Worked Solutions
Physics — Worked Solutions (Year 10 Science)
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Worked examples for Year 10 Science physics. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Watch your units and always show the formula before substituting numbers.
Example 1 — Acceleration of a cyclist
Question
A cyclist starts from rest and reaches a speed of $12\ \text{m/s}$ in $6\ \text{s}$ along a straight path. Calculate the cyclist's acceleration, and state what the answer means.
Solution
Acceleration is the change in speed divided by the time taken: $a = \dfrac{v - u}{t}$.
Starting from rest means $u = 0$, with $v = 12\ \text{m/s}$ and $t = 6\ \text{s}$.
$a = \dfrac{12 - 0}{6} = 2\ \text{m/s}^2$.
That means the speed increases by $2\ \text{m/s}$ every second. Always quote the units — $\text{m/s}^2$ is what marks acceleration apart from speed.
Let's begin with what acceleration actually is — how quickly the speed is changing. We work it out with $a = \dfrac{v - u}{t}$, where $u$ is the starting speed and $v$ is the final speed.
"Starts from rest" is a clue: it tells us $u = 0\ \text{m/s}$. We're also given $v = 12\ \text{m/s}$ and $t = 6\ \text{s}$.
Substituting in: $a = \dfrac{12 - 0}{6} = 2\ \text{m/s}^2$.
So the cyclist speeds up by $2\ \text{m/s}$ for each second that passes. The units $\text{m/s}^2$ make sense because we're measuring a change in speed (m/s) per second.
Formula: $a = \dfrac{v - u}{t}$.
- $u = 0\ \text{m/s}$ (from rest)
- $v = 12\ \text{m/s}$, $t = 6\ \text{s}$
- $a = \dfrac{12 - 0}{6} = 2\ \text{m/s}^2$
Meaning: speed rises by $2\ \text{m/s}$ each second.
Where the marks go
- 1 mark: Correct formula $a = \dfrac{v - u}{t}$ with $u = 0$ identified
- 1 mark: Correct value $a = 2\ \text{m/s}^2$ with units
- 1 mark: States the meaning (speed increases by $2\ \text{m/s}$ each second)
Key idea
Acceleration is the rate of change of speed: $a = \dfrac{v - u}{t}$, measured in $\text{m/s}^2$.
Example 2 — Kinetic energy of a skateboarder
Question
A skateboarder of mass $50\ \text{kg}$ is moving at $4\ \text{m/s}$. Calculate the skateboarder's kinetic energy, then explain what happens to that energy when she comes to a stop using the brakes.
Solution
Kinetic energy is the energy of motion: $E_k = \dfrac{1}{2}mv^2$.
With $m = 50\ \text{kg}$ and $v = 4\ \text{m/s}$: $E_k = \dfrac{1}{2}(50)(4)^2 = \dfrac{1}{2}(50)(16) = 400\ \text{J}$.
When she brakes, that $400\ \text{J}$ doesn't vanish — energy is conserved. Friction transforms it into heat (and a little sound).
Square the speed before multiplying — forgetting the square is the most common mistake here.
Kinetic energy is the energy something has because it is moving, and we find it with $E_k = \dfrac{1}{2}mv^2$.
Here $m = 50\ \text{kg}$ and $v = 4\ \text{m/s}$. The speed is squared, so $4^2 = 16$ first, then $E_k = \dfrac{1}{2} \times 50 \times 16 = 400\ \text{J}$.
Now, why does the speed get squared? Because doubling your speed gives four times the energy — that's why crashes at higher speeds are so much more dangerous.
When she brakes, energy can't be destroyed, only transformed. The law of conservation of energy tells us the $400\ \text{J}$ of motion becomes heat in the brakes and ground (with a small amount of sound).
Formula: $E_k = \dfrac{1}{2}mv^2$.
- $m = 50\ \text{kg}$, $v = 4\ \text{m/s}$
- $v^2 = 16$
- $E_k = \dfrac{1}{2}(50)(16) = 400\ \text{J}$
On braking: energy conserved → kinetic energy transformed to heat (and sound) by friction.
Where the marks go
- 1 mark: Correct formula $E_k = \dfrac{1}{2}mv^2$
- 1 mark: Correct substitution including squaring the speed ($v^2 = 16$)
- 1 mark: Correct value $E_k = 400\ \text{J}$ with units
- 1 mark: Explains energy is transformed to heat/sound (conservation of energy)
Key idea
Kinetic energy $E_k = \dfrac{1}{2}mv^2$ depends on the speed squared; energy is never destroyed, only transformed.