Forces, Newton's Laws and Momentum
Forces and dynamics cover contact and non-contact forces, resultant force, Newton's three laws of motion including F = ma, and momentum (p = mv) and its conservation in collisions and explosions. Stopping distance calculations and vehicle safety features are explained in terms of these forces and momentum principles.
Before you start
Make sure you're comfortable with these topics first:
Method
- List every force acting on the object, sort them into contact forces (friction, air resistance, tension, normal contact force) or non-contact forces (gravity, electrostatic, magnetic), then combine forces acting in the same line to find the resultant force.
- Apply Newton's Second Law, F = m a, rearranging to find whichever quantity (force, mass or acceleration) is unknown, and remember that F here means the resultant force, not just one of several forces acting.
- For a momentum question, use p = m v for a single object, or apply conservation of momentum (total momentum before an event equals total momentum after) for a collision or explosion, taking care with direction: choose one direction as positive and treat the opposite direction as negative.
- In the required practical on force, mass and acceleration, remember the runway is tilted slightly beforehand to compensate for friction, and that the total mass of the trolley-plus-hanging-masses system is kept constant while only the accelerating force is varied (masses are moved from the trolley to the hanger between repeats, not added).
- For a Newton's Third Law question, name the two different objects the pair of forces acts on, and state that the two forces are equal in size and opposite in direction, to show they do not cancel out (since they act on different objects).
- For a stopping distance question, treat thinking distance (affected by reaction time: tiredness, alcohol, drugs, distractions, and speed) and braking distance (affected by speed, mass, road and weather conditions, and brake or tyre condition) as two separate causes, and add them together for the total stopping distance.
Worked example
A go-kart and driver have a combined mass of 180 kg. The engine provides a driving force of 450 N, while friction and air resistance provide a total resistive force of 90 N. Calculate the go-kart's acceleration.
- Find the resultant force: resultant force = driving force - resistive force = 450 - 90 = 360 N.
- Write down Newton's Second Law: F = m a, rearranged to a = F / m.
- Substitute the values: a = 360 / 180.
- Calculate: a = 2.
- Final answer: the go-kart's acceleration is 2 m/s^2.
Practice questions
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Q1State Newton's First Law of Motion.Show answer
Answer: If the resultant force acting on an object is zero, the object stays at rest, or continues to move at a constant velocity.
Q2A shopping trolley has a mass of 25 kg. Calculate its weight, taking g = 9.8 N/kg.Show answer
Answer: 245 N (25 x 9.8 = 245).
Q3A resultant force of 60 N acts on a wheelbarrow of mass 15 kg. Calculate its acceleration.Show answer
Answer: 4 m/s^2 (60 / 15 = 4).
Q4Calculate the momentum of a 1200 kg car travelling at 18 m/s.Show answer
Answer: 21600 kg m/s (1200 x 18 = 21600).
Q5Explain, using Newton's Third Law, why a swimmer is able to push themselves forward by pushing backward against the water.Show answer
Answer: The swimmer exerts a backward force on the water; the water exerts an equal and opposite forward force on the swimmer. Since these two forces act on different objects (the swimmer and the water), they do not cancel out, so the swimmer accelerates forward.
Q6State two factors that increase a driver's thinking distance.Show answer
Answer: Any two of: tiredness, alcohol or drugs, distractions, or a higher speed.
Q7A skateboarder of mass 55 kg moving at 4 m/s collides with and grabs onto a stationary shopping trolley of mass 20 kg, and they move off together. Calculate their combined velocity immediately after the collision.Show answer
Answer: 2.93 m/s (3 s.f.) (momentum before = 55 x 4 = 220 kg m/s; combined mass = 75 kg; v = 220 / 75 = 2.93 m/s).
Q8Explain why a larger vehicle mass increases braking distance.Show answer
Answer: A larger mass has more kinetic energy at the same speed, so more work must be done by the brakes (more energy must be transferred as heat) to bring the vehicle to rest, which needs a greater braking distance.
Exam-style questions
Written in the style of a GCSE Science exam paper, with a full mark scheme.
In an investigation of Newton's Second Law, a trolley of total mass 0.80 kg is accelerated by a hanging mass on a string over a pulley, giving a resultant accelerating force of 2.4 N. Light gates measure the trolley's velocity as 0.60 m/s at the first gate and 1.80 m/s at the second gate, with the trolley taking 0.50 s to travel between the gates. (a) Calculate the trolley's acceleration between the two light gates. (b) Use your answer to part (a) to calculate the mass predicted by F = ma, and suggest one reason why this differs from the trolley's actual mass of 0.80 kg.
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A stationary firework rocket of total mass 2.0 kg explodes into two fragments. One fragment, of mass 0.50 kg, is thrown to the left at a velocity of 12 m/s. Calculate the velocity of the other fragment, of mass 1.5 kg, stating its direction.
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Cars are fitted with safety features such as crumple zones, seatbelts and airbags. Explain, using ideas about momentum and force, how these safety features reduce the risk of injury to passengers in a collision.
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See real GCSE Science past-paper questions, with official mark schemes →
Free printable worksheet
Want more practice on paper? Download the forces, newton's laws and momentum worksheet pack - 18 pages of exam-style questions with a full mark scheme. One email opens every download in this browser for 14 days - no account, no card. Print it for personal and classroom use.
This topic is chapter 14 of GCSE Physics Workbook, the whole course as one free printable PDF.
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