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ENGINEERING, PPT 2023 HSC 24c

A roller coaster car is at rest at point \(A\) as shown.

Determine the height, \(h\), of the roller coaster at point \(B\) if it is travelling at 30 m/s at that point. Assume no energy losses.   (3 marks)

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\(10\ \text{metres}\)

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Using the Conservation of Energy:

\(\begin{aligned}
mgh_{\text{A}}+ \dfrac{1}{2} mv_\text{A}^2 & =mgh_{\text{B}} + \dfrac{1}{2} mv_\text{B}^2 \\
550+\dfrac{1}{2} \times 0^2 & =10h_\text{B}+ \dfrac{1}{2} \times 30^2 \\
550 & =10h_{\text{B}}+450 \\
h_{\text{B}} & = \dfrac{(550-450)}{10} =10\ \text {metres }\\
\end{aligned}\)

♦ Mean mark 48%.

ENGINEERING, PPT 2018 HSC 9 MC

The diagram shows a drop hammer being used to drive a pile into the ground.
 

Which formula gives the height `h` required for the drop hammer to impact the pile at a pre-determined velocity `v` ?

  1. `h=(v^2)/(2g)`
  2. `h=(2g)/v^2`
  3. `h=(v)/(2g)`
  4. `h=(2g)/v`
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`A`

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`text{Applying conservation of energy:}`

`PE`  `=KE`  
`mgh` `= (mv^2)/2`  
`h` `= (v^2)/(2g)`  

 
`=>A`

ENGINEERING, PPT 2017 HSC 22c

A bike with a rider rolls down a hill without braking from a standing start at point `A`.
 

The combined weight of the bike and the rider is 100 kg.

Calculate the speed of the rider at point `B`. (Assume no wind resistance and 100 % efficiency.)   (2 marks)

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`20\ text{m/s}`

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`KE_text(bottom) = PE_text(top)`

`1/2 mv^2` `=mgh`  
`v^2` `=2gh`  
`v` `=sqrt(2xx10xx20)`  
  `=sqrt(400)`  
  `=20\ text{m/s}`  
Mean mark 58%.

ENGINEERING, PPT 2019 HSC 23a

A scooter and rider enter the half pipe shown at a velocity of 3 m/s. The scooter and rider have a combined mass of 55 kg.
 

Calculate the maximum height, `h`, the scooter and rider will reach above the other side of the pipe wall. Assume no loss of energy.   (3 marks)

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`0.45\ text{metres}

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  • No loss of energy so mechanical energy conserved
  • `text{KE}_2 = 0\ text{J}`
  • `g=10\ text{ms}^(-2)`
  • `h_2=2 + h`
`:.\ KE_2 + PE_2` `= KE_1 + PE_1`  
`0 + mgh_2` `= 1/2mv_1^2 + mgh_1`  
`10 xx (2 + h)` `= (1/2 xx 3^2) + (10 xx 2)`  
`20 + 10h` `= 4.5 + 20`  
`h` `= 4.5/10= 0.45\ text{metres}`