Before you read my working
Cover the solution. Spend two minutes on the question. Then uncover one step at a time. If your method is different but your physics is the same, that is fine — I care about the equation, the substitution, and the unit.
Example 1 · Speed and units · 1.1, 1.4
A coach journey
A coach travels $27\,\text{km}$ in $25\,\text{minutes}$. Calculate the average speed in m/s.
What I write first
I refuse to divide $27$ by $25$. Those units are km and minutes. Convert both.
$$d = 27 \times 1000 = 27\,000\,\text{m}$$ $$t = 25 \times 60 = 1500\,\text{s}$$ $$v = \frac{d}{t} = \frac{27000}{1500} = 18\,\text{m/s}$$Sense-check: $18\,\text{m/s} \times 3.6 = 65\,\text{km/h}$. A coach on a main road. Believable. If you had got $180\,\text{m/s}$, that would be an aircraft.
Example 2 · Acceleration · 1.6
A trolley speeding up
A trolley is moving at $0.80\,\text{m/s}$. After $2.5\,\text{s}$ its velocity is $3.3\,\text{m/s}$ in the same direction. Calculate the acceleration.
$$u = 0.80\,\text{m/s},\quad v = 3.3\,\text{m/s},\quad t = 2.5\,\text{s}$$ $$a = \frac{v-u}{t} = \frac{3.3-0.80}{2.5} = \frac{2.5}{2.5} = 1.0\,\text{m/s}^2$$I keep one more significant figure in the working if I need to, but here it lands cleanly. Direction: same as the velocity, along the track.
Example 3 · $v^2 = u^2 + 2as$ · 1.10
A cyclist with no time given
A cyclist travelling at $4.0\,\text{m/s}$ accelerates uniformly at $1.5\,\text{m/s}^2$ over $20\,\text{m}$. Calculate the final speed.
List: $u=4.0$, $a=1.5$, $s=20$, $v=?$, $t$ not given. That is the giveaway for 1.10.
$$v^2 = u^2 + 2as = (4.0)^2 + 2(1.5)(20) = 16 + 60 = 76$$ $$v = \sqrt{76} = 8.72\ldots = 8.7\,\text{m/s}\ (2\,\text{s.f.})$$Write the square-root as its own line. Students who jump to $v = 76$ lose the accuracy mark. 2 s.f. because $4.0$ and $1.5$ are 2 s.f.
Example 4 · Velocity–time graph · 1.8, 1.9
Gradient and area on one graph
A runner accelerates from rest to $8.0\,\text{m/s}$ in $4.0\,\text{s}$, runs at $8.0\,\text{m/s}$ for $6.0\,\text{s}$, then decelerates to rest in $2.0\,\text{s}$.
(a) Acceleration in the first $4.0\,\text{s}$
$$a = \frac{8.0-0}{4.0} = 2.0\,\text{m/s}^2$$(b) Deceleration in the last $2.0\,\text{s}$
$$a = \frac{0-8.0}{2.0} = -4.0\,\text{m/s}^2 \quad \text{so deceleration } 4.0\,\text{m/s}^2$$(c) Distance for the whole run
Triangle + rectangle + triangle:
$$\tfrac{1}{2}(4.0)(8.0) + (6.0)(8.0) + \tfrac{1}{2}(2.0)(8.0) = 16 + 48 + 8 = 72\,\text{m}$$If you forget the last triangle you lose $8\,\text{m}$ and usually two marks. Draw the three shapes in the margin before you calculate.
Example 5 · Resultant and $F=ma$ · 1.15, 1.17
A crate being dragged
A crate of mass $40\,\text{kg}$ is pulled to the right by $120\,\text{N}$. Friction is $40\,\text{N}$ to the left. Calculate the acceleration.
Two steps. I will not let you combine them in one messy line on a first attempt.
$$F_{\text{resultant}} = 120 - 40 = 80\,\text{N} \text{ right}$$ $$a = \frac{F}{m} = \frac{80}{40} = 2.0\,\text{m/s}^2 \text{ right}$$Weight is $400\,\text{N}$ downwards. It does not enter this calculation because it is balanced by the normal contact force from the floor. Only the horizontal unbalanced force accelerates the crate along the floor.
Example 6 · Mass and weight · 1.18
The same bag on Earth and on the Moon
A bag has mass $8.0\,\text{kg}$. Take $g_{\text{Earth}} = 10\,\text{N/kg}$ and $g_{\text{Moon}} = 1.6\,\text{N/kg}$. Calculate both weights. A student writes “the mass on the Moon is $1.28\,\text{kg}$.” What have they done wrong?
$$W_{\text{E}} = 8.0 \times 10 = 80\,\text{N}$$ $$W_{\text{M}} = 8.0 \times 1.6 = 13\,\text{N}$$Mass stays $8.0\,\text{kg}$. The student has treated $W=mg$ as if it gave a new mass. That error appears in almost every class the first time we do this.
Example 7 · Stopping distance · 1.19, 1.20
Thinking distance from speed and reaction time
A car travels at $20\,\text{m/s}$. The driver’s reaction time is $0.80\,\text{s}$. The braking distance is $30\,\text{m}$.
(a) Thinking distance $= vt = 20 \times 0.80 = 16\,\text{m}$.
(b) Stopping distance $= 16 + 30 = 46\,\text{m}$.
(c) The driver is tired and reaction time becomes $1.4\,\text{s}$, same speed, same road. New thinking distance $= 20 \times 1.4 = 28\,\text{m}$. Braking distance is still $30\,\text{m}$. New stopping distance $= 58\,\text{m}$.
The explain mark: thinking distance increased because reaction time increased; braking distance unchanged because speed, mass, brakes and road are unchanged.
Example 8 · Hooke’s law · 1.23
Using a graph point properly
In the linear region, $4.0\,\text{N}$ produces an extension of $5.0\,\text{cm}$. Calculate $k$ and the force needed for a $2.0\,\text{cm}$ extension.
$$e = 0.050\,\text{m},\quad k = \frac{F}{e} = \frac{4.0}{0.050} = 80\,\text{N/m}$$ $$F = ke = 80 \times 0.020 = 1.6\,\text{N}$$Or, because it is proportional, half the extension needs half the force: $2.0\,\text{cm}$ is $2/5$ of $5.0\,\text{cm}$, so $F = (2/5)\times 4.0 = 1.6\,\text{N}$. Either route is full marks if the unit is right.
Example 9 · Explain · 1.21
A 4-mark terminal-velocity answer, marked
Question: Explain why a skydiver reaches a terminal velocity.
What I would award 4
Weight acts downwards and stays constant [1]. When she first jumps, air resistance is small, so there is a downward resultant force and she accelerates [1]. Air resistance increases as her speed increases, so the resultant force and the acceleration decrease [1]. Eventually air resistance equals weight, the resultant force is zero, her acceleration is zero and she falls at a constant (terminal) velocity [1].
What I would award 1 or 0
“Gravity pulls her down and then air resistance stops her.” — gravity/weight is never ‘turned off’; she does not stop; air resistance becomes equal to weight. “She reaches terminal velocity when there are no forces.” — 0. Forces are balanced.
When these feel easy, sit the five chapter quizzes and then the three 25-mark papers in exam conditions.