AQA A-Level Physics Paper 1, June 2025: Question 21

🎯 What this question tests

Terminal Velocity Condition: Understanding that terminal velocity requires zero acceleration ( a = 0 ), meaning the upward resistive force (drag) equals the downward weight ( W = mg ).
Newton’s Third Law Pairs: Distinguishing between balanced forces acting on a single body (Newton's 1st/2nd Law) and action–reaction pairs acting on two different bodies (Newton's 3rd Law).
Kinematics vs. Resultant Force: Realising that an upward resultant force results in upward acceleration (deceleration of downward motion), not immediate upward movement.
Work-Energy Principle: Recognising that gravitational potential energy changes must be accounted for alongside changes in kinetic energy when work is done by resistive forces.

Question 21 • Multiple Choice [1 Mark]

"The drag force when her velocity is v₂ is equal to the drag force when her velocity is v₁."

At any terminal velocity, the resultant vertical force is zero:

F_net = W - F_drag = 0 ⇒ F_drag = W

Because the skydiver's mass has not changed, her weight W = mg is constant throughout the jump. Therefore:

Hence, F_drag(1) = F_drag(2) .

Option A: The sudden increase in drag causes an upward acceleration. Since she is travelling downward, this causes her to slow down (decelerate), not move upward. She continues to fall downward at all times.
Option C: Weight and drag act on the same body (the skydiver) and are of different physical types (gravitational vs electrostatic/fluid contact). They are balanced forces, not a Newton's 3rd law pair.
Option D: As she falls from h₁ to h₂ , she also loses gravitational potential energy ( ΔE_p = mgΔh ). By the work-energy principle, the work done against drag equals ΔE_k + ΔE_p , not just ΔE_k .

Drag force depends on both speed and surface area:

F_drag ∝ A × v²

Before opening ( v₁ ): Small effective area A , so a high speed is required to make F_drag = W .
After opening: The parachute hugely increases A . Instantly, F_drag >> W , causing upward acceleration (rapid deceleration).
New steady state ( v₂ ): As speed decreases, drag decreases until it once again matches weight ( F_drag = W ). Because A is large, this equilibrium occurs at a much lower terminal speed v₂ .

Consider the total energy conservation between height h₁ and h₂ :

Initial mechanical energy: E₁ = mg h₁ + ½ m v₁²
Final mechanical energy: E₂ = mg h₂ + ½ m v₂²
Work done against drag ( W_drag ):
W_drag = E₁ - E₂
W_drag = mg(h₁ - h₂) + ½ m(v₁² - v₂²)
Because h₁ > h₂ , mg(h₁ - h₂) > 0 . Therefore, work done by drag is substantially greater than just the change in kinetic energy.

To identify genuine Newton's third law pairs, use the 4-point test:

Equal in magnitude? Yes.
Opposite in direction? Yes.
Same type of force? Must both be gravitational, or both electrostatic contact forces. (Weight is gravitational; drag is electromagnetic/contact).
Act on two different bodies? If body A exerts force on B, B exerts force on A. (Weight is Earth on skydiver; its N3 pair is skydiver on Earth).

"The camera shot" illusion: In skydiving videos, when a parachute opens, the person appears to shoot upwards. This is an optical illusion caused by the camera operator continuing to fall at high terminal velocity while the parachutist decelerates.
Assuming drag is proportional only to speed: Students often think higher velocity always means higher drag. At terminal velocities v₁ and v₂ , the total drag forces are identical because both balance the exact same weight.

Mark Scheme Breakdown: 1 mark awarded for selecting B (Assessment Objective AO2 - Application of knowledge in unfamiliar context).

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