Dynamics of a particle moving in a straight line practice problems Edexcel IAL Mechanics 1, Chapter 3
Newton's laws, F = ma along a line, friction with F = μR, particles on inclined planes, lifts, and connected particles over pulleys. Try each question before you look: use the hint if you are stuck, then reveal the worked solution one step at a time.
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Key equations for dynamics of a particle moving in a straight line
| F = ma | resultant force = mass × acceleration |
|---|---|
| W = mg | weight, with g = 9.8 |
| F ≤ μR | friction; F = μR when moving or about to move |
| mg sin α, mg cos α | weight resolved along and perpendicular to a plane at angle α |
Take g = 9.8 m s−2, as the WME01 paper asks, and give answers to 2 or 3 significant figures. Answers within 2.5% are marked correct, so rounding along the way is fine. Revise the methods in our A level Maths mechanics guide.
Warm-up questions
Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.
Hint
F = ma.
Worked solution
- 12 = 3a.
- a = 4 m s−2.
Hint
W = mg with g = 9.8.
Worked solution
- W = 5 × 9.8 = 49 N.
Hint
Light and inextensible mean different things.
Worked solution
- Light means the mass of the string is ignored, so the tension is the same all along it.
- Inextensible is the one that means the particles share the same acceleration. The answer is B.
Exam-style questions
Questions at the level of the Edexcel IAL Mechanics 1 exam. The ones marked Challenge combine more than one idea.
Hint
Draw her forces: R up, 60g down. The resultant is upwards.
Worked solution
- Resolve upwards, F = ma: R − 60g = 60 × 1.2.
- R = 60 × 9.8 + 72 = 660 N.
Hint
First find R by resolving vertically, then friction = μR because the box is moving.
Worked solution
- Vertically: R = 2g = 19.6 N, so friction F = 0.4 × 19.6 = 7.84 N.
- Horizontally: 15 − 7.84 = 2a.
- a = 3.58 m s−2.
Hint
Resolve perpendicular to the plane for R, then along the plane. Does the mass matter?
Worked solution
- Perpendicular: R = mg cos 30°. Friction acts up the plane: F = 0.2mg cos 30°.
- Along the plane: mg sin 30° − 0.2mg cos 30° = ma, and m cancels.
- a = 9.8(0.5 − 0.2 × 0.866) = 3.20 m s−2.
Hint
Write F = ma for each particle separately, then add the equations to remove T.
Worked solution
- For A (horizontal): T = 3a. For B (downwards): 2g − T = 2a.
- Adding: 2g = 5a.
- a = 19.6/5 = 3.92 m s−2.
Hint
Find the acceleration first, from the two equations of motion, then substitute back.
Worked solution
- 5 kg (down): 5g − T = 5a. 3 kg (up): T − 3g = 3a.
- Adding: 2g = 8a, so a = 2.45 m s−2.
- T = 3(9.8 + 2.45) = 36.8 N (3 s.f.).
Exam question practice
One original structured question in the style of the WME01 paper. Answer it in full, with working, before you open the mark scheme: M marks are for a correct method, A marks for accurate answers that follow from it, and B marks for a correct statement on its own.
Structured exam question style · 6 points · about 10 minutes
A box of mass 10 kg is pulled along rough horizontal ground by a rope inclined at 30° above the horizontal. The tension in the rope is 50 N and the coefficient of friction between the box and the ground is 0.3. The box is modelled as a particle.
- (a) Find the normal reaction between the box and the ground.
- (b) Find the acceleration of the box.
- (c) State how you have used the modelling assumption that the box is a particle.
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
Vertically: R + 50 sin 30° = 10g, so R = 98 − 25 = 73 N.
Part (b)
Friction = 0.3 × 73 = 21.9 N. Horizontally: 50 cos 30° − 21.9 = 10a, so a = (43.30 − 21.9)/10 = 2.14 m s−2.
Part (c)
The box's size is ignored, so all the forces act at one point and the box does not rotate.
Your mark: 0 of 6
Common mistakes in dynamics of a particle moving in a straight line
- Taking R = mg when another force has a vertical component, as in part (a) above.
- Using F = μR when the particle is at rest and not on the point of moving: then F ≤ μR.
- In pulley questions, writing one equation for the whole system and then using it to find the tension.
- Forgetting that friction always opposes the motion, so it changes direction if the particle does.
Keep going
Revise the ideas behind these questions:
Then try real exam questions: Edexcel IAL Mechanics 1 past papers, with official scoring guidelines. · All Edexcel IAL Mechanics 1 units