Forces and momentum HL Paper 1&2

A.2 Forces and Momentum

HL examination practice: Multiple-choice questions, Extended responses and a Lab-style investigation

This original question bank develops Paper 1A multiple-choice skills, Paper 2 structured problem-solving and Paper 1B-style data analysis. It covers free-body diagrams, Newton’s laws, friction, springs, drag and buoyancy, momentum, impulse, collisions, explosions and circular motion. These are practice questions, not reproduced IB examination questions.

Instructions and data

Select one option for each multiple-choice question before opening its explanation.
For extended responses, identify the system, draw relevant free-body diagrams and state a positive direction.
Show equations, substitutions and units. Give final numerical answers to three significant figures unless otherwise stated.
Retain extra digits during calculations, particularly when comparing momenta or kinetic energies.
Use $g=9.81\,\mathrm{m\,s^{-2}}$ and degree mode for angles.
Assume negligible air resistance and external impulse only where the question says so.
Most A.2 content is shared SL/HL. The two-dimensional momentum application is identified as HL; other demanding applications are labelled “Challenge”.
Suggested marks are learning guides; equivalent correct methods are acceptable.

Equation reminders

$\sum\vec F=m\vec a,\qquad F_g=mg$

$0\leq F_s\leq\mu_sR,\qquad F_k=\mu_kR$

$F_H=-kx,\qquad F_b=\rho_{\mathrm{fluid}}Vg$

$F_d=6\pi\eta rv$

$\vec p=m\vec v,\qquad\vec J=\Delta\vec p=\vec F_{\mathrm{avg}}\Delta t$

$\sum\vec p_{\mathrm{before}}=\sum\vec p_{\mathrm{after}}$

$E_k=\frac12mv^2$

$v=\frac{2\pi r}{T}=\omega r$

$a_c=\frac{v^2}{r}=\omega^2r=\frac{4\pi^2r}{T^2},\qquad F_c=\frac{mv^2}{r}$

R denotes the normal reaction, also often written N. Friction equations give magnitudes; direction must be assigned separately. Static friction equals its maximum only at impending slip. Hooke’s law uses signed extension or compression from the natural-length position and applies within the proportional region. Stokes’ law applies to slow, laminar flow around a sphere.

The impulse relation uses the average resultant force on the chosen system. Momentum conservation requires negligible resultant external impulse over the interval. Centripetal force is the inward resultant supplied by real forces, not an extra force to add to a free-body diagram.

Common misconceptions

Newton’s third-law partners act on different objects and do not cancel on one object’s free-body diagram.
R equals mg only when the other vertical forces and vertical acceleration permit that balance.
Static friction adjusts to the required value up to its maximum.
Momentum is conserved in an isolated inelastic collision even though kinetic energy decreases.
Terminal velocity means zero resultant force, not zero individual forces.
Constant speed in circular motion still involves acceleration because velocity changes direction.

Section A — Paper 1A-style multiple choice

15 questions × 1 mark = 15 marks.

Multiple choice 1 — Newton’s first law [1 mark]

A particle moves in a straight line with constant non-zero velocity. What must be true?

No forces act on it.
The vector sum of the forces on it is zero.
A constant resultant force acts along its velocity.
The resultant force is opposite its velocity.

Show answer and explanation

Correct option: B.

Constant velocity means zero acceleration. Newton’s second law therefore requires zero resultant force. Individual forces may be present and balanced.

Multiple choice 2 — Newton’s third law [1 mark]

A heavy truck collides with a lighter car. Assume their contact forces dominate during the collision. Which comparison is correct?

The truck exerts a larger force and has larger acceleration magnitude.
The car exerts a larger force and has smaller acceleration magnitude.
The forces have equal magnitudes and both vehicles have equal acceleration magnitudes.
The forces have equal magnitudes and the car has larger acceleration magnitude.

Show answer and explanation

Correct option: D.

The interaction forces are equal and opposite. For these contact-force contributions, acceleration magnitude is F/m, so the lighter car has the larger acceleration magnitude. Assume other forces are negligible during contact.

Multiple choice 3 — Actual static friction [1 mark]

A 5.00 kg box is at rest on a horizontal floor with coefficient of static friction 0.600. A horizontal force of 15.0 N is applied. What is the frictional force on the box?

15.0 N opposite the applied force.
29.4 N opposite the applied force.
15.0 N along the applied force.
Zero.

Show answer and explanation

Correct option: A.

$F_{s,\max}=0.600(5.00)(9.81)=29.4\,\mathrm{N}$

The applied force is below the maximum. Static friction takes the required value 15.0 N in the opposite direction, maintaining rest.

Multiple choice 4 — Apparent weight [1 mark]

A 60.0 kg passenger stands on a scale in a lift accelerating downward at 2.00 m/s². What does the scale read?

709 N.
589 N.
469 N.
120 N.

Show answer and explanation

Correct option: C.

Take upward positive. The scale reads the normal force R, not the resultant.

