Evaluate — 2 marks
A car manufacturer is testing two different braking systems for their new vehicle. System A applies a constant braking force of 5000 N, while System B applies a constant braking force of 8000 N. Both systems are tested on identical cars with a mass of 1000 kg, all travelling at the same initial speed.
-
(a) Calculate the deceleration produced by System A using Newton's Second Law (F = ma).
[1 mark]
-
(b) Evaluate which braking system would be more effective at stopping the car quickly. Justify your answer using Newton's Second Law.
[1 mark]
Show mark scheme
- (a) Correct rearrangement of F = ma to a = F/m
- (a) Correct substitution: a = 5000 ÷ 1000
- (a) Correct answer: 5 m/s² (or 5 m s⁻²)
- (b) System B is more effective / produces greater deceleration
- (b) System B applies a larger force (8000 N > 5000 N) OR System B produces greater deceleration (a = 8 m/s² compared to 5 m/s²)
- (b) Valid reference to Newton's Second Law: larger force produces larger acceleration/deceleration for the same mass
Suggest — 5 marks
A stunt driver is planning a scene where a car must accelerate rapidly from rest and then brake suddenly to avoid an obstacle. The car has a mass of 1200 kg. During acceleration, the engine provides a driving force, and during braking, friction acts on the wheels. The driver is concerned about passenger safety and the structural integrity of the vehicle during both phases of motion.
-
(a) During the acceleration phase, the driving force from the engine is 4800 N and the resistance forces total 1200 N. Calculate the acceleration of the car.
[2 marks]
-
(b) Suggest why the passengers experience a backward force during the acceleration phase, referring to Newton's laws in your answer.
[2 marks]
-
(c) The car reaches a velocity of 20 m/s before braking. The braking force is 6000 N. Suggest what additional safety consideration the stunt coordinator should make regarding the magnitude of the braking force and explain your reasoning using Newton's second law.
[1 mark]
Show mark scheme
- (a) Net force = 4800 - 1200 = 3600 N (1 mark)
- (a) Acceleration = F/m = 3600/1200 = 3 m/s² (1 mark)
- (b) Passengers have inertia and tend to remain at rest (Newton's first law) (1 mark)
- (b) The car accelerates forward beneath them, so they appear to move backward relative to the car / the seat must exert a forward force to accelerate the passengers (1 mark)
- (c) Excessive braking force could cause deceleration that is too large / a = F/m = 6000/1200 = 5 m/s², which could cause injury to passengers or damage to the vehicle structure / the passengers would experience a large inertial force forward (1 mark)
Define — 4 marks
A delivery driver is transporting a parcel in the back of a van. When the van suddenly brakes, the parcel slides forwards across the floor, even though the driver applied the brakes to slow the van down.
-
(a) Define what is meant by inertia.
[1 mark]
-
(b) Define Newton's first law of motion.
[2 marks]
-
(c) Using Newton's first law, explain why the parcel slides forwards when the van brakes suddenly.
[1 mark]
Show mark scheme
- (a) The resistance of an object to change in velocity / the tendency of an object to continue in its state of motion unless acted upon by a force. Accept: 'the property that makes objects resist acceleration' or similar.
- (b) An object will remain at rest or continue moving at constant velocity / in a straight line (1 mark) unless acted upon by a resultant/net force (1 mark). Accept: 'stationary objects stay stationary and moving objects continue moving in straight lines at constant speed unless a force acts on them.'
- (c) The parcel has inertia and continues to move forwards at constant velocity (1 mark) because there is no resultant force acting on it (or friction is insufficient) to decelerate it with the van / the parcel tends to maintain its state of motion. Accept: 'no force acts on the parcel to slow it down' or 'the friction force is not large enough.'
Show — 4 marks
A delivery driver pushes a 25 kg box across a warehouse floor. The box accelerates from rest to 2 m/s in 4 seconds. The driver applies a constant horizontal force, and friction acts against the motion.
-
(a) Calculate the acceleration of the box.
[1 mark]
-
(b) Show that the resultant force acting on the box is 12.5 N. (You may assume friction is negligible.)
[2 marks]
-
(c) In reality, friction acts on the box with a force of 37.5 N. Show that the driver must apply a force of 50 N to achieve the same acceleration as calculated in part (a).
[1 mark]
Show mark scheme
- (a) a = Δv / Δt = 2 / 4 = 0.5 m/s² (1 mark)
- (b) Use F = ma (1 mark)
- (b) F = 25 × 0.5 = 12.5 N (1 mark)
- (c) Applied force − friction = resultant force; F − 37.5 = 12.5; F = 50 N (1 mark)
State — 4 marks
A skydiver jumps from an aircraft at high altitude. During the initial phase of the jump, the skydiver accelerates downwards. After several seconds, the skydiver reaches terminal velocity where the downward acceleration becomes zero. Later, the skydiver deploys their parachute, causing a large upward force that rapidly reduces their speed.
-
(a) State Newton's second law of motion.
[1 mark]
-
(b) State the condition that must be satisfied for the skydiver to be in equilibrium (zero acceleration) at terminal velocity, in terms of the forces acting on the skydiver.
[1 mark]
-
(c) State which of Newton's three laws of motion explains why the skydiver experiences a reaction force from the air resistance acting downwards on the atmosphere.
[1 mark]
-
(d) State the relationship between the net force on the skydiver and their acceleration immediately after the parachute deploys, when the upward force from the parachute is greater than the skydiver's weight.
[1 mark]
Show mark scheme
- (a) Force equals mass times acceleration (F = ma) OR The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass OR Net force is proportional to the rate of change of momentum
- (b) The upward force (air resistance/drag) equals the downward force (weight) OR The net force is zero OR Resultant force = 0
- (c) Newton's third law (of motion) OR Newton's third law of motion states that forces act in equal and opposite pairs
- (d) The net force is upward and the acceleration is upward (in the direction of the net force) OR Acceleration is in the same direction as the net force OR The skydiver decelerates (upward acceleration)
GCSE Perfect