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Mechanics, EXT2 M1 2023 HSC 13c

A particle of mass 1 kg is projected from the origin with speed 40 m s\( ^{-1}\) at an angle 30° to the horizontal plane.

  1. Use the information above to show that the initial velocity of the particle is
  2.     \(\mathbf{v}(0)={\displaystyle\left(\begin{array}{cc}20 \sqrt{3} \\ 20\end{array}\right)} \).   (1 mark)

The forces acting on the particle are gravity and air resistance. The air resistance is proportional to the velocity vector with a constant of proportionality 4 . Let the acceleration due to gravity be 10 m s \( ^{-2}\).

The position vector of the particle, at time \(t\) seconds after the particle is projected, is \(\mathbf{r}(t)\) and the velocity vector is \(\mathbf{v}(t)\).
 

  1. Show that  \(\mathbf{v}(t)={\displaystyle \left(\begin{array}{cc}20 \sqrt{3} e^{-4 t} \\ \dfrac{45}{2} e^{-4 t}-\dfrac{5}{2}\end{array}\right)}\)  (3 marks)

  2. Show that  \(\mathbf{r}(t)=\left(\begin{array}{c}5 \sqrt{3}\left(1-e^{-4 t}\right) \\ \dfrac{45}{8}\left(1-e^{-4 t}\right)-\dfrac{5}{2} t\end{array}\right)\)  (2 marks)

  3. The graphs  \(y=1-e^{-4 x}\)  and  \(y=\dfrac{4 x}{9}\) are given in the diagram below.
     
     
  4. Using the diagram, find the horizontal range of the particle, giving your answer rounded to one decimal place.  (2 marks)

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  1. \(\text{Proof (See Worked Solutions)}\)
  2. \(\text{Proof (See Worked Solutions)}\)
  3. \(\text{Proof (See Worked Solutions)}\)
  4. \(8.7\ \text{metres}\)

Show Worked Solution

i. 

\(\underset{\sim}{v}(0)={\displaystyle\left(\begin{array}{cc} 40 \cos\ 30° \\ 40 \sin\ 30°\end{array}\right)} = {\displaystyle\left(\begin{array}{cc} 40 \times \frac{\sqrt3}{2} \\ 40 \times \frac{1}{2}\end{array}\right)}  = {\displaystyle\left(\begin{array}{cc}20 \sqrt{3} \\ 20\end{array}\right)} \)
 

ii.   \(\text{Air resistance:} \)

\(\underset{\sim}{F} = -4\underset{\sim}{v} = {\displaystyle\left(\begin{array}{cc} -4\dot{x} \\ -4\dot{y} \end{array}\right)} \)

\(\text{Horizontally:}\)

\(1 \times \ddot{x} \) \(=-4 \dot{x} \)  
\(\dfrac{d\dot{x}}{dt}\) \(=-4\dot{x}\)  
\(\dfrac{dt}{d\dot{x}}\) \(= -\dfrac{1}{4\dot{x}} \)  
\(t\) \(=-\dfrac{1}{4} \displaystyle \int \dfrac{1}{\dot{x}} \ d\dot{x} \)  
\(-4t\) \(=\ln |\dot{x}|+c \)  

 
\(\text{When}\ \ t=0, \ \dot{x}=20\sqrt3 \ \ \Rightarrow\ \ c=-\ln{20\sqrt3} \)

\(-4t\) \(=\ln|\dot{x}|-\ln 20\sqrt3 \)  
\(-4t\) \(=\ln\Bigg{|}\dfrac{\dot{x}}{20\sqrt{3}} \Bigg{|} \)  
\(\dfrac{\dot{x}}{20\sqrt{3}} \) \(=e^{-4t} \)  
\(\dot{x}\) \(=20\sqrt{3}e^{-4t}\)  

 
\(\text{Vertically:} \)

