1. 1
   A skydiver jumps from a stationary hot air balloon high above the ground and falls, reaching increasing speed before opening a parachute later in the descent.

   (a)

   In the first few seconds after jumping, before drag becomes significant, what is the resultant force acting on the skydiver?

   [1 mark]

   • AZero, since the skydiver is in free fall
   • BEqual to the skydiver's weight, acting downwards
   • CEqual to the skydiver's weight, acting upwards
   • DEqual to air resistance only

   (b)

   As the skydiver falls faster, air resistance increases. What happens to the skydiver's acceleration as their speed increases towards terminal velocity?

   [1 mark]

   • AIt stays constant throughout the fall
   • BIt increases as speed increases
   • CIt decreases, reaching zero at terminal velocity
   • DIt reverses direction before terminal velocity is reached

   (c)

   Explain what is meant by 'terminal velocity', and describe how the forces of weight and air resistance compare once the skydiver reaches terminal velocity.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   Later in the descent, the same skydiver opens a parachute while still falling at a high speed, well above their eventual terminal velocity with the parachute open.

   (a)

   Later in the descent, the skydiver opens a parachute. What immediate effect does opening the parachute have on the forces acting on the skydiver?

   [1 mark]

   • AIt greatly increases air resistance
   • BIt greatly increases the skydiver's weight
   • CIt removes air resistance entirely
   • DIt has no effect on the forces acting on the skydiver

   (b)

   Immediately after the parachute opens, air resistance is much greater than the skydiver's weight. What effect does this have on the skydiver's motion at that instant?

   [1 mark]

   • AThe skydiver accelerates downwards even faster
   • BThe skydiver begins moving upwards
   • CThe skydiver's speed stays exactly the same
   • DThe skydiver decelerates, since the resultant force now acts upwards

   (c)

   Explain why the skydiver's new, lower terminal velocity (reached after the parachute opens) is much smaller than their terminal velocity before the parachute opened, and describe how the parachute achieves this effect.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   A scientist drops a small steel ball bearing into a tall, vertical tube of clear, viscous oil and records its velocity at regular time intervals as it sinks to the bottom of the tube.

   (a)

   A scientist drops a small steel ball bearing into a tall tube of clear, viscous oil and records its velocity at regular time intervals as it sinks. Sketch (in words) how a velocity–time graph for the ball bearing would look, describing what happens to its velocity from the moment it is dropped until it reaches terminal velocity within the oil.

   [3 marks]

   (b)

   The scientist repeats the experiment with a larger, heavier steel ball bearing of the same shape dropped into the same oil. Explain how you would expect the terminal velocity of the larger ball bearing to compare with that of the smaller one, and explain your reasoning in terms of the forces of weight and drag acting on each ball bearing.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   A car manufacturer is developing a new electric car and wants to maximise how far it can travel on a single battery charge. The design team is considering changing the shape of the car's body to reduce drag at motorway speeds, alongside other efficiency improvements.

   (a)

   Explain, using ideas about air resistance, terminal velocity and streamlining, how changing the car's shape could improve its efficiency at motorway speeds, even though a car does not fall like a skydiver.

   [6 marks]

   (b)

   The design team also tests the car at a range of constant motorway speeds in a wind tunnel, measuring the drag force at each speed. Explain how you would expect the drag force to change as the car's constant test speed increases, and explain why doubling the car's speed does not simply double the energy needed to overcome drag over a given distance.

   [6 marks]

   Total for question 4: 12 marks

End of questions