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Mechanics, EXT2 EQ-Bank 36

An experimental rocket is at a height of 5000 m, ascending at a speed of \(50\sqrt{2}\) m s\(^{-1}\) at an angle of 45° to the horizontal, when its engine stops. The rocket is then subject to gravity and to air resistance proportional to its velocity. Take \(g\) = 10 m s\(^{-2}\).

The velocity vector of the rocket, \(t\) seconds after the engine stops, is

\(\mathbf{v}(t) = 50e^{-0.2t}\,\mathbf{i} + (100e^{-0.2t}-50)\mathbf{j}.\)   (Do NOT prove this.)
 

  1. Show that the rocket reaches its greatest height when  \(t =5\ln 2\)  seconds, and calculate its greatest height.   (3 marks)

  2. The pilot can only operate the ejection seat while the rocket is descending at an angle between 45° and 60° to the horizontal. Find the earliest and latest times at which the pilot can eject.   (3 marks)

  3. As the rocket continues to fall, its speed approaches a limiting value. Find this terminal speed, justifying your answer.   (1 mark)

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a.    \(\text{See Worked Solutions}\)

b.    \(\text{Pilot can eject between 5.5 and 6.6 seconds.}\)

c.    \(\text{Terminal speed}=50 \ \text{ms}^{-1}\)

Show Worked Solution

a.    \(\mathbf{v}(t)=50 e^{-0.2 t}\,\mathbf{i}+\left(100 e^{-0.2 t}-50\right)\mathbf{j}\)

\(\text{Max height occurs when} \ \ \dot{y}=0:\)

\(100 e^{-0.2 t}-50\) \(=0\)
\(e^{-0.2 t}\) \(=\dfrac{1}{2}\)
\(-0.2 t\) \(=-\ln 2\)
\(t\) \(=5 \ln 2\)

 
\(\text{Find} \ y  \ \text{when}\ \  t=5 \ln 2:\)

\(y(t)=\displaystyle \int 100 e^{-0.2 t}-50\, d t=-500 e^{-0.2 t}-50 t+c\)

\(\text{When} \ \ t=0, y=5000:\)

\(5000=-500 e^{\circ}+c \ \Rightarrow \ c=5500\)

\(y=5500-500 e^{-0.2 t}-50 t\)
 

\(\text{At} \ \ t=5\ln 2:\)

\(y=5500-500 e^{-\ln 2}-50 \times 5 \ln 2=5076.71 \ldots=5077 \ \text{m}\).
 

b.    \(\text {On descent,} \ \ \dot{y}<0.\)

\(\text{Let} \ \ \theta=\text{angle below the horizontal}\)

\(\tan \theta=\dfrac{\abs{\dot{y}}}{\dot{x}}\)

\(\text{Let}\ \  a=e^{-0.2 t}\)

\(\tan \theta=\dfrac{50-100 a}{50 a}=\dfrac{1}{a}-2 \ \Rightarrow \ \theta=\tan ^{-1}\left(\dfrac{1}{a}-2\right)\)
 

\(\text{When} \ 45^{\circ} \ \text {is reached:}\)

\(\dfrac{1}{a}-2=1 \ \Rightarrow \ a=3\)

\(e^{-0.2 t}=\dfrac{1}{3} \ \Rightarrow \ t=\dfrac{\ln 3}{0.2} \approx 5.5 \ \text{s  (1 d.p.)}\)
 

\(\text{When} \ 60^{\circ} \ \text {is reached:}\)

\(\dfrac{1}{a}-2=\sqrt{3} \ \Rightarrow \ a=\dfrac{1}{2+\sqrt{3}}=2-\sqrt{3}\)

\(e^{-0.2 t}\) \(=2-\sqrt{3}\)
\(-0.2 t\) \(=\ln (2-\sqrt{3})\)
\(t\) \(=-5\ln (2-\sqrt{3}) \approx 6.6 \ \text{s  (1 d.p.)}\)

 
\(\therefore \ \text{Pilot can eject between 5.5 and 6.6 seconds.}\)
 

c.    \(\text{As} \ \ t \rightarrow \infty:\)

\(e^{-0.2 t} \rightarrow 0\ \ \Rightarrow\ \ \dot{x} \rightarrow 0, \ \ \dot{y} \rightarrow -50\)