$R-mg=-ma,\qquad R=60.0(9.81-2.00)=469\,\mathrm{N}$

Multiple choice 5 — Normal reaction under an angled pull [1 mark]

A 10.0 kg box on a horizontal floor is pulled by a 40.0 N force at 30.0° above horizontal. There is no vertical acceleration. What is the normal reaction?

98.1 N.
78.1 N.
118 N.
63.5 N.

Show answer and explanation

Correct option: B.

$R=mg-F\sin30.0^\circ=98.1-20.0=78.1\,\mathrm{N}$

The upward component of the pull reduces the normal reaction.

Multiple choice 6 — Signed spring force [1 mark]

A spring has k = 200 N/m. Its signed extension from the natural-length position is x = −0.0300 m. What is the restoring force?

+6.00 N.
−6.00 N.
+0.150 N.
−0.150 N.

Show answer and explanation

Correct option: A.

$F_H=-kx=-200(-0.0300)=+6.00\,\mathrm{N}$

The restoring force points opposite the signed displacement.

Multiple choice 7 — Terminal velocity [1 mark]

A sphere falls at terminal speed in a liquid. Weight, buoyancy and drag are the only forces. Which relation is correct?

Weight is zero.
Drag equals weight and buoyancy is zero.
Weight plus buoyancy equals drag.
Weight equals buoyancy plus drag.

Show answer and explanation

Correct option: D.

The acceleration is zero, so upward buoyancy and drag balance downward weight.

$mg=F_b+F_d$

Multiple choice 8 — Impulse during a rebound [1 mark]

A 0.150 kg ball moves at +20.0 m/s and rebounds at −10.0 m/s. What impulse acts on the ball?

+4.50 N s.
−1.50 N s.
−4.50 N s.
+1.50 N s.

Show answer and explanation

Correct option: C.

$J=m(v-u)=0.150(-10.0-20.0)=-4.50\,\mathrm{N\,s}$

The signed final velocity must be used; the ball reverses direction.

Multiple choice 9 — Force–time area [1 mark]

A resultant force–time graph is a triangle of base 0.0400 s and peak force +100 N. What impulse does it represent?

4.00 N s.
2.00 N s.
2500 N s.
0.0200 N s.

Show answer and explanation

Correct option: B.

$J=\frac12(0.0400)(100)=2.00\,\mathrm{N\,s}$

Impulse is the signed area, not peak force multiplied by the whole contact time.

Multiple choice 10 — A sticking collision [1 mark]

A 2.00 kg trolley moves at +3.00 m/s and collides with a 1.00 kg trolley moving at −1.00 m/s. They stick. External impulse is negligible. What is their common velocity?

+1.67 m/s.
+2.33 m/s.
+1.00 m/s.
−1.67 m/s.

Show answer and explanation

Correct option: A.

$v=\frac{(2.00)(3.00)+(1.00)(-1.00)}{3.00}=+1.67\,\mathrm{m\,s^{-1}}$

Multiple choice 11 — Explosion from rest [1 mark]

An isolated object initially at rest splits into fragments of masses 2.00 kg and 3.00 kg. The 2.00 kg fragment moves at −6.00 m/s. What is the other fragment’s velocity?

−4.00 m/s.
+9.00 m/s.
+4.00 m/s.
−9.00 m/s.

Show answer and explanation

Correct option: C.

$0=(2.00)(-6.00)+(3.00)v,\qquad v=+4.00\,\mathrm{m\,s^{-1}}$

The fragments have equal and opposite momenta, not necessarily equal speeds.

Multiple choice 12 — Elastic collision of identical masses [1 mark]

An identical-mass trolley moving at +4.00 m/s has a head-on elastic collision with a stationary trolley. External impulse is negligible. What are the velocities afterward, in the order moving trolley then stationary trolley?

+2.00 m/s; +2.00 m/s.
−4.00 m/s; 0.
0; −4.00 m/s.
0; +4.00 m/s.

Show answer and explanation

Correct option: D.

In this one-dimensional elastic collision of identical masses, the velocities are exchanged. This preserves both total momentum and kinetic energy. The pair of +2.00 m/s velocities preserves momentum but loses kinetic energy.

Multiple choice 13 — Centripetal acceleration scaling [1 mark]

A particle’s circular-motion speed is doubled and its radius is doubled. How does centripetal acceleration change?

It doubles.
It halves.
It is unchanged.
It quadruples.

Show answer and explanation

Correct option: A.

$\frac{a_{\mathrm{new}}}{a_{\mathrm{old}}}=\frac{(2v)^2/(2r)}{v^2/r}=2$

Multiple choice 14 — Loss of the inward force [1 mark]

A puck moves on a frictionless horizontal surface, constrained to a circle by a string. The string breaks. What is its initial path afterward?

Radially outward.
Along the tangent at the breaking point.
Along the same circle.
Radially inward.

Show answer and explanation

Correct option: B.

