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

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

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

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

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

\(\text{Magnitude of acceleration}\)

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

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

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

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)

Show Answers Only

  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 2022 HSC 8 MC

As a projectile of mass `m` kilograms travels through air, it experiences a frictional force. The magnitude of this force is proportional to the square of the speed `v` of the projectile. The constant of proportionality is the positive number `k`. The position of the particle at time `t` is denoted by `([x],[y])`. The acceleration due to gravity is `g \ text{m s}^(-2)`.

Based on Newton's laws of motion, which equation models the motion of this projectile?

  1. `([0],[-mg])+kv([overset.x],[overset.y])=m([overset{..}x],[overset{..}y])`
  2. `([0],[-mg])-kv([overset.x],[overset.y])=m([overset{..}x],[overset{..}y])`
  3. `([0],[-mg])+kv^(2)([overset.x],[overset.y])=m([overset{..}x],[overset{..}y])`
  4. `([0],[-mg])-kv^(2)([overset.x],[overset.y])=m([overset{..}x],[overset{..}y])`
Show Answers Only

`B`

Show Worked Solution

`text{Friction}\ (F)\ text{works against velocity}`

`:.\ F prop v^2\ \ =>\ \ F=-kv^2\ \ (k>0)`

`text{→ Eliminate A and C}`
 


♦♦♦ Mean mark 26%.

`text{S}text{ince}\ \ v=abs(((dotx),(doty))):`

`- abs(kv((dotx),(doty))) = -kv abs(((dotx),(doty)))=-kv^2`

`=>B`