\(1 \times \ddot{y} \) \(=-1 \times 10-4 \dot{y} \)  
\(\dfrac{d\dot{y}}{dt}\) \(=-(10+4\dot{y})\)  
\(\dfrac{dt}{d\dot{y}}\) \(= -\dfrac{1}{10+4\dot{y}} \)  
\(t\) \(=- \displaystyle \int \dfrac{1}{10+4\dot{y}} \ d\dot{y} \)  
\(-4t\) \(=- \displaystyle \int \dfrac{4}{10+4\dot{y}} \ d\dot{y} \)  
\(-4t\) \(=\ln |10+4\dot{y}|+c \)  

 
\(\text{When}\ \ t=0, \ \dot{y}=20 \ \ \Rightarrow\ \ c=-\ln{90} \)

\(-4t\) \(=\ln|10+4\dot{y}|-\ln 90 \)  
\(-4t\) \(=\ln\Bigg{|}\dfrac{10+4\dot{y}}{\ln{90}} \Bigg{|} \)  
\(\dfrac{10+4\dot{y}}{90} \) \(=e^{-4t} \)  
\(4\dot{y}\) \(=90e^{-4t}-10\)  
\(\dot{y}\) \(=\dfrac{45}{2} e^{-4t}-\dfrac{5}{2} \)  

 
\(\therefore \underset{\sim}v={\displaystyle \left(\begin{array}{cc}20 \sqrt{3} e^{-4 t} \\ \dfrac{45}{2} e^{-4 t}-\dfrac{5}{2}\end{array}\right)}\) 

 
iii.   \(\text{Horizontally:}\)

\(x\) \(= \displaystyle \int \dot{x}\ dx\)  
  \(= \displaystyle \int 20\sqrt3 e^{-4t}\ dt \)  
  \(=-5\sqrt3 e^{-4t}+c \)  

 
\(\text{When}\ \ t=0, \ x=0\ \ \Rightarrow\ \ c=5\sqrt3 \)

\(x\) \(=5\sqrt3-5\sqrt3 e^{-4t} \)  
  \(=5\sqrt3(1-e^{-4t}) \)  

 
\(\text{Vertically:}\)

\(y\) \(= \displaystyle \int \dot{y}\ dx\)  
  \(= \displaystyle \int \dfrac{45}{2} e^{-4t}-\dfrac{5}{2}\ dt \)  
  \(=-\dfrac{45}{8}e^{-4t}-\dfrac{5}{2}t+c \)  

 
\(\text{When}\ \ t=0, \ y=0\ \ \Rightarrow\ \ c= \dfrac{45}{8} \)

\(y\) \(=\dfrac{45}{8}-\dfrac{45}{8} e^{-4t}-\dfrac{5}{2}t \)  
  \(=\dfrac{45}{8}(1-e^{-4t})-\dfrac{5}{2} \)  

 
\(\therefore \underset{\sim}{r}=\left(\begin{array}{c}5 \sqrt{3}\left(1-e^{-4 t}\right) \\ \dfrac{45}{8}\left(1-e^{-4 t}\right)-\dfrac{5}{2} t\end{array}\right)\)
 

iv.   \(\text{Range}\ \Rightarrow\ \text{Find}\ \ t\ \ \text{when}\ \ y=0: \)

\(\dfrac{45}{8}(1-e^{-4t})-\dfrac{5}{2}t \) \(=0\)  
\(\dfrac{45}{8}(1-e^{-4t}) \) \(=\dfrac{5}{2}t \)  
\(1-e^{-4t}\) \(=\dfrac{4}{9}t \)  

 
\(\text{Graph shows intersection of these two graphs.}\)

\(\Rightarrow \text{Solution when}\ \ t\approx 2.25\)

\(\therefore\ \text{Range}\) \(=5\sqrt3(1-e^{(-4 \times 2.25)}) \)  
  \(=8.659…\)  
  \(=8.7\ \text{metres (to 1 d.p.)}\)  

♦ Mean mark (iv) 43%.
 

Mechanics, EXT2 M1 EQ-Bank 4

A torpedo with a mass of 80 kilograms has a propeller system that delivers a force of `F` on the torpedo, at maximum power. The water exerts a resistance on the torpedo proportional to the square of the torpedo's velocity `v`.

