A.1 Kinematics – IB Physics exam questions with mark scheme

1Standard5 marks

A stone is thrown horizontally at $15\ \mathrm{m\,s^{-1}}$ from the top of a cliff 45 m above the sea. Ignore air resistance and use $g = 9.81\ \mathrm{m\,s^{-2}}$.

(a) Determine the time taken for the stone to reach the sea. [2]

(b) Calculate the horizontal distance from the foot of the cliff at which the stone lands. [1]

Vertical motion from rest: $s = \tfrac12 g t^2$ (initial vertical velocity is zero)(1)
$t = \sqrt{2\times45/9.81} = 3.03\ \mathrm{s}$ (3.0 s)(1)
$x = u t = 15\times3.03 = 45\ \mathrm{m}$ (45 m; allow 45.4 m)(1)
Vertical speed $v_y = gt = 29.7\ \mathrm{m\,s^{-1}}$ (29.7)(1)
Speed $=\sqrt{15^2+29.7^2} = 33\ \mathrm{m\,s^{-1}}$ (33)(1)
Answer: (a) 3.03 s (b) 45 m (c) 33 m s$^{-1}$
Examiner tip: Treat horizontal and vertical motion separately and combine the components as vectors at the end.

2Core4 marks

An athlete runs exactly half a lap of a circular running track. The full lap has a length of 400 m. She takes 31.0 s.

(a) Distinguish between distance and displacement. [1]

(b) Calculate the average speed of the athlete. [1]

(a) distance is the total length of path travelled (scalar); displacement is the change in position / straight-line distance from start to finish in a stated direction (vector). Must convey both ideas: direction / vector for displacement(1)
(b) $\dfrac{200}{31.0} = 6.45\ \mathrm{m\,s^{-1}}$ (allow 6.5)(1)
(c) displacement = diameter of the track $= \dfrac{400}{\pi} = 127\ \mathrm{m}$ (allow $2r$ with $r = 400/2\pi = 63.7$ m)(1)
$\dfrac{127}{31.0} = 4.1\ \mathrm{m\,s^{-1}}$ (accept 4.11; ECF from wrong displacement; award [0] for (c) if 200 m is used)(1)
Answer: (a) see mark scheme (b) 6.45 m s$^{-1}$ (c) 4.1 m s$^{-1}$
Examiner tip: Average velocity uses the straight-line displacement (the diameter), not the 200 m run around the bend.

3Core4 marks

A driver travelling at $24\ \mathrm{m\,s^{-1}}$ sees a hazard ahead. Her reaction time is 0.65 s, during which the car keeps moving at constant speed. She then brakes with a uniform deceleration of $6.5\ \mathrm{m\,s^{-2}}$ until the car stops.

(a) Calculate the distance travelled during the reaction time. [1]

(b) Calculate the total stopping distance of the car. [3]

(a) $24 \times 0.65 = 15.6\ \mathrm{m}$ (accept 16 m)(1)
(b) use of $v^2 = u^2 + 2as$ with $v = 0$, $u = 24$, $a = -6.5$ (sign of $a$ may be implied)(1)
braking distance $= \dfrac{24^2}{2\times6.5} = 44.3\ \mathrm{m}$(1)
total $= 15.6 + 44.3 = 60\ \mathrm{m}$ (accept 59.9; ECF from (a))(1)
Answer: (a) 15.6 m (b) 60 m (59.9 m)
Examiner tip: Add the thinking distance at constant speed to the braking distance; suvat only applies once the deceleration starts.

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4Standard6 marks

The graph shows how the velocity $v$ of a trolley moving along a straight track varies with time $t$ over 16 s. The trolley starts from the origin.

(a) Calculate the acceleration of the trolley during the first 4.0 s. [1]

(b) State the time at which the trolley is furthest from its starting point. Give a reason. [1]

(i) the total distance travelled by the trolley. [2]

(ii) the magnitude of the average velocity of the trolley. [2]

(a) $a = \dfrac{6}{4} = 1.5\ \mathrm{m\,s^{-2}}$(1)
(b) $t = 14\ \mathrm{s}$ AND velocity becomes zero / changes sign then (the trolley reverses direction after 14 s)(1)
(c)(i) areas: $12 + 36 + 12 = 60\ \mathrm{m}$ (forward, 0–14 s) and $3\ \mathrm{m}$ (backward, 14–16 s) seen(1)
distance $= 60 + 3 = 63\ \mathrm{m}$(1)
(c)(ii) displacement $= 60 - 3 = 57\ \mathrm{m}$ (area below the axis subtracted)(1)
average velocity $= \dfrac{57}{16} = 3.6\ \mathrm{m\,s^{-1}}$ (accept 3.56; ECF from displacement; $63/16 = 3.9$ scores [0] for (ii))(1)
Answer: (a) 1.5 m s$^{-2}$ (b) 14 s (c)(i) 63 m (ii) 3.6 m s$^{-1}$
Examiner tip: Area below the time axis counts as positive distance but negative displacement; do not mix the two.

