A.2 Forces and momentum – IB Physics exam questions with mark scheme

1Core4 marks

A car of mass 900 kg travels at a constant speed of $10\ \mathrm{m\,s^{-1}}$ around a circular roundabout of radius 25 m.

(a) Calculate the angular speed of the car. [1]

(b) Calculate the centripetal acceleration of the car. [1]

(d) State the direction of the resultant force. [1]

$\omega = v/r = 10/25 = 0.40\ \mathrm{rad\,s^{-1}}$ (0.40)(1)
$a = v^2/r = 4.0\ \mathrm{m\,s^{-2}}$ (allow $\omega^2 r$)(1)
$F = ma = 900 \times 4.0 = 3600\ \mathrm{N}$ (3600 N; allow $3.6\times10^3$ N)(1)
Towards the centre of the circle (allow: perpendicular to velocity / radially inward)(1)
Answer: (a) 0.40 rad s$^{-1}$ (b) 4.0 m s$^{-2}$ (c) 3600 N (d) towards the centre of the circle
Examiner tip: In (c) use the mass in kilograms and the centripetal acceleration; do not add a separate 'centrifugal force'.

2Standard5 marks

A block of mass 5.0 kg slides down a rough plane inclined at $30^\circ$ to the horizontal. The coefficient of dynamic friction between the block and the plane is 0.20. Use $g=9.81\ \mathrm{m\,s^{-2}}$.

(a) Calculate the component of the weight of the block parallel to the plane. [1]

(b) Determine the frictional force acting on the block. [2]

$mg\sin30^\circ = 5.0\times9.81\times\sin30^\circ = 24.5\ \mathrm{N}$ (24.5 N)(1)
Normal reaction $R = mg\cos30^\circ = 42.5\ \mathrm{N}$ (42.5 N)(1)
$F = \mu R = 0.20\times42.5 = 8.5\ \mathrm{N}$ (8.5 N)(1)
Resultant force down the plane $= 24.5-8.5 = 16.0\ \mathrm{N}$ (16 N)(1)
$a = F/m = 3.2\ \mathrm{m\,s^{-2}}$ (3.2)(1)
Answer: (a) 24.5 N (b) 8.5 N (c) 3.2 m s$^{-2}$
Examiner tip: Friction depends on the normal reaction, which is $mg\cos\theta$ on a slope, not $mg$.

3Standard5 marks

A trolley of mass 0.80 kg moves at $1.5\ \mathrm{m\,s^{-1}}$ along a frictionless track and collides with a stationary trolley of mass 0.40 kg. The trolleys stick together.

(a) Determine the common speed of the trolleys immediately after the collision. [2]

(b) Calculate the kinetic energy lost in the collision. [2]

Total momentum is conserved: $0.80\times1.5 = (0.80+0.40)v$(1)
$v = 1.0\ \mathrm{m\,s^{-1}}$ (1.0)(1)
Initial KE $=\tfrac12(0.80)(1.5)^2 = 0.90$ J and final KE $=\tfrac12(1.2)(1.0)^2=0.60$ J(1)
Loss $= 0.30\ \mathrm{J}$ (0.30 J)(1)
Inelastic, because kinetic energy is not conserved (some KE transferred to thermal/sound energy)(1)
Answer: (a) 1.0 m s$^{-1}$ (b) 0.30 J (c) inelastic
Examiner tip: Quote the conservation principle in words or symbols; a bare answer for speed can lose the first mark.

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4Standard4 marks

A footballer kicks a stationary ball of mass 0.16 kg. The graph shows how the force exerted by the foot on the ball varies with time while they are in contact.

(a) Determine the impulse given to the ball. [2]

(b) Calculate the speed of the ball as it leaves the foot. [1]

Impulse equals the area under the force–time graph (e.g. two triangles plus a rectangle)(1)
Impulse $= 0.4+0.8+0.4 = 1.6\ \mathrm{N\,s}$ (1.6)(1)
$v = \Delta p/m = 1.6/0.16 = 10\ \mathrm{m\,s^{-1}}$ (10)(1)
$F_{av} = J/\Delta t = 1.6/0.06 = 27\ \mathrm{N}$ (27 N; allow 26.7)(1)
Answer: (a) 1.6 N s (b) 10 m s$^{-1}$ (c) 27 N
Examiner tip: Work out areas from the axes values, and remember a ball starting at rest has zero initial momentum.