  1. Explain why  `(dv)/(dt) = 1/80 (F - kv^2)`
     
    where `k` is a positive constant.  (1 mark)

  2. If the torpedo increases its velocity from  `text(10 ms)\ ^(−1)`  to  `text(20 ms)\ ^(−1)`, show that the distance it travels in this time, `d`, is given by
     
         `d = 40/k log_e((F - 100k)/(F - 400k))`  (3 marks)

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
Show Worked Solution

i.   `R ∝ v^2`

`R = −kv^2\ \ (k\ text{is positive constant})`

`text(Newton’s 2nd Law:)`

`text(Net Force)= mddotx` `= F – R`
`80ddotx` `= F – kv^2`
`ddotx` `= 1/80 (F – kv^2)`
`(dv)/(dt)` `= 1/80 (F – kv^2)`

 

ii.    `v · (dv)/(dx)` `= 1/80 (F – kv^2)`
  `(dv)/(dx)` `= (F – kv^2)/(80v)`
  `(dx)/(dv)` `= (80v)/(F – kv^2)`
  `x` `= −40 int (−2v)/(F – kv^2)\ dv`
    `= −40/k log_e (F – kv^2) + C`

 

`text(When)\ \ v = 10:`

`x_1 = −40/k log_e (F – 100k) + C`
 

`text(When)\ \ v = 20:`

`x_2 = −40/k log_e (F – 400k) + C`
 

`d` `= x_2 – x_1`
  `= −40/k log_e (F – 400k) + 40/k log_e (F – 100k)`
  `= 40/k log_e ((F – 100k)/(F – 400k))`

Mechanics, EXT2 M1 EQ-Bank 1

A canon ball of mass 9 kilograms is dropped from the top of a castle at a height of `h` metres above the ground.

The canon ball experiences a resistance force due to air resistance equivalent to  `(v^2)/500`, where `v` is the speed of the canon ball in metres per second. Let  `g=9.8\ text(ms)^-2`  and the displacement, `x` metres at time `t` seconds, be measured in a downward direction.

  1. Show the equation of motion is given by
     
         `ddotx = g - (v^2)/4500`  (1 mark)

  2. Show, by integrating using partial fractions, that
     
         `v = 210((e^(7/75 t) - 1)/(e^(7/75 t) + 1))`  (5 marks)

     

  3. If the canon hits the ground after 4 seconds, calculate the height of the castle, to the nearest metre.  (3 marks)

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `78\ text(metres)`
Show Worked Solution

i.   `text(Newton’s 2nd Law:)`

`text(Net Force)` `= mddotx`
`mddotx` `= mg – (v^2)/500`
`9ddotx` `= 9g – (v^2)/500`
`ddotx` `= g – (v^2)/4500`

 

ii.    `(dv)/(dt)` `= g – (v^2)/4500`
    `= (44\ 100 – v^2)/4500`

 
`(dt)/(dv) = 4500/(44\ 100 – v^2)`
 

`text(Using Partial Fractions:)`

`1/(44\ 100 – v^2) = A/(210- v) + B/(210 – v)`

`A(210 – v) + B(210 + v) = 1`
 

`text(If)\ \ v = 210,`

`420B = 1 \ => \ B = 1/420`
 

`text(If)\ \ v = −210,`

`420A = 1 \ => \ A = 1/420`
 

`t` `= int 4500/(210^2 – v^2)\ dv`
  `= 4500/420 int 1/(210 + v) + 1/(210 – v)\ dv`
  `= 75/7 [ln(210 + v) – ln(210 – v)] + c`
  `= 75/7 ln((210 + v)/(210 – v)) + c`

 

`text(When)\ \ t = 0, v = 0:`

`0` `= 75/7 ln(210/210) + c`
`:. c` `= 0`

 