The puck retains its instantaneous tangential velocity. With no subsequent horizontal resultant force, its path is a straight tangent.

Multiple choice 15 — Friction-limited circular motion [1 mark]

A car travels around a flat circular track of radius 40.0 m. The coefficient of static friction is 0.500. Neglect aerodynamic forces. What is its maximum speed without sliding?

19.8 m/s.
9.81 m/s.
14.0 m/s.
28.0 m/s.

Show answer and explanation

Correct option: C.

$\frac{mv_{\max}^2}{r}=\mu_smg,\qquad v_{\max}=\sqrt{\mu_sgr}=14.0\,\mathrm{m\,s^{-1}}$

Static friction supplies the inward resultant; it reaches its maximum at impending sideways slip.

Section B — Paper 2-style extended responses

Show full working and label every force in a free-body diagram. The marking points below describe the required diagrams as well as the calculations.

Extended response 1 — Resultant force and acceleration [4 marks]

A 6.00 kg crate slides right along a horizontal floor. A horizontal force of 20.0 N pulls it right, while kinetic friction has magnitude 8.00 N. There is no vertical acceleration.

Draw and label the free-body diagram. [2]
Determine the horizontal resultant force. [1]
Determine acceleration. [1]

Show worked answer

(a) Draw weight mg downward and normal reaction R upward. [1] Draw applied force right and friction left. [1] All four arrows act on the crate.

(b) Take right as positive:

$F_{\mathrm{net}}=20.0-8.00=+12.0\,\mathrm{N}$

Correct resultant and direction. [1]

(c)

$a=F_{\mathrm{net}}/m=12.0/6.00=2.00\,\mathrm{m\,s^{-2}}$

Correct acceleration. [1]

Answer: Resultant 12.0 N right; acceleration 2.00 m/s² right.

Check: The applied force alone is not the resultant.

Extended response 2 — Static friction below its maximum [6 marks]

A 10.0 kg box rests on a horizontal surface with coefficient of static friction 0.400. A horizontal force of 35.0 N is applied.

Determine normal reaction. [1]
Determine maximum static friction. [2]
State the actual frictional force. [1]
State whether the box moves and explain. [2]

Show worked answer

(a) With no other vertical force and no vertical acceleration:

$R=mg=10.0(9.81)=98.1\,\mathrm{N}$

Correct reaction. [1]

(b)

$F_{s,\max}=\mu_sR=0.400(98.1)=39.2\,\mathrm{N}$

Equation [1]; maximum [1].

(c) Actual friction is 35.0 N opposite the pull. [1]

(d) The box remains at rest [1] because 35.0 N is below the maximum and static friction can balance the applied force [1].

Answer: R = 98.1 N; maximum static friction 39.2 N; actual friction 35.0 N; no motion.

Check: Using 39.2 N as the actual friction here would incorrectly predict a backward resultant.

Extended response 3 — Weight, normal force and third-law partners [6 marks]

A person of mass 70.0 kg stands still on a horizontal floor. Consider the person as the system.

Determine the person’s weight. [1]
Determine the normal force on the person, stating direction. [2]
Identify the third-law pair associated with the normal force. [2]
Explain why weight and normal force on the person are not a third-law pair. [1]

Show worked answer

(a)

$F_g=mg=70.0(9.81)=687\,\mathrm{N}$

Weight is downward. [1]

(b) Zero vertical acceleration requires R − mg = 0. [1] Thus R = 687 N upward. [1]

(c) The floor pushes the person upward [1]; the person pushes the floor downward with equal magnitude [1].

(d) Weight and normal force both act on the person. Third-law partners act on different objects. [1]

Answer: Weight 687 N down; normal force 687 N up; the normal-force pair is floor-on-person and person-on-floor.

Check: The partner to the person’s weight is the gravitational force exerted by the person on Earth.

Extended response 4 — Hooke’s law and signed displacement [5 marks]

A spring obeys Hooke’s law with spring constant 250 N/m. Its signed extension from the natural-length position is x = +3.60 cm.

Convert x into metres. [1]
Determine the signed restoring force. [2]
State its direction relative to displacement. [1]
State the condition under which Hooke’s law applies. [1]

Show worked answer

(a)

$x=3.60\,\mathrm{cm}=0.0360\,\mathrm{m}$

Correct conversion. [1]

(b)

$F_H=-kx=-250(0.0360)=-9.00\,\mathrm{N}$

Equation [1]; signed force [1].

(c) The force acts opposite to the displacement, toward equilibrium. [1]

(d) Extension must remain within the proportional region, below the limit of proportionality. [1]

Answer: Restoring force −9.00 N; magnitude 9.00 N, opposite the displacement.

Extended response 5 — Rebound, impulse and average force [7 marks]

A 0.200 kg ball approaches a vertical wall horizontally at +12.0 m/s and rebounds at −8.00 m/s. Contact lasts 0.0500 s. Positive is toward the wall. Neglect other horizontal forces.

