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ACSEE (Form 6) Physics — practice by topic

Browse 35 questions across 9 Physics topics. Choose what you want to revise, then open a question to check its answer and worked solution.

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Worked practice questions

For $P=\dfrac{F}{\pi R^2}$ with percentage errors of $\pm2$ in $F$ and $\pm1$ in $R$, the percentage error in $P$ is:

  1. $\pm1\%$
  2. $\pm4\%$
  3. $\pm3\%$
  4. $\pm5\%$
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Correct answer: B. $\pm4\%$

1. Take logarithms: $\ln P=\ln F-\ln\pi-2\ln R$. 2. Differentiating gives $\dfrac{\Delta P}{P}=\dfrac{\Delta F}{F}+2\dfrac{\Delta R}{R}$ — the terms ADD when errors are maximised. 3. Multiply through by 100%: $\pm(2\%+2\times1\%)$. 4. That gives $\pm4\%$. 5. Subtracting instead of adding gives $-1\%$, which was the recorded error and shows a misunderstanding of maximising errors. _Source: NECTA ACSEE 2023 Physics 131/1, Question 1: Measurement_

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The correct way to write the dimensions of area is:

  1. $\text{Area}=[L]^2$
  2. $[\text{Area}]=L^2$
  3. $[\text{Area}]=L^2[\;]$
  4. $[\text{Area}]=[L][L][\;]$
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Correct answer: B. $[\text{Area}]=L^2$

1. Square brackets denote 'the dimensions of' the quantity inside them. 2. So the bracket belongs around the quantity, on the left of the equation. 3. Area is a length multiplied by a length, giving $L^2$ on the right. 4. The correct notation is therefore $[\text{Area}]=L^2$. 5. Writing $\text{Area}=[L]^2$ or trailing an empty bracket misplaces the notation, both of which were recorded errors. _Source: NECTA ACSEE 2023 Physics 131/1, Question 1: Measurement_

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A package is released from an aircraft flying level. Its time of fall is found from:

  1. $t=\sqrt{\dfrac{2y}{g}}$
  2. $t=\dfrac{v_0\sin\theta}{g}$
  3. $t=\dfrac{2v_0}{g}$
  4. $t=\dfrac{y}{v_0}$
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Correct answer: A. $t=\sqrt{\dfrac{2y}{g}}$

1. Released from level flight, the package starts with zero vertical velocity, so $\theta=0$. 2. The vertical equation $y=v_0\sin\theta\,t+\tfrac12 gt^2$ therefore reduces to $y=\tfrac12 gt^2$. 3. Rearranging for $t$ gives $t=\sqrt{\dfrac{2y}{g}}$. 4. The formula $t=\dfrac{v_0\sin\theta}{g}$ applies to a projectile launched at an angle, and here it evaluates to zero. 5. Missing that a horizontal projection has $\theta=0$ was the recorded error on this question. _Source: NECTA ACSEE 2023 Physics 131/1, Question 2: Mechanics (Projectile Motion)_

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