` t` `= 75/7 ln((210 + v)/(210 – v))`
`7/75 t` `= ln((210 + v)/(210 – v))`
`e^(7/75 t)` `= (210 + v)/(210 – v)`
`e^(7/75 t) (210 – v)` `= 210 + v`
`210e^(7/75 t) – 210` `= ve^(7/75 t) + v`
`210(e^(7/75 t) – 1)` `= v(e^(7/75 t) + 1)`
`:. v` `= 210((e^(7/75 t) – 1)/(e^(7/75 t) + 1))`

 

iii.    `v · (dv)/(dx)` `= (44\ 100 – v^2)/4500`
  `(dx)/(dv)` `= (4500v)/(44\ 100 – v^2)`
  `int (dx)/(dv)\ dv` `= −4500/2 int (−2v)/(44\ 100 – v^2)\ dv`
  `x` `= −2550 log_e(44\ 100 – v^2) + c`

 
`text(When)\ \ x = 0, v = 0:`

`0` `= −2250 log_e(44\ 100) + c`
`c` `= 2250 log_e(44\ 100)`
`x` `= −2250 log_e(44\ 100 – v^2) + 2250 log_e44\ 100`
  `= 2250 log_e((44\ 100)/(44\ 100 – v^2))`

 
`text(When)\ \ t = 4:`

`v` `= 210((e^(28/75) – 1)/(e^(28/75) + 1))`
  `= 38.7509…`

 

`:. h` `= 2250 log_e((44\ 100)/(44\ 100 – 38.751^2))`
  `= 77.94…`
  `= 78\ text(metres)`

Mechanics, EXT2 M1 2019 HSC 14b

A parachutist jumps from a plane, falls freely for a short time and then opens the parachute. Let t be the time in seconds after the parachute opens, `x(t)`  be the distance in metres travelled after the parachute opens, and  `v(t)`  be the velocity of the parachutist in `text(ms)^(-1)`.

The acceleration of the parachutist after the parachute opens is given by

`ddot x = g - kv,`

where `g\ text(ms)^(-2)` is the acceleration due to gravity and `k` is a positive constant.

  1. With an open parachute the parachutist has a terminal velocity of  `w\ text(ms)^(-1)`.

     

    Show that  `w = g/k`.  (1 mark)

    At the time the parachute opens, the speed of descent is `1.6 w\ text(ms)^(-1)`.

  2. Show that it takes  `1/k log_e 6`  seconds to slow down to a speed of `1.1w\ text(ms)^(-1)`.  (4 marks)

  3. Let  `D`  be the distance the parachutist travels between opening the parachute and reaching the speed `1.1w\ text(ms)^(-1)`.

     

     

    Show that  `D = g/k^2 (1/2 + log_e 6)`.  (3 marks)

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  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `text(Proof)\ text{(See Worked Solutions)}`
Show Worked Solution

i.   `v_T=w\ \  text(when)\ \ ddot x = 0`

`0` `= g – kw`
`w` `= g/k`

 

ii.   `text(Show)\ \ t = 1/k log_e 6\ \ text(when)\ \ v = 1.1w`

`(dv)/(dt)` `= g – kv`
`(dt)/(dv)` `= 1/(g – kv)`
`t` `= int 1/(g – kv)\ dv`
  `= -1/k ln(g – kv) + C`

 
`text(When)\ \ t = 0,\ \ v = 1.6w`

`0` `= -1/k ln(g – 1.6 kw) + C`
`C` `= 1/k ln(g – 1.6 kw)`
`t` `= 1/k ln (g – 1.6kw) – 1/k ln(g – kv)`
  `= 1/k ln((g – 1.6 kw)/(g – kv))`

 
`text(Find)\ \ t\ \ text(when)\ \ v = 1.1w`

`t` `= 1/k ln((g – 1.6 k xx g/k)/(g – 1.1k xx g/k))`
  `=1/k ln((g – 1.6 g)/(g – 1.1g))`
  `=1/k((-0.6g)/(-0.1g))`
  `= 1/k ln 6`