Determine the ball’s change in horizontal momentum. [2]
Determine the horizontal impulse on the ball. [1]
Determine the average horizontal force of the wall on the ball. [2]
State the corresponding average force of the ball on the wall. [1]
For the same velocity change, explain what happens to average force magnitude if contact time doubles. [1]

Show worked answer

(a)

$\Delta p=m(v-u)=0.200(-8.00-12.0)=-4.00\,\mathrm{kg\,m\,s^{-1}}$

Signed momentum-change method [1]; answer [1].

(b)

$J=\Delta p=-4.00\,\mathrm{N\,s}$

Correct impulse. [1]

(c)

$F_{\mathrm{avg}}=J/\Delta t=-4.00/0.0500=-80.0\,\mathrm{N}$

Equation [1]; force away from the wall [1].

(d) The ball exerts +80.0 N on the wall on average, equal and opposite to the wall’s force on the ball. [1]

(e) The average magnitude halves to 40.0 N because the impulse is unchanged. [1]

Answer: Δp = −4.00 kg m/s; J = −4.00 N s; average wall-on-ball force −80.0 N.

Extended response 6 — Perfectly inelastic collision [7 marks]

A 1.50 kg trolley moving at +2.40 m/s collides with a stationary 2.00 kg trolley. They stick. The track is horizontal and the external horizontal impulse during contact is negligible.

Identify the collision type. [1]
Determine the common velocity. [2]
Calculate the loss of kinetic energy. [3]
Explain what happens to this energy. [1]

Show worked answer

(a) Perfectly inelastic: the trolleys stick. [1]

(b) Treat both trolleys as one system:

$(1.50)(2.40)+(2.00)(0)=(3.50)v$

$v=3.60/3.50=+1.03\,\mathrm{m\,s^{-1}}$

Momentum equation [1]; common velocity [1].

(c) Keep the common velocity unrounded:

$E_i=\frac12(1.50)(2.40)^2=4.32\,\mathrm{J}$

$E_f=\frac12(3.50)\left(\frac{3.60}{3.50}\right)^2=1.85143\,\mathrm{J}$

$E_{\mathrm{lost}}=E_i-E_f=2.47\,\mathrm{J}$

Initial energy [1]; final energy [1]; loss [1].

(d) The lost kinetic energy becomes internal energy, deformation and sound; total energy is conserved. [1]

Answer: Common velocity +1.03 m/s; kinetic-energy loss 2.47 J.

Check: Momentum conservation does not imply kinetic-energy conservation.

Extended response 7 — Uniform horizontal circular motion [7 marks]

A 0.600 kg puck moves in a horizontal circle of radius 0.800 m on a frictionless table. A horizontal string supplies the inward force. The period is 1.60 s.

Determine speed. [2]
Determine centripetal acceleration. [2]
Determine the inward resultant force. [2]
Identify the real force supplying this resultant. [1]

Show worked answer

(a)

$v=\frac{2\pi r}{T}=\frac{2\pi(0.800)}{1.60}=3.14\,\mathrm{m\,s^{-1}}$

Equation [1]; speed [1].

(b)

$a_c=\frac{v^2}{r}=\frac{\pi^2}{0.800}=12.3\,\mathrm{m\,s^{-2}}$

Equation [1]; inward acceleration [1].

(c)

$F_c=ma_c=0.600\frac{\pi^2}{0.800}=7.40\,\mathrm{N}$

Equation [1]; inward force [1].

(d) String tension supplies the horizontal inward resultant. [1]

Answer: Speed 3.14 m/s; acceleration 12.3 m/s² inward; tension 7.40 N.

Check: Retaining π rather than rounded intermediate values gives 7.40 N. Weight and normal reaction balance vertically.

Extended response 8 — Terminal velocity and model consistency [8 marks]

A sphere of mass 2.00 × 10⁻³ kg falls in a viscous liquid at a reported terminal speed of 0.400 m/s. Its buoyant force is 8.00 × 10⁻³ N. The sphere radius is 1.50 × 10⁻³ m and liquid viscosity is 0.500 Pa s. Use the supplied Stokes model F_d = 6πηrv.

Draw or describe the free-body diagram and write terminal force balance. [2]
Determine the drag required by that balance. [2]
Calculate the drag predicted by Stokes’ law at the reported speed. [2]
Assess consistency and find the terminal speed predicted by Stokes’ law with the given parameters. [2]

Show worked answer

(a) Weight acts down; buoyancy and drag act up. [1] At terminal velocity, acceleration is zero:

$mg=F_b+F_d$

Correct force balance. [1]

(b)

$F_d=(2.00\times10^{-3})(9.81)-8.00\times10^{-3}=1.162\times10^{-2}\,\mathrm{N}$

Method [1]; required drag [1].

(c)

$F_{d,\mathrm{Stokes}}=6\pi(0.500)(1.50\times10^{-3})(0.400)=5.65\times10^{-3}\,\mathrm{N}$

Substitution [1]; prediction [1].