5Standard5 marks

A ball is thrown vertically upwards at $12.0\ \mathrm{m\,s^{-1}}$ from a balcony 18.0 m above the ground. It misses the balcony on the way down and lands on the ground below. Air resistance is negligible.

(a) Calculate the maximum height of the ball above the ground. [2]

(b) Determine the time between the ball being thrown and it hitting the ground. [3]

(a) height above balcony $= \dfrac{u^2}{2g} = \dfrac{12.0^2}{2\times9.81} = 7.34\ \mathrm{m}$(1)
height above ground $= 18.0 + 7.34 = 25.3\ \mathrm{m}$(1)
(b) $-18.0 = 12.0t - \tfrac12(9.81)t^2$ OR equivalent with consistent signs (displacement is negative / below the start)(1)
solving the quadratic / valid two-stage method (e.g. 1.22 s up, then fall of 25.3 m taking 2.27 s)(1)
$t = 3.5\ \mathrm{s}$ (accept 3.50; reject the negative root; award [1 max] for (b) if $+18.0$ is used)(1)
Answer: (a) 25.3 m (b) 3.5 s
Examiner tip: Take one direction as positive for the whole flight: the final displacement is −18.0 m, not +18.0 m.

6Standard5 marks

A football is kicked from ground level with a velocity of $18.0\ \mathrm{m\,s^{-1}}$ at $40^\circ$ above the horizontal. A wall of height 4.0 m stands 25.0 m from the point where the ball is kicked. Air resistance is negligible.

(a) Calculate the time taken for the ball to reach the wall. [2]

(b) Determine whether the ball passes over the wall. [2]

(a) horizontal component $= 18.0\cos40^\circ = 13.8\ \mathrm{m\,s^{-1}}$(1)
$t = \dfrac{25.0}{13.8} = 1.81\ \mathrm{s}$(1)
(b) $y = (18.0\sin40^\circ)(1.81) - \tfrac12(9.81)(1.81)^2$(1)
$y = 4.9\ \mathrm{m}$ (accept 4.8–4.9), greater than 4.0 m so the ball passes over (by about 0.9 m); conclusion must be consistent with the calculated height(1)
(c) falling: the time to reach maximum height is $\dfrac{18.0\sin40^\circ}{9.81} = 1.18\ \mathrm{s}$, which is less than 1.81 s / vertical velocity at the wall is $11.6 - 9.81\times1.81 = -6.2\ \mathrm{m\,s^{-1}}$ (downwards)(1)
Answer: (a) 1.81 s (b) height 4.9 m at the wall, so it clears (c) falling
Examiner tip: The horizontal velocity stays constant, so the time to the wall comes from the horizontal motion only.

7Hard5 marks

A cyclist moving along a straight road at $3.0\ \mathrm{m\,s^{-1}}$ starts to pedal harder at $t = 0$. The graph shows how her acceleration $a$ varies with time $t$ for the next 8.0 s.

(a) Explain why the equations for uniformly accelerated motion cannot be used to find her velocity at $t = 6.0\ \mathrm{s}$. [1]

(b) Determine her speed at $t = 6.0\ \mathrm{s}$. [2]

(c) Describe how her velocity changes from $t = 0$ to $t = 8.0\ \mathrm{s}$, referring to the shape of a velocity–time graph. [2]

(a) the acceleration is not constant / changes with time (decreases from 4.0 to 0), and the suvat equations need uniform acceleration(1)
(b) change in velocity = area under the a–t graph $= \tfrac12 \times 6.0 \times 4.0 = 12\ \mathrm{m\,s^{-1}}$(1)
$v = 3.0 + 12 = 15\ \mathrm{m\,s^{-1}}$ (12 m s$^{-1}$ alone scores [1])(1)
(c) 0–6 s: velocity increases but at a decreasing rate, so the v–t graph is a curve whose gradient decreases (to zero at 6 s)(1)
6–8 s: velocity constant (15 m s$^{-1}$), so the v–t graph is a horizontal straight line(1)
Answer: (a) acceleration not uniform (b) 15 m s$^{-1}$ (c) increases at decreasing rate, then constant
Examiner tip: Area under an acceleration–time graph is the CHANGE in velocity: add the initial 3.0 m s$^{-1}$.

8Hard6 marks

A golf ball is hit from level ground at an angle above the horizontal. Discuss how the motion of the ball when air resistance acts differs from its motion when air resistance is negligible. In your answer refer to the components of velocity and acceleration, and to the shape of the trajectory. [6]

Air resistance acts opposite to the velocity of the ball, and increases with speed(1)
Horizontal component of velocity is no longer constant: it decreases throughout the flight (there is a horizontal deceleration)(1)
On the way up, weight and the vertical component of drag both act downwards, so the vertical deceleration is greater than $g$(1)
On the way down, drag has an upward component, so the downward acceleration is less than $g$; the ball could approach a terminal speed if it fell far enough(1)
Maximum height is lower and is reached sooner; the horizontal range is shorter(1)
Trajectory is no longer a symmetrical parabola: the descent is steeper than the ascent / the highest point is closer to the landing point than to the start(1)
Time of descent is longer than time of ascent; total time of flight is (usually) shorter(1)
Ball lands with a smaller speed than its launch speed (energy transferred to the air)(1)