 

iii.    `v ⋅ (dv)/(dx)` `= g – kv`
  `(dv)/(dx)` `= (g – kv)/v`
  `(dx)/(dv)` `= v/(g – kv)`
  `x` `= int v/(g – kv)\ dv`
    `= 1/k int (kv)/(g – kv)\ dv`
    `= -1/k int 1 – g/(g – kv)\ dv`

 

`:. D` `= -1/k int_(1.6w)^(1.1w) 1 – g/(g – kv)\ dv`
  `= 1/k int_(1.1w)^(1.6w) 1 – g/(g – kv)\ dv`
  `= 1/k[v + g/k ln (g – kv)]_(1.1w)^(1.6w)`
  `= g/k^2[(kv)/g + ln (g – kv)]_(1.1w)^(1.6w)`
  `= g/k^2[((1.6kw)/g + ln (g – 1.6kw)) – ((1.1 kw)/g + ln (g – 1.1kw))]`
  `= g/k^2[1.6 + ln ((g – 1.6kw)/(g – 1.1kw)) – 1.1]`
  `= g/k^2(0.5 + ln 6)`

Mechanics, EXT2 M1 2012 HSC 13a

An object on the surface of a liquid is released at time  `t = 0`  and immediately sinks. Let  `x`  be its displacement in metres in a downward direction from the surface at time  `t`  seconds.

The equation of motion is given by

`(dv)/(dt) = 10 − (v^2)/40`,

where  `v`  is the velocity of the object.  

  1. Show that  `v = (20(e^t − 1))/(e^t + 1)`.   (4 marks)

  2. Use  `(dv)/(dt) = v (dv)/(dx)`  to show that  
     
         `x = 20\ log_e(400/(400 − v^2))`   (2 marks)

     

  3. How far does the object sink in the first 4 seconds?   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `text(See Worked Solutions.)`
  3. `40log_e((e^4 + 1)/(2e^2))\ text(m)`
Show Worked Solution
i.   `(dv)/(dt)` `= 10 − (v^2)/40`
  `(dv)/(dt)` `= (400 − v^2)/40`
  `dt` `= 40/(400 −v^2)\ dv`
  `int dt` `=int 40/(400 −v^2)\ dv`
  `t` `= int (1/(20 + v) + 1/(20 − v))\ dv`
    `= log_e(20 + v) − log_e(20 − v) + c`

 

`text(When)\ \ t = 0, v = 0\ \ => \ c = 0`

`t=` ` log_e((20 + v)/(20 − v))`
`e^t=` ` (20 + v)/(20 − v)`
`20e^t-ve^t=` ` 20 + v`
`v+ ve^t=` ` 20e^t − 20`
`v(1+e^t)=` `20(e^t − 1)`
`v=` ` (20(e^t − 1))/(e^t + 1)\ \ \ \ text(… as required)`

 

ii.   `v (dv)/(dx)` `= 10 − (v^2)/40`
  `(40v\ dv)/(400 − v^2)` `= dx`
`int dx` `= int (40v)/(400 − v^2)\ dv`
 `x` `= -20log_e(400 − v^2) + c`

 

`text(When)\ \ x = 0, v = 0\ \ \ => c = 20log_e 400`

`:.x` `= 20log_e400 − 20log_e(400 − v^2)`
  `= 20log_e((400)/(400 − v^2))\ \ \ \ text(… as required)`

 

iii.  `text(When)\ \ t = 4,\  v = (20(e^4 − 1))/(e^4 + 1)`

`x` `= 20log_e[400/(400 −((20(e^4 − 1))/(e^4 + 1))^2)]`
  `= 20log_e[((e^4 + 1)^2)/((e^4 + 1)^2 − (e^4 − 1)^2)]`
  `= 20log_e(((e^4 + 1)^2)/((e^4+1 + e^4 − 1)(e^4 + 1 − e^4 + 1)))`
  `= 20log_e(((e^4 + 1)^2)/(4e^4))`
  `= 40log_e\ (e^4 + 1)/(2e^2)\ \ text(metres)`