\(\therefore \ \text{Terminal speed}=\sqrt{0^2+50^2}=50 \ \text{ms}^{-1}\)

Mechanics, EXT2 EQ-Bank 31

A particle is projected from the origin with speed, \(V\), at an angle of \(\alpha\) above the horizontal. It is subject to both gravity and an air resistance proportional to its velocity, so that its horizontal and vertical components of acceleration while it is rising are given by

\(\ddot{x}=-k\dot{x}\)  and  \(\ddot{y} = -g-k\dot{y}\)

  1. Show that  \(\dot{x} = V\cos\,\alpha\ e^{-kt}\)  and  \(\dot{y} = \left( \dfrac{g}{k} + V\sin\,\alpha\right)e^{-kt}-\dfrac{g}{k} \)   (2 marks)

  2. Show that when the particle reaches its greatest height, it has travelled a horizontal distance of
  3.      \(\dfrac{V^2\sin\,2\alpha}{2(g+Vk\,\sin\,\alpha)}\).   (3 marks)

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a.    \(\text{See Worked Solutions}\)

b.    \(\text{See Worked Solutions}\)

Show Worked Solution

a.    \(\ddot{x}=-k \dot{x} \ \Rightarrow \ \dfrac{d \dot{x}}{d t}=-k \dot{x}\)

\(\text{Solving 1st order differential equation:}\)

\(\dot{x}=A e^{-k t}\)

\(\text{Since}\ \ \dot{x}=V \cos \alpha\ \ \text{when} \ \ t=0\ \ \Rightarrow\ \ A=V \cos \alpha\)

\(\dot{x}=V \cos \alpha e^{-k t}\)
 

\(\ddot{y}=-g-k \dot{y}\ \Rightarrow \ \dfrac{d \dot{y}}{d t}+k \dot{y}=-g\)

\(\text{Solving 1st order differential equation:}\)

\(\dot{y}=B e^{-k t}-\dfrac{g}{k}\)

\(\text{Since} \ \ \dot{y}=V \sin \alpha \ \ \text{when} \ \ t=0:\)

\(V \sin \alpha=B e^{-k t}-\dfrac{g}{k} \ \Rightarrow \ B=\dfrac{g}{k}+V \sin \alpha\)

\(\dot{y}=\left(V \sin \alpha+\dfrac{g}{k}\right) e^{-k t}-\dfrac{g}{k}\)
 

b.    \(\text{At max height,} \ \ \dot{y}=0\)

\(\left(V \sin \alpha+\dfrac{g}{k}\right) e^{-k t}-\dfrac{g}{k}=0 \ \Rightarrow \ e^{-k t}=\dfrac{g}{g+V k \, \sin \alpha}\ \ldots\ (1)\)

\(\text{Find horizontal distance} \ (x):\)

\(x\) \(=\displaystyle \int_0^t V \cos \alpha\, e^{-k t}\, d t\)
  \(=-\dfrac{V \cos \alpha}{k}\big[e^{-k t}\big]_0^t\)
  \(=\dfrac{V \cos \alpha}{k}\left(1-e^{-k t}\right)\)
  \(=\dfrac{V \cos \alpha}{k}\left(1-\dfrac{g}{g+Vk\, \sin \alpha}\right)\ \ \ \text{(using (1) above)}\)
  \(=\dfrac{V \cos \alpha}{k}\left(\dfrac{g+Vk\, \sin \alpha-g}{g+Vk\, \sin \alpha}\right)\)
  \(=\dfrac{V^2 \sin \alpha\, \cos \alpha}{g+Vk\, \sin \alpha}\)
  \(=\dfrac{V^2 \sin 2 \alpha}{2(g+Vk\, \sin \alpha)}\)

Mechanics, EXT2 EQ-Bank 30

In a circus act, an 8 kg cannon ball is projected from the origin into the air with an initial velocity of 26 m s\(^{-1}\) and at an angle of 67.4° to the horizontal. The ball is caught at the top of its trajectory by a performer who is at the position \((A, B)\).

The velocity vector, \(\mathbf{v} (t)\), of the ball at time \(t\) seconds after launch is given by

\(\mathbf{v}(t)=10 e^{-0.8 t} \mathbf{i} +\left[36.5 e^{-0.8 t}-12.5\right] \mathbf{j}\).   (Do NOT Prove this.)

