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Mechanics, EXT2 M1 SM-Bank 5

A student throws a ball for his dog to retrieve. The position vector of the ball, relative to an origin \(O\) at ground level \(t\) seconds after release, is given by  \(  \underset{\sim}{\text{r}}{}_\text{B} (t)=5 t \underset{\sim}{\text{i}}+7 t \underset{\sim}{\text{j}}+(15 t-4.9 t^2+1.5) \underset{\sim}{\text{k}} \). Displacement components are measured in metres, where \(\underset{\sim}{\text{i}}\) is a unit vector to the east, \(\underset{\sim}{\text{j}}\) is a unit vector to the north and \( \underset{\sim} {\text{k}}\) is a unit vector vertically up.

Calculate the total vertical distance, in metres, travelled by the ball before it hits the ground. Give your answer correct to one decimal place.   (3 marks)

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\(24.5\ \text{m} \)

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\(\text{Upwards distance}\ (z) = 15t-4.9t^2+1.5\)

\(\dfrac{dz}{dt}=15-9.8t\)

\(\text{Find}\ t\ \text{when}\ \dfrac{dz}{dt}=0\ \text{(vertical max):} \)

\(15-9.8t=0\ \ \Rightarrow \ \ t=\dfrac{15}{9.8} \)

\(z\Big{(}t=\frac{15}{9.8}\Big{)} = 15 \times \Big{(}\dfrac{15}{9.8}\Big{)}-4.9 \times \Big{(}\dfrac{15}{9.8}\Big{)}^2 + 1.5 \approx 12.98\ \text{m} \)
  

\(\text{At}\ \ t=0, \ z=1.5 \)

\(\therefore \text{Total vertical distance}\ = (12.98-1.5)+12.98 = 24.5\ \text{m (1 d.p.)} \)

Mechanics, EXT2 M1 2023 HSC 9 MC

A particle travels along a curve from \(O\) to \(E\) in the \(x y\)-plane, as shown in the diagram.
 

The position vector of the particle is \(r\), its velocity is \( v \), and its acceleration is \( a \).

While travelling from \(O\) to \(E\), the particle is always slowing down.

Which of the following is consistent with the motion of the particle?

  1. \( r \cdot v \leq 0\)  and  \( a \cdot v \geq 0\)
  2. \( r \cdot  v \leq 0\)  and  \( a \cdot v \leq 0\)
  3. \( r \cdot v \geq 0\)  and  \( a \cdot v \geq 0\)
  4. \( r \cdot v \geq 0\)  and  \(a \cdot v \leq 0\)
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\(D\)

Show Worked Solution

\(\text{By elimination:}\)

\(x\ \text{and}\ y\ \text{components of vector}\ r\ \text{are both positive.}\)

\(\text{The velocity of vector}\ v\ \text{is tangential to the curve}\ OE\ \text{and}\)

\(x\ \text{and}\ y\ \text{components of vector}\ v\ \text{are also both positive.}\)

\( r \cdot v \geq 0\ \ \text{(eliminate}\ A\ \text{and}\ B).\)

\(\text{Particle is slowing down so acceleration must be acting, in part,}\)

\(\text{in the opposite direction to velocity.}\)

\( a \cdot v \leq 0\ \ \text{(eliminate}\ C).\)

\(\Rightarrow D\)

Mechanics, EXT2 M1 2023 HSC 12c

An object with mass \(m\) kilograms slides down a smooth inclined plane with velocity \( \underset{\sim}{v}(t)\), where \(t\) is the time in seconds after the object started sliding down the plane. The inclined plane makes an angle \(\theta\) with the horizontal, as shown in the diagram. The normal reaction force is \(\underset{\sim}{R}\). The acceleration due to gravity is \(\underset{\sim}{g}\) and has magnitude \(g\). No other forces act on the object.

The vectors \(\underset{\sim}{i}\) and \( \underset{\sim}{j} \) are unit vectors parallel and perpendicular, respectively, to the plane, as shown in the diagram.
 

  1. Show that the resultant force on the object is  \(\underset{\sim}{F}=-(m g \ \sin \theta) \underset{\sim}{i}\).  (2 marks)

  2. Given that the object is initially at rest, find its velocity \(\underset{\sim}{v}(t)\) in terms of \(g\), \(\theta, t\) and \(\underset{\sim}{i}\).  (2 marks)

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i.    \(\text{Proof (See Worked Solution)} \)

ii.   \(\underset{\sim}{v}=-gt\ \sin \theta \ \underset{\sim}{i} \)

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i.       
         

\(\text{Resolving forces in}\ \underset{\sim}{j} \ \text{direction:} \)

\( {\underset{\sim}{F}}_\underset{\sim}{j} = \underset{\sim}{R} + m\underset{\sim}{g}\ \cos \theta = 0\ \ \text{(in equilibrium)} \)

\(\text{Resolving forces in}\ \underset{\sim}{i} \ \text{direction:} \)

\( {\underset{\sim}{F}}_\underset{\sim}{i} = -m\underset{\sim}{g} \ \sin \theta \ \ \ \text{(down slope)} \)

\(\therefore \text{Resultant force:}\ \ \underset{\sim}{F}=-(m g \ \sin \theta) \underset{\sim}{i} \)
 

♦ Mean mark (i) 50%.

ii.   \(\text{Using}\ \ \underset{\sim}{F}=m \underset{\sim}{a}: \)

\(m \underset{\sim}{a}\) \(=-mg\ \sin \theta \ \underset{\sim}{i} \)  
\(\underset{\sim}{a}\) \(=-g\ \sin \theta \ \underset{\sim}{i} \)  
\(\underset{\sim}{v}\) \(= \displaystyle \int \underset{\sim}{a}\ dt \)  
  \(=-gt\ \sin \theta +c \)  

 
\(\text{When}\ \ t=0,\ \ \underset{\sim}{v}=0\ \ \Rightarrow \ \ c=0 \)

\(\therefore \underset{\sim}{v}=-gt\ \sin \theta \ \underset{\sim}{i} \)

Mechanics, EXT2 V1 2021 HSC 8 MC

A particle is travelling from `A` on the curve joining `A` to `B` . At a particular time, the particle is at point `P` and has velocity  `underset~v` , as shown in the diagram. 
 
The speed of the particle is increasing.
 
Which of the following diagrams shows an acceleration,  `underset~a`, which would allow the particle to follow the curve to `B`?
 

 
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`D`

Show Worked Solution

`text{The horizontal velocity of the particle}`

`text{is increasing as the curve moves from}`

`A \ text{to} \ B.`

`=> D`