5Core5 marks

A crate of mass 40 kg rests on the flat, horizontal load bed of a truck. The coefficient of static friction between the crate and the load bed is 0.45. The truck starts to accelerate forwards along a straight, level road and the crate does not slide.

(a) Identify the horizontal force that accelerates the crate, and state its direction. [1]

(b) State the force that forms a Newton's third law pair with the force in (a). [1]

(d) The truck actually accelerates at $2.0\ \mathrm{m\,s^{-2}}$. A student says the friction force on the crate is $\mu_s F_N = 177\ \mathrm{N}$. Explain why the student is wrong. [1]

(static) friction exerted by the load bed/truck on the crate, acting forwards / in the direction of the acceleration(1)
friction exerted by the crate on the load bed/truck, (equal in size and) acting backwards; must name both bodies; do not accept the weight or normal force(1)
$\mu_s m g = ma$ / $F_{f,\max} = 0.45\times40\times9.81 = 177\ \mathrm{N}$(1)
$a_{\max} = \mu_s g = 4.4\ \mathrm{m\,s^{-2}}$ (allow 4.41; allow 4.5 with $g = 10$)(1)
$\mu_s F_N$ is only the maximum/limiting static friction; the actual friction is just what is needed: $F = ma = 40\times2.0 = 80\ \mathrm{N}$ (OWTTE; award for either idea)(1)
Answer: (a) static friction from the load bed on the crate, forwards (b) friction from the crate on the load bed, backwards (c) $4.4\ \mathrm{m\,s^{-2}}$ (d) $F_f \le \mu_s F_N$; actual friction $= 80\ \mathrm{N}$
Examiner tip: $F_f \le \mu_s F_N$ is an inequality: static friction only reaches $\mu_s F_N$ when the object is just about to slip.

6Standard7 marks

A student uses falling-ball viscometry to measure the viscosity of glycerol. A steel ball of diameter 3.00 mm is released at the surface of glycerol in a tall tube. The student times the ball between two markers 0.200 m apart and measures a time of 8.6 s.

Density of steel $= 7.80\times10^{3}\ \mathrm{kg\,m^{-3}}$; density of glycerol $= 1.26\times10^{3}\ \mathrm{kg\,m^{-3}}$.

(a) The ball moves at terminal speed between the markers. Identify the three forces acting on the ball and state the relationship between them. [2]

(b) Show that the buoyancy force on the ball is about $1.7\times10^{-4}\ \mathrm{N}$. [1]

(d) Suggest why marker 1 is placed some distance below the surface of the glycerol. [1]

weight/gravitational force (down), buoyancy/upthrust (up) and viscous drag (up); all three needed(1)
weight = buoyancy + drag / resultant force is zero(1)
$F_b = \rho V g = 1260\times\tfrac{4}{3}\pi(1.50\times10^{-3})^3\times9.81 = 1.75\times10^{-4}\ \mathrm{N}$; answer to at least 3 s.f. or full substitution needed(1)
$v = 0.200/8.6 = 0.0233\ \mathrm{m\,s^{-1}}$(1)
$F_d = W - F_b = 1.08\times10^{-3} - 1.75\times10^{-4} = 9.07\times10^{-4}\ \mathrm{N}$(1)
$\eta = \dfrac{F_d}{6\pi r v} = 1.4\ \mathrm{Pa\,s}$ (allow 1.38; ECF for $v$ and $F_d$; award [2 max] if the diameter 3.00 mm is used for $r$ in $6\pi\eta r v$, giving 0.69 Pa s)(1)
so the ball has reached terminal speed (before timing starts) / the ball is still accelerating near the surface (OWTTE)(1)
Answer: (a) weight = buoyancy + drag (b) $1.75\times10^{-4}\ \mathrm{N}$ (c) $\eta \approx 1.4\ \mathrm{Pa\,s}$ (d) so the ball is at terminal speed when timing starts
Examiner tip: Use the radius, not the diameter, in both the volume and in $6\pi\eta r v$, and remember the drag is weight minus buoyancy, not the weight alone.