  1. Show that the ball reaches the performer at  \(t=1.339\) (to three decimal places).   (2 marks)

  2. Find the values of \(A\) and \(B\).   (3 marks)

  3. What is the speed of the ball when it reaches the performer?   (1 mark)

  4. What is the magnitude of the force on the ball when it reaches the performer?   (2 marks)

Show Answers Only

a.    \(t=1.339 \ \text{s}\)

b.    \(A=8.217\ \text{m},\ \ B=13.257\ \text{m}\)

c.    \(\text{Speed}\ = 3.426\ \text{m s}^{-1}\)

d.    \(82.951 \ \text{N}\)

Show Worked Solution

a.    \(\text{At top of trajectory:}\)

\(36.5 e^{-0.8 t}-12.5\) \(=0\)
\(36.5 e^{-0.8 t}\) \(=12.5\)
\(e^{-0.8 t}\) \(=\dfrac{12.5}{36.5}\)
\(-0.8 t\) \(=\ln \dfrac{12.5}{36.5}\)
\( t\) \(=\dfrac{\ln\frac{12.5}{36.5}}{-0.8}=1.339 \ \text{s (3 d.p.)}\)

 

b.     \(\text{Horizontal velocity}\ =10 e^{-0.8 t}\)

\(x=\displaystyle \int 10 e^{-0.8 t}\,d t=\dfrac{10}{-0.8} e^{-0.8 t}+c_1=-12.5 e^{-0.8 t}+c_1\) 

\(\text{When} \ \ t=0, \ x=0:\)

\(0=-12.5 e^0+c_1\ \ \Rightarrow\ \ c_1=12.5\)

\(x=-12.5 e^{-0.8 t}+12.5\)
 

\(\text{When } t=1.339, \ x=A:\)

\(A=-12.5 e^{-0.8 \times 1.339}+12.5=8.217\ \text{m (3 d.p.)}\)
 

\(\text {Vertical velocity}\ =36.5 e^{-0.8 t}-12.5\)

\(y=\displaystyle \int\left(36.5 e^{-0.8 t}-12.5\right)\,d t=\dfrac{36.5}{-0.8} e^{-0.8 t}-12.5 t+c_2=-45.625 e^{-0.8 t}-12.5 t+c_2\)

\(\text{When} \ \ t=0, \ y=0:\)

\(0=-45.625+c_2\ \ \Rightarrow\ \ c_2=45.625\)

\(y=-45.625 e^{-0.8 t}-12.5 t+45.625\)
 

\(\text{When}\ \ t=1.339, \ y=B:\)

\(B=-45.625 e^{-0.8 \times 1.339}-12.5 \times 1.339+45.625=13.257\ \text{m (3 d.p.)}\)
 

c.    \(\text{At top of trajectory,}\ \mathbf{j} \text{-component of velocity = 0.}\)

\(\Rightarrow\ \text{Speed at top is the}\ \mathbf{i} \text{-component of velocity at}\ t=1.339:\)

\(\text{Speed}\ =10e^{-0.8 \times 1.339} = 3.426\ \text{m s}^{-1}\ \text{(3 d.p.)}\)
 

d.    \(\text{Using}\ \ F=m \ddot{x}:\)

\(\mathbf{v}(t)=10 e^{-0.8 t}\mathbf{i} +\left(36.5 e^{-0.8 t}-12.5\right) \mathbf{j}\)

\(\mathbf{a} =\dfrac{d v }{d t}=-8 e^{-0.8 t} \mathbf{i} -29.2 e^{-0.8 t} \mathbf{j}\)

\(\text{When} \ \ t=1.339:\)

\(\mathbf{a}=-8 e^{-0.8 \times 1.339}\mathbf{i}-29.2 e^{-0.8 \times 1.339}\mathbf{j}=-2.74 1\,\mathbf{i} -10\, \mathbf{j}\)
 

\(\text{Magnitude of acceleration}\)

\(=\sqrt{(2.741)^2+(10)^2}=10.36885 \ldots\ \text{ms}^{-2}\)
 

\(\therefore \ \text{Magnitude of force}\)

\(=8 \times 10.36885 \ldots =82.951 \ \text{N (3 d.p.)}\)