ball is \(mk\) times its velocity relative to the wind velocity, where \(k\) is a positive constant. Show that the equation of motion of the ball can be written in the form \[ \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}+k\mathbf{v}=\mathbf{g}+k\mathbf{w}, \] where \(\mathbf{v}\) is the ball's velocity relative to the ground, and \(\mathbf{g}\) is the acceleration due to gravity. By writing down horizontal and vertical equations of motion for the ball, or otherwise, find its position at time \(t\) after it was kicked. On the assumption that the goalkeeper moves out of the way, show that if \(\tan\alpha=\left|\mathbf{g}\right|/(k\left|\mathbf{w}\right|),\) then the goalkeeper scores an own goal.
Solution: Applying \(\mathbf{F} = m\mathbf{a} = m \frac{\d \mathbf{v}}{dt}\) we have: \begin{align*} && m \frac{\d \mathbf{v}}{d t} &= m\mathbf{g} - mk(\mathbf{v} - \mathbf{w}) \\ \Rightarrow && \frac{\d \mathbf{v}}{d t} +k \mathbf{v} &= \mathbf{g} + k \mathbf{w} \\ \\ \Rightarrow && e^{k t} \l \frac{\d \mathbf{v}}{d t} +k \mathbf{v} \r &= e^{kt} ( \mathbf{g} + k \mathbf{w}) \\ \Rightarrow && \frac{\d}{\d t} \l e^{kt} \mathbf{v} \r &= e^{kt}( \mathbf{g} + k \mathbf{w}) \\ \Rightarrow && e^{kt} \mathbf{v} &= \frac{1}ke^{kt}( \mathbf{g} + k \mathbf{w}) + c \\ \Rightarrow && \mathbf{v}_0 &= \frac{1}{k} ( \mathbf{g} + k \mathbf{w})+c \\ \Rightarrow && \mathbf{v} &= e^{-kt} \l \mathbf{v_0} - \frac{1}{k}\mathbf{g} - \mathbf{w} \r + \frac{1}{k} \mathbf{g} + \mathbf{w} \\ \Rightarrow && \mathbf{x} &= -\frac{1}{k}e^{-kt} \l \mathbf{v_0} - \frac{1}{k}\mathbf{g} - \mathbf{w} \r + \frac{1}{k} \mathbf{g}t + \mathbf{w}t+C \\ \Rightarrow && \mathbf{0} &= -\frac{1}{k} \l \mathbf{v_0} - \frac{1}{k}\mathbf{g} - \mathbf{w} \r + C \\ \Rightarrow && \mathbf{x} &= \frac1{k}\l 1- e^{-kt} \r\l \mathbf{v_0} - \frac{1}{k}\mathbf{g} - \mathbf{w} \r + \frac{1}{k} \mathbf{g}t + \mathbf{w}t \end{align*} Position at time \(t\) is: \begin{align*} && x_x &= \frac1{k} ( 1-e^{-kt})(u_x - w)+wt \\ && x_y &= \frac1{k} ( 1-e^{-kt})(u_x \frac{g}{kw} - \frac{g}{k})+\frac{1}{k}gt \\ &&&= \frac{g}{kw} \left ( ( 1-e^{-kt})(u_x - w)+wt \right) \\ &&&= \frac{g}{kw} x_x \end{align*} Therefore if \(x_x\) is ever \(0\) then \(x_y\) will also be zero. But the ball must eventually hit the ground, and when it does, it will be in the process of scoring an own goal.