Unit 5 / Practice Drills

Unit 5 Practice Drills

Kinetics

Newton second law, work-energy, and impact. This unit scores well if you pick the right method family for the question.

5 drills5 openfree through June 30, 2026Newton on an incline plus direct impulse-momentumFree beta active

Drill 1 - Newton on an Incline

Problem

An 18 kg block slides down a rough plane inclined at 25° to the horizontal. The coefficient of kinetic friction between the block and the plane is 0.20. Find the acceleration of the block.

Figure 1A

Block on a 25° rough incline

Step 1 - Resolve weight parallel and perpendicular to the plane.

Weight component down the plane = 18 × 9.81 × sin 25° = 74.63 N

Normal reaction = 18 × 9.81 × cos 25° = 160.04 N

Step 2 - Find the friction force.

Friction = μN = 0.20 × 160.04 = 32.01 N

Step 3 - Apply F = ma along the plane.

Net force down the plane = 74.63 − 32.01 = 42.62 N

42.62 = 18a

a = 2.37 m/s² down the plane

Worked Diagram Hidden

Reveal solution figure

Open this only after you have attempted the setup yourself.

Figure 1B

Worked incline resolution and net force

Where students lose marks

Do not write friction as μmg. On an incline the friction force is μN, and the normal reaction is mg cos θ, not the full weight.

Drill 2 - Work-Energy

Problem

A 12 kg crate is pulled along a rough horizontal floor by a constant 90 N force through a distance of 6 m. The coefficient of kinetic friction is 0.20. Starting from rest, find the speed of the crate after it has moved 6 m.

Figure 1A

Crate pulled 6 m across a rough floor

Step 1 - Write the initial and final kinetic-energy terms.

Initial KE = 0 because the crate starts from rest

Final KE = ½mv²

Step 2 - Find the work done by the pull and by friction.

Normal reaction = 12 × 9.81 = 117.72 N

Friction = 0.20 × 117.72 = 23.544 N

Work by pull = 90 × 6 = 540 J

Work by friction = -23.544 × 6 = -141.264 J

Step 3 - Apply the work-energy equation.

540 - 141.264 = ½ × 12 × v²

398.736 = 6v²

v² = 66.456

v = 8.15 m/s

Worked Diagram Hidden

Reveal solution figure

Open this only after you have attempted the setup yourself.

Figure 1B

Worked energy balance for the crate

Where students lose marks

Students often add the friction work instead of subtracting it. Friction opposes the displacement here, so its work term is negative.

Drill 3 - Impact and Restitution

Problem

A 2 kg ball moving at 6 m/s catches up with a 3 kg ball moving at 1 m/s in the same straight line. If the coefficient of restitution is 0.5, find the velocities of both balls after impact.

Figure 1A

Two balls moving in the same line before impact

Step 1 - Write the momentum equation.

Take the original direction of motion as positive.

2 × 6 + 3 × 1 = 2v₁ + 3v₂

15 = 2v₁ + 3v₂

Step 2 - Write the restitution equation.

e = (v₂ - v₁) / (u₁ - u₂)

0.5 = (v₂ - v₁) / (6 - 1)

v₂ - v₁ = 2.5

Step 3 - Solve the two equations together.

v₂ = v₁ + 2.5

15 = 2v₁ + 3(v₁ + 2.5)

15 = 5v₁ + 7.5

v₁ = 1.5 m/s, v₂ = 4.0 m/s

Worked Diagram Hidden

Reveal solution figure

Open this only after you have attempted the setup yourself.

Figure 1B

Worked impact result with restitution

Where students lose marks

Momentum and restitution are separate equations. Students often try to fold restitution into momentum directly and lose the sign convention. Keep one positive direction from the start.

Drill 4 - Man on Rotating Platform

Problem

A person stands 1.5 m from the centre of a horizontal rotating platform. The platform starts from rest with constant angular acceleration 0.8 rad/s². If the coefficient of static friction is 0.40, find the time at which the person just begins to slip.

Figure 1A

Person on a rotating platform at radius 1.5 m

Step 1 - Write the tangential acceleration demand.

aₜ = rα = 1.5 × 0.8 = 1.2 m/s²

Step 2 - Write the friction limit and the normal acceleration demand.

Maximum available friction acceleration = μₛg = 0.4 × 9.81 = 3.924 m/s²

ω = αt = 0.8t

aₙ = rω² = 1.5(0.8t)² = 0.96t²

Step 3 - Combine the perpendicular acceleration demands.

Slip begins when √(aₜ² + aₙ²) = μₛg

√(1.2² + (0.96t²)²) = 3.924

1.44 + 0.9216t⁴ = 15.3978

t⁴ = 15.144

t = 1.97 s

Worked Diagram Hidden

Reveal solution figure

Open this only after you have attempted the setup yourself.