(d) The Stokes prediction is about half the required drag, so the reported speed and supplied parameters are inconsistent with this model. This is not a small rounding discrepancy. [1]

$v_{t,\mathrm{model}}=\frac{mg-F_b}{6\pi\eta r}=0.822\,\mathrm{m\,s^{-1}}$

Correct predicted speed. [1]

Answer: Required drag 0.0116 N; predicted drag at 0.400 m/s is 0.00565 N; Stokes-model terminal speed 0.822 m/s.

Check: Check measured parameters and the slow-laminar-flow assumption before treating the model as valid.

Extended response 9 — Challenge: sliding down a rough incline [8 marks]

A 5.00 kg block slides down a straight incline at 30.0° to horizontal. The coefficient of kinetic friction is 0.200. Its initial downhill speed is 1.00 m/s. Neglect air resistance.

Draw and label its free-body diagram. [2]
Determine normal reaction. [2]
Determine friction magnitude. [1]
Determine downhill acceleration. [2]
Determine speed after 2.00 s. [1]

Show worked answer

(a) Draw weight vertically down, normal reaction perpendicular outward from the plane, and kinetic friction uphill. Weight and reaction correctly shown [1]; friction correctly shown [1]. Do not add the weight components as extra forces if weight itself is already drawn.

(b) Perpendicular acceleration is zero:

$R=mg\cos30.0^\circ=42.5\,\mathrm{N}$

Equation [1]; reaction [1].

(c)

$F_k=\mu_kR=8.50\,\mathrm{N}$

Correct friction. [1]

(d) Take downhill positive:

$mg\sin30.0^\circ-\mu_kmg\cos30.0^\circ=ma$

$a=g(\sin30.0^\circ-0.200\cos30.0^\circ)=3.21\,\mathrm{m\,s^{-2}}$

Signed force equation [1]; acceleration [1].

(e)

$v=1.00+a(2.00)=7.41\,\mathrm{m\,s^{-1}}$

Correct speed using unrounded acceleration. [1]

Answer: R = 42.5 N; friction 8.50 N uphill; acceleration 3.21 m/s² downhill; speed 7.41 m/s.

Extended response 10 — Challenge: angled pull and kinetic friction [8 marks]

An 8.00 kg crate slides right on a horizontal floor. A force of 35.0 N pulls at 30.0° above horizontal. The coefficient of kinetic friction is 0.250. There is no vertical acceleration.

Draw and label its free-body diagram. [2]
Determine normal reaction. [2]
Determine kinetic friction. [1]
Determine horizontal acceleration. [2]
Explain why the angled pull reduces friction compared with a horizontal pull of the same magnitude. [1]

Show worked answer

(a) Draw weight down, normal reaction up, applied force up-right at 30°, and friction left. Vertical forces [1]; pull and friction [1].

(b)

$R+F\sin30.0^\circ-mg=0$

$R=8.00(9.81)-35.0\sin30.0^\circ=60.98\,\mathrm{N}$

Vertical balance [1]; reaction [1].

(c)

$F_k=0.250(60.98)=15.245\,\mathrm{N}$

Correct friction. [1]

(d)

$a=\frac{35.0\cos30.0^\circ-15.245}{8.00}=1.88\,\mathrm{m\,s^{-2}}$

Horizontal resultant method [1]; acceleration [1].

(e) The upward component reduces R, so μ_kR is smaller. [1] This alone does not imply that the horizontal acceleration must be greater, since the horizontal pull component also decreases.

Answer: Normal reaction 61.0 N; friction 15.2 N; acceleration 1.88 m/s² right.

Extended response 11 — Challenge: two connected hanging masses [8 marks]

Masses 3.00 kg and 2.00 kg are connected by a light, inextensible string over a frictionless pulley with negligible rotational inertia. The masses are released from rest and the string remains taut.

Draw a free-body diagram for each mass. [2]
Write Newton’s second-law equations, taking the direction of each mass’s acceleration as positive. [2]
Determine the common acceleration magnitude. [2]
Determine string tension. [2]

Show worked answer

(a) Each mass has upward tension and downward weight. One mark per correctly labelled diagram. [2]

(b) The 3.00 kg mass accelerates down and the 2.00 kg mass up:

$3.00g-T=3.00a,\qquad T-2.00g=2.00a$

One mark per force equation. [2]

(c) Add equations to eliminate tension:

$a=\frac{(3.00-2.00)g}{3.00+2.00}=1.962\,\mathrm{m\,s^{-2}}$

Elimination/method [1]; acceleration [1].

(d)

$T=2.00(g+a)=23.544\,\mathrm{N}$

Substitution into a force equation [1]; tension [1].

Answer: Acceleration 1.96 m/s²; tension 23.5 N.

Check: Tension lies between the two weights; it is not equal to either weight while the masses accelerate.

Extended response 12 — Impulse from a non-constant force [7 marks]

A 0.300 kg ball initially moves at −5.00 m/s. During contact with a bat, the resultant horizontal force follows the graph defined below. Positive is right.

