Newton's Laws of Motion, Momentum and Stopping Distances
Newton's three laws of motion describe how forces affect the motion of objects: an object stays at rest or at constant velocity unless a resultant force acts on it (first law); a resultant force causes acceleration according to F = m x a (second law); and forces always occur in equal and opposite interaction pairs (third law). Momentum is a property of a moving object given by momentum = mass x velocity, and in any collision or explosion the total momentum before equals the total momentum after, provided no external force acts (conservation of momentum). Stopping distance is the sum of thinking distance (the distance travelled during the driver's reaction time) and braking distance (the distance travelled while the brakes are working), extending beyond the core GCSE course to include quantitative momentum and impact-force calculations.
Before you start
Make sure you're comfortable with these topics first:
Method
- Identify all the forces acting on the object and find the resultant (net) force by adding forces in the same direction and subtracting forces in opposite directions.
- If the resultant force is zero, the object is in equilibrium: it stays at rest or continues at constant velocity (Newton's first law).
- If there is a resultant force, use F = m x a to calculate the force, mass or acceleration, rearranging as needed and keeping units in newtons, kilograms and m/s^2.
- For momentum questions, calculate momentum = mass x velocity for each object before and after the event, and apply conservation of momentum: total momentum before = total momentum after.
- For a collision or explosion, set up an equation with total momentum before equal to total momentum after, then solve for the unknown mass or velocity.
- For stopping distance questions, calculate thinking distance and braking distance separately, then add them for the total stopping distance; identify factors like speed, reaction time, tiredness, alcohol or drugs (thinking distance), and road surface, weather, tyre condition or brake condition (braking distance).
- For impact-force questions, use force = change in momentum / time taken, showing why a longer collision time (as in an airbag or crumple zone) reduces the force on the occupants.
Worked example
A trolley of mass 2.0 kg moving at 3.0 m/s collides with a stationary trolley of mass 4.0 kg. The trolleys stick together after the collision. Calculate their common velocity immediately after the collision.
- Calculate the total momentum before the collision: momentum = mass x velocity for the moving trolley = 2.0 x 3.0 = 6.0 kg m/s. The stationary trolley has zero momentum.
- Total momentum before the collision = 6.0 + 0 = 6.0 kg m/s.
- By conservation of momentum, total momentum after the collision is also 6.0 kg m/s.
- After the collision the two trolleys move together, so their combined mass is 2.0 + 4.0 = 6.0 kg.
- Rearrange momentum = mass x velocity to find velocity: velocity = momentum / mass = 6.0 / 6.0.
- State the final answer with its unit: common velocity = 1.0 m/s, in the same direction as the moving trolley.
Practice questions
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Q1State Newton's first law of motion.Show answer
Answer: An object remains at rest or moving at constant velocity unless a resultant force acts on it.
Q2A resultant force of 20 N acts on a mass of 5.0 kg. Calculate the acceleration produced. Use F = m x a.Show answer
Answer: 4.0 m/s^2 (20/5.0).
Q3Calculate the momentum of a car of mass 1200 kg travelling at 15 m/s.Show answer
Answer: 18000 kg m/s (1200 x 15), or 1.8 x 10^4 kg m/s in standard form.
Q4State the two components that make up total stopping distance.Show answer
Answer: Thinking distance and braking distance.
Q5Name one factor that increases braking distance.Show answer
Answer: A wet or icy road surface (or worn tyres, or worn brakes) reduces friction and increases braking distance.
Q6A ball of mass 0.50 kg hits a wall at 8.0 m/s and rebounds at the same speed. The collision lasts 0.020 s. Calculate the force exerted on the ball. Use force = change in momentum / time.Show answer
Answer: 400 N; change in momentum = (0.50 x 8.0) - (-0.50 x 8.0) = 4.0 - (-4.0) = 8.0 kg m/s, so force = 8.0/0.020 = 400 N.
Q7State Newton's third law of motion.Show answer
Answer: When two objects interact, they exert equal and opposite forces on each other.
Exam-style questions
Written in the style of a IGCSE Science exam paper, with a full mark scheme.
A resultant force of 1500 N acts on a car of mass 750 kg, initially at rest. Calculate the velocity of the car after it has accelerated for 4.0 s.
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A car of mass 1000 kg is travelling at 20 m/s when the driver sees a hazard and brakes. The driver's reaction time is 0.60 s, and once the brakes are applied the car decelerates uniformly at 5.0 m/s^2 to rest. (a) Calculate the thinking distance. (2 marks) (b) Calculate the braking distance, using v^2 = u^2 + 2as. (2 marks) (c) Explain, in terms of forces, why a larger mass would increase the braking distance for the same braking force. (2 marks)
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Free printable worksheet
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This topic is chapter 23 of IGCSE Science Workbook, the whole course as one free printable PDF.
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