Figure 1B

Worked friction-limit check for impending slip

Where students lose marks

This is not a one-component friction check. Static friction must supply both the tangential acceleration and the normal acceleration at the same time, so compare the vector resultant with μₛg.

Unit 5 Practice Drills

Drill 5 - Curvilinear Motion: Normal and Tangential Acceleration

5 questions. Covers circular path problems, speed increasing on a curve, and recurring exam-family variants.

This drill set is about identifying whether a problem changes speed, changes direction, or does both. Once that is clear, the method is mechanical: find aₜ, find aₙ, combine them with Pythagoras, then interpret the direction.

Figure 0A

Reference: tangential, normal, and total acceleration

Key formulas

Quantity	Formula
Normal (centripetal) acceleration	aₙ = v² / r
Tangential acceleration	aₜ = dv/dt
Total acceleration	a = √(aₙ² + aₜ²)
Direction of total acceleration	θ = tan⁻¹(aₙ / aₜ)
When aₜ = f(s)	v dv/ds = aₜ → integrate
When aₜ = constant	v² = u² + 2aₜs

Normal acceleration points toward the centre of curvature. Tangential acceleration is along the path. They are always perpendicular.

Question 1 - Repeat-family variant

Problem

A truck enters a circular road of radius 50 m. Starting from rest, its tangential acceleration is given by aₜ = 0.05s m/s², where s is the distance travelled in metres. Find the magnitude and direction of the total acceleration after the truck has travelled s = 10 m.

Adapted from a curvilinear problem family that appeared in May/Jun 2024 and April 2025.

Figure 1A

Truck on circular road with aₙ and aₜ at P

Step 1 - Find aₜ at s = 10 m.

aₜ = 0.05 × 10 = 0.5 m/s²

Step 2 - Find velocity at s = 10 m by integrating.

v dv/ds = aₜ = 0.05s

∫v dv = ∫0.05s ds

v²/2 = 0.025s² + C

At s = 0, v = 0, so C = 0

v² = 0.05s²

At s = 10: v² = 0.05 × 100 = 5 m²/s²

Step 3 - Find normal acceleration.

aₙ = v²/r = 5/50 = 0.1 m/s²

Step 4 - Find total acceleration.

a = √(aₙ² + aₜ²) = √(0.1² + 0.5²) = √0.26 = 0.510 m/s²

Step 5 - Find direction.

θ = tan⁻¹(aₙ/aₜ) = tan⁻¹(0.1/0.5) = 11.3° from the tangential direction toward the centre

Worked Diagram Hidden

Reveal solution figure

Open this only after you have attempted the setup yourself.

Figure 1B

Acceleration triangle for Q1

Where students lose marks

Using v² = u² + 2aₜs here is wrong because aₜ = 0.05s is not constant. You must integrate v dv/ds = 0.05s. Substituting directly gives v² = 10, which corrupts aₙ and the final answer.

Question 2 - Constant tangential acceleration on a circular track

Problem

A runner on a circular track of diameter 126 m increases their speed uniformly from 4.2 m/s to 7.2 m/s over a distance of 28.5 m along the track. Find the total acceleration of the runner at the instant their speed reaches 7.2 m/s.

Adapted from a recurring problem family seen in May/Jun 2022 and Nov/Dec 2025.

Figure 2A

Runner on circular track from A to B

Step 1 - Identify radius.

Diameter = 126 m, so r = 63 m

Step 2 - Find aₜ using v² = u² + 2aₜs.

Here aₜ is constant because the speed increases uniformly.

(7.2)² = (4.2)² + 2 × aₜ × 28.5

51.84 = 17.64 + 57aₜ

aₜ = 0.6 m/s²

Step 3 - Find aₙ at v = 7.2 m/s.

aₙ = v²/r = 51.84/63 = 0.823 m/s²

Step 4 - Find total acceleration.

a = √(aₙ² + aₜ²) = √(0.823² + 0.6²) = √1.037 = 1.018 m/s²

Step 5 - Find direction.

θ = tan⁻¹(aₙ/aₜ) = tan⁻¹(0.823/0.6) = 53.9° from the tangential direction

Worked Diagram Hidden

Reveal solution figure

Open this only after you have attempted the setup yourself.

Figure 2B

Acceleration triangle for Q2

Where students lose marks

The problem gives diameter, not radius. Always halve the diameter before using aₙ = v²/r. Using 126 instead of 63 halves the normal acceleration.