Level	Marks	What it looks like
Level 3	5–6	Effects on both velocity components and on the acceleration are explained with reasons linked to the direction of the drag force, and at least two consequences for the trajectory (e.g. lower peak, shorter range, steeper descent / asymmetry) are clearly stated. Reasoning is logical and uses correct physics terms.
Level 2	3–4	Several correct effects are described, including at least one on a velocity or acceleration component, but the reasoning is partly missing or the asymmetry of the path is not addressed.
Level 1	1–2	One or two simple, relevant statements, such as the range or height being smaller, with little or no explanation.
Level 0	0	No relevant content.

Answer: Drag reduces the horizontal velocity, makes the upward deceleration greater than $g$ and the downward acceleration less than $g$; lower peak, shorter range, asymmetric path with a steeper descent.
Examiner tip: Saying only 'it goes less far' is not enough: say what happens to each velocity component and why the path becomes asymmetric.

9Hard7 marks

A student investigates free fall. A steel ball is held by an electromagnet a height $h$ above a trapdoor. Switching off the electromagnet starts a timer and the ball opening the trapdoor stops it, giving the fall time $t$. The student plots $\sqrt{h}$ against $t$ for six heights, as shown.

(a) Show that, for a ball released from rest, the gradient of the graph should be $\sqrt{g/2}$. [1]

(b) Determine a value for $g$ using the graph. [3]

(c) The line of best fit does not pass through the origin. Identify the type of error this shows and suggest a cause. [2]

(d) Another student uses a single reading, $h = 1.000\ \mathrm{m}$ and $t = 0.472\ \mathrm{s}$, with $g = 2h/t^2$. Explain why this gives a less accurate value of $g$ than the graphical method. [1]

(a) $h = \tfrac12 g t^2$ so $\sqrt{h} = \sqrt{g/2}\,t$ (a straight line through the origin with gradient $\sqrt{g/2}$)(1)
(b) gradient found from two well-separated points on the line, e.g. $\dfrac{1.095 - 0.447}{0.515 - 0.222}$(1)
gradient $= 2.21$ (allow 2.17 to 2.25) in units of m$^{1/2}$ s$^{-1}$(1)
$g = 2 \times 2.21^2 = 9.8\ \mathrm{m\,s^{-2}}$ (allow 9.4 to 10.1 consistent with their gradient; do not allow use of a single data point with the origin)(1)
(c) systematic error (accept zero error / timing offset)(1)
the timer starts before the ball is actually released, e.g. the magnetism of the electromagnet takes time to decay (residual magnetism), so every $t$ is too large by about 0.02 s (intercept on the $t$-axis)(1)
(d) the single reading includes the systematic delay of about 0.02 s, making $t$ too large and $g$ too small (8.98 m s$^{-2}$); the gradient is unaffected by a constant offset (OR the graph averages random error over several readings)(1)
Answer: (b) $g \approx 9.8\ \mathrm{m\,s^{-2}}$ (c) systematic error from release delay (d) offset in $t$ cancels in the gradient
Examiner tip: A constant offset in t shifts the line but leaves its gradient unchanged, so always take g from the gradient, not from one point.

10Stretch6 marks

A car travelling at a constant $25.0\ \mathrm{m\,s^{-1}}$ passes a stationary police car. 1.50 s later the police car sets off from rest along the same straight road, accelerating uniformly at $4.00\ \mathrm{m\,s^{-2}}$.

(a) Determine the time, measured from the moment the car passes the police car, at which the police car catches up with the car. [4]

(b) Calculate the distance travelled by the police car before it catches up. [1]

(a) car: $s = 25.0t$(1)
police car: $s = \tfrac12(4.00)(t - 1.50)^2$ (time measured from when the car passes; allow equivalent with time $T$ from the police car starting, $25.0(T+1.50) = 2.00T^2$)(1)
equating and forming the quadratic: $2t^2 - 31t + 4.5 = 0$ (or $2T^2 - 25T - 37.5 = 0$)(1)
$t = 15.4\ \mathrm{s}$ (accept 15.35; reject the other root 0.15 s as it is before the police car starts; if $T$ is used, $T = 13.9$ s must be converted to 15.4 s)(1)
(b) $s = 25.0 \times 15.35 = 384\ \mathrm{m}$ (accept 380–385 m; ECF from (a))(1)
(c) not realistic: the police car would be moving at $4.00 \times 13.9 = 55\ \mathrm{m\,s^{-1}}$ (about 200 km h$^{-1}$), so its acceleration would decrease well before then (due to air resistance / engine limits / speed limits on safety)(1)
Answer: (a) 15.4 s (b) 384 m (c) not realistic: final speed about 55 m s$^{-1}$
Examiner tip: Write both displacements as functions of the same time variable before equating, and check which root is physically possible.

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