7Core5 marks

Two ice skaters stand at rest facing each other on smooth, horizontal ice. Skater A has a mass of 72 kg and skater B has a mass of 48 kg. They push against each other's hands and move apart. Skater B moves away at $2.1\ \mathrm{m\,s^{-1}}$.

(a) Explain why the total momentum of the two skaters is zero after they push apart. [1]

(b) Calculate the speed of skater A. [2]

(d) State where this kinetic energy comes from. [1]

total momentum was zero before / at rest initially and no (resultant) external (horizontal) force acts, so momentum is conserved(1)
$72\,v_A = 48\times2.1$ / $0 = 72v_A - 48\times2.1$(1)
$v_A = 1.4\ \mathrm{m\,s^{-1}}$ (ignore sign/direction)(1)
$\tfrac12\times72\times1.4^2 + \tfrac12\times48\times2.1^2 = 176\ \mathrm{J}$ (allow 180; ECF from (b))(1)
(chemical energy in the) skaters' muscles / work done by the skaters (OWTTE)(1)
Answer: (b) $1.4\ \mathrm{m\,s^{-1}}$ (c) 176 J (d) chemical energy in the skaters' muscles
Examiner tip: In an explosion momentum is conserved but kinetic energy is not: the KE is created from another store, so never set KE before equal to KE after.

8Hard8 marks

On a horizontal air track, glider A of mass 0.30 kg moves at $0.60\ \mathrm{m\,s^{-1}}$ towards glider B of mass 0.20 kg, which is at rest. A light spring bumper of spring constant $48\ \mathrm{N\,m^{-1}}$ is fixed to the front of glider B. After the collision, light gates record that A moves at $0.12\ \mathrm{m\,s^{-1}}$ and B moves at $0.72\ \mathrm{m\,s^{-1}}$, both in A's original direction. Friction is negligible.

(a) Show that the data are consistent with conservation of momentum. [2]

(b) Deduce whether the collision is elastic. [2]

(d) At the instant of maximum compression of the spring, both gliders move with the same velocity. Determine the maximum compression of the spring. [3]

momentum before $= 0.30\times0.60 = 0.18\ \mathrm{kg\,m\,s^{-1}}$(1)
momentum after $= 0.30\times0.12 + 0.20\times0.72 = 0.036 + 0.144 = 0.18\ \mathrm{kg\,m\,s^{-1}}$, so the two are equal(1)
KE before $= \tfrac12\times0.30\times0.60^2 = 0.054\ \mathrm{J}$ and KE after $= 0.00216 + 0.0518 = 0.054\ \mathrm{J}$(1)
total KE unchanged so the collision is elastic; conclusion must follow from the comparison of KE values(1)
$F = \Delta p/\Delta t = 0.144/0.16 = 0.90\ \mathrm{N}$ (allow 0.9)(1)
common velocity $= 0.18/0.50 = 0.36\ \mathrm{m\,s^{-1}}$(1)
energy stored in spring $= 0.054 - \tfrac12\times0.50\times0.36^2 = 0.0216\ \mathrm{J}$(1)
$\tfrac12 k x^2 = 0.0216 \Rightarrow x = \sqrt{2\times0.0216/48} = 0.030\ \mathrm{m}$ (ECF from common velocity)(1)
Answer: (a) 0.18 kg m s$^{-1}$ before and after (b) elastic: KE = 0.054 J before and after (c) 0.90 N (d) 0.030 m (3.0 cm)
Examiner tip: At maximum compression the gliders move together, so the spring energy is the KE lost at that instant, not all of the initial KE.