Resultant force–time graph; join adjacent coordinates with straight lines
t / s	F / N
0	0
0.0200	+120
0.0500	+120
0.0700	0

Determine the impulse by calculating the graph area. [3]
Determine average resultant force during contact. [1]
Determine final velocity. [2]
State how the horizontal contact impulse on the bat compares with that on the ball, assuming other horizontal forces on the ball are negligible during contact. [1]

Show worked answer

(a) The two triangles each contribute 1.20 N s and the rectangle contributes 3.60 N s:

$J=\frac12(0.0200)(120)+(0.0300)(120)+\frac12(0.0200)(120)=6.00\,\mathrm{N\,s}$

Triangular areas [1]; rectangular area [1]; total impulse [1].

(b)

$F_{\mathrm{avg}}=6.00/0.0700=85.7\,\mathrm{N}$

Correct average, smaller than the peak. [1]

(c)

$J=m(v-u),\qquad v=u+J/m=-5.00+6.00/0.300=+15.0\,\mathrm{m\,s^{-1}}$

Impulse–momentum method [1]; signed final velocity [1].

(d) The ball’s contact impulse on the bat is −6.00 N s, equal and opposite to the bat’s contact impulse on the ball. [1]

Answer: Impulse +6.00 N s; average force +85.7 N; final velocity +15.0 m/s.

Extended response 13 — Recoil and energy in an explosion [7 marks]

Two trolleys of masses 1.20 kg and 0.800 kg are initially at rest on a horizontal track, with a compressed light spring between them. Upon release, the 1.20 kg trolley moves left at 1.50 m/s. External horizontal impulse is negligible and energy losses are negligible.

Determine the other trolley’s velocity. [2]
Determine the total final kinetic energy. [2]
Identify the source of this kinetic energy. [1]
Compare the interaction forces during release. [1]
Compare the magnitudes of their accelerations during release. [1]

Show worked answer

(a) Take right positive:

$0=(1.20)(-1.50)+(0.800)v$

$v=+2.25\,\mathrm{m\,s^{-1}}$

Momentum equation [1]; signed velocity [1].

(b)

$E_k=\frac12(1.20)(1.50)^2+\frac12(0.800)(2.25)^2=3.38\,\mathrm{J}$

Both energy contributions [1]; total [1].

(c) Elastic potential energy stored in the spring becomes kinetic energy. [1]

(d) The interaction forces have equal magnitudes and opposite directions. [1]

(e) The lighter trolley’s acceleration magnitude is 1.20/0.800 = 1.50 times that of the heavier trolley. [1]

Answer: Other trolley velocity +2.25 m/s; total kinetic energy 3.38 J; lighter trolley acceleration is 1.50 times larger.

Check: Total momentum remains zero while kinetic energy increases.

Extended response 14 — Challenge: forces in a vertical circle [8 marks]

A 0.250 kg ball attached to a light string moves in a vertical circle of radius 0.800 m. Its measured speed is 3.50 m/s at the top and 6.61 m/s at the bottom. The string is taut. Neglect drag.

Draw the free-body diagram at the top and bottom. [2]
Determine tension at the top. [2]
Determine tension at the bottom. [2]
Find the minimum top speed for the string to remain just taut. [2]

Show worked answer

(a) At the top, both weight and tension point down toward the centre. [1] At the bottom, tension points up toward the centre and weight points down. [1]

(b) Take inward positive at the top:

$T_{\mathrm{top}}+mg=\frac{mv_{\mathrm{top}}^2}{r}$

$T_{\mathrm{top}}=\frac{0.250(3.50)^2}{0.800}-0.250(9.81)=1.38\,\mathrm{N}$

Radial force equation [1]; tension [1].

(c) At the bottom:

$T_{\mathrm{bottom}}-mg=\frac{mv_{\mathrm{bottom}}^2}{r}$

$T_{\mathrm{bottom}}=\frac{0.250(6.61)^2}{0.800}+0.250(9.81)=16.1\,\mathrm{N}$

Radial force equation [1]; tension [1].

(d) At the limiting top speed, T = 0:

$mg=mv_{\min}^2/r,\qquad v_{\min}=\sqrt{gr}=2.80\,\mathrm{m\,s^{-1}}$

Zero-tension condition [1]; speed [1].

Answer: Top tension 1.38 N; bottom tension 16.1 N; minimum top speed 2.80 m/s.

Check: A string can pull but cannot supply a negative tension. The ball’s speed need not be constant around a vertical circle.

Extended response 15 — HL: two-dimensional collision [9 marks]

Two pucks, each of mass 0.500 kg, collide on a frictionless horizontal surface. Puck A initially moves east at 8.00 m/s; puck B is initially stationary. Afterward, A moves at 4.00 m/s at 30.0° north of east. External horizontal impulse is negligible.

