📘 GCSE Physics · Notes & worksheet
Forces and Motion.
Notes
Key equations
| Quantity | Equation | Units |
|---|---|---|
| Resultant force | F = ma | N, kg, m/s² |
| Acceleration | a = (v − u) ÷ t | m/s² |
| Constant acceleration | v² − u² = 2as | m/s, m |
| Weight | W = mg (g = 9.8 N/kg) | N |
| Momentum | p = mv | kg m/s |
| Force and momentum | F = (mv − mu) ÷ t | N |
| Moment | force × perpendicular distance | N m |
Newton's laws
- An object stays at rest or moves at constant velocity unless a resultant force acts.
- Acceleration is proportional to resultant force and inversely proportional to mass (F = ma).
- When two objects interact, the forces are equal and opposite (and act on different objects).
Terminal velocity
A skydiver: (1) weight is bigger than air resistance and they accelerate; (2) air resistance rises with speed; (3) when air resistance equals weight, the resultant force is zero and speed is constant. Opening the parachute increases air resistance, which decelerates the skydiver to a new, lower terminal velocity.
Stopping distance
Stopping distance = thinking distance + braking distance.
- Thinking distance increases with reaction time (tiredness, alcohol, distractions) and with speed.
- Braking distance increases with speed (it is proportional to v²), poor brakes, worn tyres and wet or icy roads.
Momentum
In a closed system, total momentum before = total momentum after. In collisions that stick together: m₁u₁ + m₂u₂ = (m₁ + m₂)v. Kinetic energy is usually not conserved in a collision.
Moments
For equilibrium: clockwise moments = anticlockwise moments about any pivot.
Common mistakes
- Using mass in kg for weight without multiplying by g.
- Forgetting the direction (sign) of velocity in momentum problems.
- Saying "objects stop when forces stop". Constant velocity needs zero resultant force.
- Measuring distance to a pivot along the beam when the force is at an angle.
Exam tips
- Draw a free-body diagram first and label every force.
- For "explain" questions about terminal velocity, always say "resultant force = 0".
Worksheet
Prints the questions only, without answers. Choose “Save as PDF” in the print window to download.
Total: 31 marks. Use g = 9.8 N/kg.
- A car of mass 1200 kg accelerates uniformly from rest to 25 m/s in 8.0 s. Calculate the resultant force. (4)
- An object is dropped from rest and falls 20 m. Ignoring air resistance, calculate its speed on landing. (3)
- A 0.50 kg ball moving at 12 m/s is brought to rest by a goalkeeper's gloves in 0.040 s. Calculate the average force on the ball. (3)
- Trolley A (2.0 kg) moves at 3.0 m/s and collides with stationary trolley B (1.0 kg). They stick together.
(a) Calculate their velocity after the collision. (3)
(b) Show that kinetic energy is not conserved. (3) - Describe and explain the motion of a skydiver from the moment of jumping until landing with the parachute open. (5)
- A uniform plank of length 4.0 m is pivoted at its midpoint. A child of mass 25 kg sits 1.5 m from the pivot. Where must a 40 kg child sit on the other side to balance the plank? (3)
- A driver has a reaction time of 0.60 s and is travelling at 30 m/s. The car decelerates uniformly at 6.0 m/s² when braking.
(a) Calculate the thinking distance. (2)
(b) Calculate the braking distance. (2)
(c) Calculate the total stopping distance. (1) - State Newton's third law and identify the pair of forces when a book rests on a table. (2)
Show mark scheme and answers (tutor copy)
Try the worksheet first, then check your work here.
- a = (25 − 0) ÷ 8.0 (M1) = 3.125 m/s² (A1). F = ma = 1200 × 3.125 (M1) = 3750 N (A1).
- v² = u² + 2as (M1), with u = 0: v² = 2 × 9.8 × 20 = 392 (M1). v = 19.8 m/s (A1).
- Change in momentum = 0.50 × 12 = 6.0 kg m/s (M1). F = 6.0 ÷ 0.040 (M1) = 150 N (A1).
- (a) Momentum before = 2.0 × 3.0 = 6.0 kg m/s (M1). 6.0 = 3.0 × v (M1), so v = 2.0 m/s (A1).
(b) KE before = ½ × 2.0 × 3.0² = 9.0 J (M1). KE after = ½ × 3.0 × 2.0² = 6.0 J (M1). KE is lower by 3.0 J, so it is not conserved; the energy is transferred to heat and sound (A1). - Initially the weight is greater than air resistance, so the skydiver accelerates downwards (B1). As speed increases, air resistance increases (B1), so the resultant force decreases and acceleration decreases (B1). When air resistance = weight, the resultant force is zero and the skydiver falls at constant terminal velocity (B1). When the parachute opens, air resistance suddenly exceeds weight, the skydiver decelerates, and eventually reaches a new, lower terminal velocity until landing (B1).
- Anticlockwise moment = 25 × 9.8 × 1.5 (M1). Balanced by 40 × 9.8 × d (M1), so d = 37.5 ÷ 40 = 0.94 m (A1) from the pivot (the plank's weight acts at the pivot so has no moment).
- (a) Thinking distance = 30 × 0.60 = 18 m. (M1, A1)
(b) v² = u² + 2as, so 0 = 900 − 2 × 6.0 × s (M1), giving s = 900 ÷ 12 = 75 m. (A1)
(c) 18 + 75 = 93 m. (B1) - When two objects interact, the forces they exert on each other are equal in size and opposite in direction (B1). Example: the book pushes down on the table and the table pushes up on the book with equal force. (B1)