9Standard6 marks

A car travels at constant speed around a level, unbanked bend that is part of a horizontal circle. A passenger on a smooth leather seat slides across the seat towards the door on the outside of the bend, until the door stops her. The passenger says: "A force pushed me outwards, away from the centre of the bend."

Explain, with reference to Newton's laws of motion and circular motion, the motion of the car and of the passenger. In your answer, identify the forces that act in each case and comment on the passenger's statement. [6]

the car's velocity changes direction so it accelerates (even though its speed is constant)(1)
the acceleration is towards the centre of the circle, $a = v^2/r$, so a resultant (centripetal) force acts towards the centre (Newton's second law)(1)
for the car this force is the (static) friction of the road on the tyres, directed towards the centre(1)
the centripetal force is perpendicular to the velocity, so it does no work and the speed is unchanged(1)
the smooth seat exerts (almost) no sideways friction on the passenger, so no resultant horizontal force acts on her(1)
by Newton's first law she continues in a straight line (tangent to the circle) while the car turns beneath her, so she moves outwards relative to the car(1)
when she reaches the door, the normal force from the door acts on her towards the centre and provides her centripetal force(1)
the statement is wrong: no outward force acts in the road's (inertial) frame; the effect is her inertia (OWTTE); weight and vertical normal force from the seat balance(1)
Newton's third law: she pushes outward on the door / tyres push outward on the road; these forces act on the other body, not on her/the car(1)

Level	Marks	What it looks like
Level 3	5–6	Clear, correct account that explains the centripetal acceleration of the car and names friction as its source, and uses Newton's first law to explain the passenger's motion relative to the car, naming the door's normal force as her centripetal force. Rejects the outward force clearly. Physics terms are used accurately throughout.
Level 2	3–4	Correct explanation of either the car or the passenger in detail, with some relevant points about the other. A centripetal force towards the centre is identified, but the source of the force or the role of Newton's first law may be missing or unclear. Minor errors in terminology.
Level 1	1–2	Isolated correct statements, e.g. that there is a force towards the centre or that the passenger keeps moving in a straight line. Reasoning is fragmentary, and the answer may accept the outward force.
Level 0	0	No relevant physics.

Answer: The car accelerates towards the centre because its direction changes; road friction provides the centripetal force. The smooth seat cannot supply a centripetal force to the passenger, so by Newton's first law she continues in a straight line while the car turns, until the door's normal force pushes her towards the centre. There is no outward force on her.
Examiner tip: Answers that accept a 'centrifugal force' cannot get past Level 1–2: say what provides the force towards the centre and use Newton's first law for the outward slide.

10Stretch6 marks

A small bob of mass 0.25 kg hangs from a light string of length 0.80 m. The bob moves at constant speed in a horizontal circle so that the string makes a constant angle of $35^\circ$ with the vertical (a conical pendulum). Air resistance is negligible.

(a) Show that the tension in the string is about 3.0 N. [2]

(b) Determine the speed of the bob. [3]

vertical forces balance: $T\cos35^\circ = mg$(1)
$T = \dfrac{0.25\times9.81}{\cos35^\circ} = 2.99\ \mathrm{N}$; answer to at least 3 s.f. or full substitution needed(1)
radius $r = 0.80\sin35^\circ = 0.459\ \mathrm{m}$(1)
$T\sin35^\circ = \dfrac{mv^2}{r}$ / $mg\tan35^\circ = \dfrac{mv^2}{r}$(1)
$v = \sqrt{g r\tan35^\circ} = 1.8\ \mathrm{m\,s^{-1}}$ (allow 1.78; award [2 max] if $r = 0.80$ m is used, giving 2.3 m s$^{-1}$)(1)
$T_{\text{period}} = 2\pi r/v = 2\pi\times0.459/1.78 = 1.6\ \mathrm{s}$ (allow 1.62; ECF from (b))(1)
Answer: (a) 2.99 N (b) $1.8\ \mathrm{m\,s^{-1}}$ (c) 1.6 s
Examiner tip: The radius of the circle is $L\sin\theta$, not the string length; resolve vertically for the tension and horizontally for the centripetal force.

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