Determine A’s final eastward and northward velocity components. [2]
Apply momentum conservation separately to find B’s final velocity components. [3]
Determine B’s speed and direction. [2]
Calculate the total final kinetic energy and determine whether the collision is elastic. [2]

Show worked answer

(a) Take east as +x and north as +y:

$v_{Ax}=4.00\cos30.0^\circ=3.46410\,\mathrm{m\,s^{-1}},\qquad v_{Ay}=2.00\,\mathrm{m\,s^{-1}}$

One mark per component. [2]

(b) Equal masses cancel in each momentum equation:

$8.00=v_{Ax}+v_{Bx},\qquad 0=v_{Ay}+v_{By}$

$v_{Bx}=4.53590\,\mathrm{m\,s^{-1}},\qquad v_{By}=-2.00\,\mathrm{m\,s^{-1}}$

Component conservation method [1]; eastward component [1]; signed northward component [1].

(c)

$v_B=\sqrt{4.53590^2+(-2.00)^2}=4.96\,\mathrm{m\,s^{-1}}$

$\theta=\tan^{-1}(2.00/4.53590)=23.8^\circ$

Speed [1]; direction south of east [1].

(d)

$E_i=\frac12(0.500)(8.00)^2=16.0\,\mathrm{J}$

$E_f=\frac12(0.500)(4.00)^2+\frac12(0.500)v_B^2=10.1\,\mathrm{J}$

Correct final energy [1]. The collision is inelastic because final kinetic energy is lower; about 5.86 J becomes other energy forms. [1]

Answer: B moves at 4.96 m/s, 23.8° south of east; final total kinetic energy 10.1 J; collision inelastic.

Check: Northward and southward momentum components cancel. Adding speed magnitudes alone would not conserve vector momentum.

Section C — Lab-style investigation: momentum conservation

Sticking trolley collisions [20 suggested marks]

Research question: How does the common velocity after a sticking collision depend on the incoming velocity when the two trolley masses remain fixed?

Use a level, low-friction track. Trolley A has mass $m_A=(0.300\pm0.001)\,\mathrm{kg}$ and trolley B has mass $m_B=(0.200\pm0.001)\,\mathrm{kg}$ . B is stationary before each collision. Velcro bumpers make them stick. Measure A’s velocity immediately before contact and the common velocity immediately afterward using calibrated motion sensors or light gates. Keep clear of moving trolleys and use end stops.

These are illustrative practice data. Each row represents mean velocities from repeated trials. For this exercise, take the absolute uncertainty of each mean velocity as ±0.005 m/s. Positive is along the initial motion of A.

Mean velocities and processing columns
Trial setting	u / m s⁻¹	v / m s⁻¹	Initial momentum / kg m s⁻¹	Final momentum / kg m s⁻¹
1	0.500	0.298	________	________
2	0.600	0.362	________	________
3	0.700	0.418	________	________
4	0.800	0.481	________	________
5	0.900	0.539	________	________

Identify independent and dependent variables. [2]
Derive the predicted linear relationship between v and u, and calculate its gradient. [3]
Calculate initial momentum, final momentum and their difference for each row. [3]
State graph axes and use the first and last points to estimate the gradient. [2]
For the first row, evaluate whether the momentum difference is significant using the supplied conservative uncertainty model below. [4]
Calculate the first row’s initial and final kinetic energies and explain their difference. [2]
Identify two limitations and give a specific improvement for each. [4]

For part (e), use $\Delta p=(m_A+m_B)v-m_Au$ and the supplied first-order absolute-uncertainty model:

$\delta(\Delta p)\approx |v-u|\delta m_A+|v|\delta m_B+(m_A+m_B)\delta v+m_A\delta u$

This treats the same measured masses consistently in both initial and final momentum and gives a conservative approximate bound.

Show investigation answers and processed table

(a) Independent variable: A’s incoming velocity u. [1] Dependent variable: common outgoing velocity v. [1] The masses and initial rest condition of B are controlled.

(b) With negligible external impulse:

$m_Au=(m_A+m_B)v$

$v=\frac{m_A}{m_A+m_B}u=0.600u$

Conservation equation [1]; rearrangement [1]; predicted dimensionless gradient 0.600 [1]. The ideal intercept is zero.

(c) Use $p_i=0.300u$ , $p_f=0.500v$ and $\Delta p=p_f-p_i$ . Keep extra digits for the comparison. Method [1]; all initial/final values [1]; differences [1].

Processed momentum data; extra digits retained
Setting	pᵢ / kg m s⁻¹	p𝒻 / kg m s⁻¹	Δp / kg m s⁻¹
1	0.1500	0.1490	−0.0010
2	0.1800	0.1810	+0.0010
3	0.2100	0.2090	−0.0010
4	0.2400	0.2405	+0.0005
5	0.2700	0.2695	−0.0005

(d) Plot v vertically against u horizontally. [1]

$b_{\mathrm{endpoint}}=\frac{0.539-0.298}{0.900-0.500}=0.6025\approx0.603$

Correct gradient. [1] It is close to the prediction 0.600. A complete analysis should fit all points with uncertainty bars on both axes, allow an intercept initially and estimate gradient uncertainty before making a quantitative agreement claim.

