Dynamics of a particle moving in a straight line or plane
Edexcel International A-Level MathematicsΒ· 2018 Specification Issue 3, M1 Β§4.1-4.4Β· 25 min read
1. Newton's Laws of Motion for Constant Accelerationβ β ββββ± 5 min
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Newton's Second Law of Motion
The resultant force acting on a particle equals the product of its mass and acceleration: (F_{\text{net}} = ma). For 2D motion, this applies separately to i and j components.
Example:
A 2kg particle with resultant force (6i - 4j) N has acceleration (3i - 2j) m sβ»Β².
A 3kg particle is acted on by two forces: (F_1 = (2i + 5j)) N and (F_2 = (7i - 11j)) N. Calculate the magnitude of the particle's acceleration, to 2 decimal places.
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Step 1: Calculate resultant force by adding vector components
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Step 2: Apply (F_{net}=ma) to find acceleration vector
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Step 3: Calculate magnitude using Pythagoras
Exam tip:
Always resolve forces into perpendicular components (horizontal/vertical or parallel/perpendicular to an incline) before applying Newton's second law.
2. Connected Particles and Inclined Planesβ β β βββ± 7 min
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Connected particle problems involve two or more masses linked by light inextensible strings, often over smooth fixed pulleys. For rough inclined planes, include kinetic friction if the particle is moving.
Light inextensible string
String with negligible mass that does not stretch, meaning tension is equal throughout the string, and connected particles accelerate at the same magnitude.
Example:
Two masses hanging over a pulley accelerate at the same rate, with equal tension on both sides of the pulley.
A 2kg block rests on a rough horizontal table, connected by a light inextensible string over a smooth fixed pulley to a 3kg mass hanging vertically. The coefficient of kinetic friction between the block and table is 0.4. Calculate the acceleration of the system, to 2 decimal places.
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Step 1: Assign positive direction as the hanging mass moving downward, so both masses accelerate at rate (a)
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Step 2: Write force equation for hanging mass
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Step 3: Calculate normal reaction for block: (R = 2g), so friction (F = \mu R = 0.8g). Write force equation for block:
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Step 4: Add equations to eliminate T
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Step 5: Solve for a
Exam tip:
If a connected particle hits the ground, the string goes slack so tension becomes zero, and the remaining particle moves under new forces (e.g. friction) with a new acceleration.
3. Impulse and 1D Conservation of Momentumβ β β βββ± 6 min
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Impulse-Momentum Principle
Impulse equals the change in momentum of a particle: (I = mv - mu), where (u) is initial velocity, (v) is final velocity, and (m) is mass. Momentum is conserved for 1D collisions with no external forces: total momentum before = total momentum after.
A 0.5kg ball moving at 4 m sβ»ΒΉ right collides directly with a 0.3kg ball moving at 2 m sβ»ΒΉ left. After collision, the 0.3kg ball moves at 3 m sβ»ΒΉ right. Calculate the final velocity of the 0.5kg ball, stating direction.
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Step 1: Assign positive direction as right, so (u_1 = 4) m sβ»ΒΉ, (u_2 = -2) m sβ»ΒΉ, (v_2 = 3) m sβ»ΒΉ
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Step 2: Apply conservation of momentum
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Step 3: Substitute values
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Step 4: Final result: (v_1 = 1) m sβ»ΒΉ to the right.
Exam tip:
Always state your chosen positive direction for velocity in momentum problems to avoid sign errors and show clear reasoning to examiners.
4. Kinetic Friction Calculationsβ β β β ββ± 5 min
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Kinetic Friction
Friction force acting on a moving particle, given by (F = \mu R), where (\mu) is the coefficient of kinetic friction and (R) is the normal reaction force. Friction always acts opposite to the direction of motion.
A 5kg particle slides down a rough plane inclined at 30Β° to the horizontal. The coefficient of kinetic friction is 0.2. Calculate the acceleration of the particle, to 2 decimal places.
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Step 1: Resolve perpendicular to plane to find normal reaction
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Step 2: Calculate friction force
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Step 3: Resolve parallel to plane, apply (F_{net}=ma)
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Step 4: Solve for a
Exam tip:
Use (\sin(\theta)) for the parallel component of weight on an incline, and (\cos(\theta)) for the perpendicular component. Mixing these is a very common error.
5. Common Pitfalls
Wrong move:
Using F=ΞΌR for stationary particles
Why:
F=ΞΌR only applies to kinetic friction for moving particles. Static friction is not covered in M1 and this relationship does not hold for stationary objects.
Correct move:
Only use F=ΞΌR when the question explicitly states the particle is moving, or you have confirmed it is in motion.
Wrong move:
Ignoring velocity direction in momentum calculations
Why:
Momentum is a vector quantity, so particles moving opposite to your chosen positive direction have negative velocity values. Omitting signs leads to incorrect total momentum sums.
Correct move:
Define a positive direction at the start of all momentum problems, and assign negative values to velocities opposite to this direction.
Wrong move:
Using g=10 m sβ»Β² instead of 9.8 m sβ»Β²
Why:
Edexcel IAL M1 explicitly requires g=9.8 m sβ»Β². Using g=10 will lead to incorrect answers and lose accuracy marks.
Correct move:
Always use g=9.8 m sβ»Β² in all calculations unless the question explicitly states otherwise.
Wrong move:
Assuming tension differs on either side of a smooth pulley
Why:
Smooth fixed pulleys have no friction, so tension is identical throughout a light inextensible string passing over them.
Correct move:
Use the same tension value for both sides of a smooth fixed pulley when writing force equations for connected particles.
Wrong move:
Applying conservation of momentum when external forces act
Why:
Conservation of momentum only holds if there are no external forces (e.g. friction, applied forces) acting on the system of colliding particles.
Correct move:
Only use conservation of momentum for direct 1D collisions between two particles with no external forces acting during impact.
6. Quick Reference Cheatsheet
Concept | Formula/Rule | Key Note |
|---|---|---|
Newton's Second Law (scalar) | (F_{\text{net}} = ma) | Apply to resolved force components |
Newton's Second Law (vector) | (\mathbf{F}_{\text{net}} = m\mathbf{a}) | Apply separately to i and j components |
Kinetic Friction | (F = \mu R) | Only use for moving particles |
Impulse | (I = mv - mu) | Vector quantity, sign depends on direction |
Conservation of Momentum | (m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2) | 1D only, no external forces, assign positive direction first |
Connected Particles (smooth pulley) | Tension equal on both sides, acceleration magnitude identical | If one particle hits ground, tension becomes zero |
7. Frequently Asked
When do I use F = ΞΌR?
You only use (F = \mu R) for kinetic friction, when the particle is moving along the contact surface. If the particle is stationary, this relationship does not apply in M1.
Do I need to consider velocity direction in 1D collision problems?
Yes! Always assign a positive direction for velocity first. Particles moving opposite to this direction have negative velocity values in your calculations.
Going deeper
What's Next
Now that you have mastered core M1 dynamics concepts, you are ready to practice full exam-style questions and move on to subsequent M1 topics. Dynamics accounts for ~25% of marks in most M1 papers, so practicing mixed problem sets will help you build speed and accuracy for your exam. Make sure you also review vector applications for 2D force problems, as these are often combined with dynamics in longer 6-8 mark exam questions. Once you are comfortable with dynamics, you can move on to the next M1 topic: moments.