Question 3 - Finding speed at a given total acceleration

Problem

A racing car travels on a circular track of radius 100 m. Its tangential acceleration is constant at aₜ = 7 m/s². The car starts from rest. Find the time at which the total acceleration of the car equals 8 m/s².

Adapted from a recurring problem family seen in May/Jun 2023 and Nov/Dec 2023.

Figure 3A

Known total acceleration and tangential acceleration

Step 1 - Find aₙ using the total acceleration.

a² = aₙ² + aₜ²

8² = aₙ² + 7²

64 = aₙ² + 49

aₙ = √15 = 3.873 m/s²

Step 2 - Find the speed at that instant.

aₙ = v²/r → v² = aₙ × r = 3.873 × 100 = 387.3

v = 19.68 m/s

Step 3 - Find the time to reach that speed.

Car starts from rest, so u = 0.

v = u + aₜt

19.68 = 7t

t = 2.81 s

Worked Diagram Hidden

Reveal solution figure

Open this only after you have attempted the setup yourself.

Figure 3B

Acceleration triangle for Q3

Where students lose marks

Work backward. The total acceleration is given, so isolate aₙ first from a² = aₙ² + aₜ², then find v from aₙ = v²/r, then find t from v = u + aₜt.

Question 4 - Normal acceleration only, constant speed on a curve

Problem

A car travels at a constant speed of 72 km/h along a curved section of road. The radius of curvature is 80 m. Find: (a) the normal acceleration, (b) the total acceleration, (c) the minimum coefficient of friction required between the tyres and road to maintain this speed without skidding.

Figure 4A

Constant-speed motion on a curve

Step 1 - Convert speed to m/s.

v = 72 km/h = 72/3.6 = 20 m/s

Step 2 - Find normal acceleration.

aₙ = v²/r = 400/80 = 5 m/s²

Step 3 - Find total acceleration.

aₜ = 0, so a = √(aₙ² + 0²) = 5 m/s²

Step 4 - Find minimum coefficient of friction.

For no skidding, μmg ≥ maₙ

μ ≥ aₙ/g = 5/9.81 = 0.51

Minimum μ = 0.51

Worked Diagram Hidden

Reveal solution figure

Open this only after you have attempted the setup yourself.

Figure 4B

Free body diagram: friction provides centripetal force

Where students lose marks

Constant speed means aₜ = 0. In this case total acceleration equals aₙ only. Do not invent a tangential component that the problem does not have.

Question 5 - Combined: speed increasing on a curve, find position where total acceleration reaches a limit

Problem

A motorcycle enters a circular bend of radius 120 m at a speed of 10 m/s. Its engine provides a constant tangential acceleration of 2 m/s². The road surface can provide a maximum total acceleration of 6 m/s² before the motorcycle skids. Find the distance along the bend at which the motorcycle reaches this limit.

Figure 5A

Motorcycle reaching the total-acceleration limit

Step 1 - Find aₙ at the limit.

6² = aₙ² + 2²

36 = aₙ² + 4

aₙ = √32 = 5.657 m/s²

Step 2 - Find speed at that point.

aₙ = v²/r → v² = aₙ × r = 5.657 × 120 = 678.9 m²/s²

Step 3 - Find distance using v² = u² + 2aₜs.

aₜ is constant at 2 m/s².

v² = u² + 2aₜs

678.9 = 10² + 2 × 2 × s

678.9 = 100 + 4s

s = 144.7 m

Worked Diagram Hidden

Reveal solution figure

Open this only after you have attempted the setup yourself.

Figure 5B

Solved diagram for Q5

Where students lose marks

aₜ is constant here, so use v² = u² + 2aₜs directly. Save the integration method for questions like Q1 where aₜ depends on s.

Figure summary for Drill 5

Figure	Type	Purpose
0A	Annotated curved path	Reference: aₙ, aₜ, resultant a, angle θ
1A	Top-down circular road	Truck position with aₙ and aₜ at P
1B	Acceleration vector triangle	Q1 values filled in
2A	Circular track with arc AB	Runner positions and speed change
2B	Acceleration vector triangle	Q2 values filled in
3A	Circular track with unknowns	Known total a with unknown aₙ
3B	Acceleration vector triangle	Q3 values filled in
4A	Car on curved road	Constant speed so only aₙ matters
4B	Free body diagram	Weight, normal reaction, friction toward centre
5A	Circular road with limit point	Arc s left blank, a = 6 given
5B	Same diagram with values	Arc 144.7 m and labeled triangle

The most important figures here are 0A and 1A. Figure 0A stays visible because students consistently confuse the directions of aₙ and aₜ. Figure 1A makes the split between the integration case and the constant-aₜ case visible before any algebra begins.