(e) For the first row:

$\Delta p=(0.500)(0.298)-(0.300)(0.500)=-0.0010\,\mathrm{kg\,m\,s^{-1}}$

Correct signed difference. [1] Apply the supplied model [1]:

$\delta(\Delta p)\approx(0.202)(0.001)+(0.298)(0.001)+(0.500)(0.005)+(0.300)(0.005)$

$\delta(\Delta p)\approx0.0045\,\mathrm{kg\,m\,s^{-1}}$

Correct uncertainty. [1] Zero lies within the approximate interval −0.0010 ± 0.0045 kg m/s, so the difference is not significant at the stated uncertainty level. [1]

Conclusion: The first trial is consistent with momentum conservation within the supplied measurement limits. Agreement does not establish exact equality or eliminate all systematic errors.

(f)

$E_i=\frac12(0.300)(0.500)^2=0.0375\,\mathrm{J}$

$E_f=\frac12(0.500)(0.298)^2=0.0222\,\mathrm{J}$

Correct energies [1]. About 0.0153 J becomes deformation, internal energy and sound in the sticking collision; kinetic energy is not conserved even though momentum is approximately conserved. [1]

(g) Accept two distinct limitation–improvement pairs, one mark for each limitation and one for its matching improvement. [4]

External friction or a tilted track: Level the track carefully, use low-friction bearings or an air track, and measure velocities close to contact so external impulse is minimized.
Sensor zero or velocity calibration error: Calibrate using known distances and timing, verify the zero reading and check any light-gate flag length.
B moves before contact or collisions are not straight: Verify B is stationary and guide alignment so the motion remains one-dimensional.
Random variation between trials: Repeat each incoming-speed setting, calculate means and inspect the scatter instead of relying on a single collision.

Evaluation: The measured trend supports the predicted proportional relation. A fuller investigation would determine uncertainties for every momentum difference and compare a fitted gradient with its mass-based prediction.

Mastery checklist

Draw forces acting on the chosen object only.
Apply Newton’s laws with consistent directions and signs.
Distinguish actual static friction from its maximum.
Resolve angled forces before finding normal reaction or acceleration.
Use signed momentum and graph areas to calculate impulse.
State why external impulse is negligible before conserving momentum.
Compare kinetic energy to classify collisions.
For HL two-dimensional collisions, conserve momentum separately in each direction.
Identify which physical forces provide the inward resultant in circular motion.
Evaluate measurements using uncertainty rather than exact numerical equality.

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Forces and momentum HL Paper 1&2

Instructions and data

Equation reminders

Common misconceptions

Section A — Paper 1A-style multiple choice

Multiple choice 1 — Newton’s first law [1 mark]

Multiple choice 2 — Newton’s third law [1 mark]

Multiple choice 3 — Actual static friction [1 mark]

Multiple choice 4 — Apparent weight [1 mark]

Multiple choice 5 — Normal reaction under an angled pull [1 mark]

Multiple choice 6 — Signed spring force [1 mark]

Multiple choice 7 — Terminal velocity [1 mark]

Multiple choice 8 — Impulse during a rebound [1 mark]

Multiple choice 9 — Force–time area [1 mark]

Multiple choice 10 — A sticking collision [1 mark]

Multiple choice 11 — Explosion from rest [1 mark]

Multiple choice 12 — Elastic collision of identical masses [1 mark]

Multiple choice 13 — Centripetal acceleration scaling [1 mark]

Multiple choice 14 — Loss of the inward force [1 mark]

Multiple choice 15 — Friction-limited circular motion [1 mark]

Section B — Paper 2-style extended responses

Extended response 1 — Resultant force and acceleration [4 marks]

Extended response 2 — Static friction below its maximum [6 marks]

Extended response 3 — Weight, normal force and third-law partners [6 marks]

Extended response 4 — Hooke’s law and signed displacement [5 marks]

Extended response 5 — Rebound, impulse and average force [7 marks]

Extended response 6 — Perfectly inelastic collision [7 marks]

Extended response 7 — Uniform horizontal circular motion [7 marks]

Extended response 8 — Terminal velocity and model consistency [8 marks]

Extended response 9 — Challenge: sliding down a rough incline [8 marks]

Extended response 10 — Challenge: angled pull and kinetic friction [8 marks]

Extended response 11 — Challenge: two connected hanging masses [8 marks]

Extended response 12 — Impulse from a non-constant force [7 marks]

Extended response 13 — Recoil and energy in an explosion [7 marks]

Extended response 14 — Challenge: forces in a vertical circle [8 marks]

Extended response 15 — HL: two-dimensional collision [9 marks]

Section C — Lab-style investigation: momentum conservation

Sticking trolley collisions [20 suggested marks]

Mastery checklist

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