Matter: Density & Pressure

Reading cue. W is drawn from the dot at her centre of mass and F from the surface meeting the air; v is a velocity, not a third force, and W is drawn longer than F for a reason.

1. aDeduce Deduce from the relative lengths of the arrows W and F in Fig. 7.1 whether the skydiver has yet reached terminal velocity.

   recall2 marks

   Check answer · 2 marks

   1. F is drawn shorter than W, so there is a resultant force downwards
   2. she is therefore still accelerating and has not reached terminal velocity

2. bCalculate At the instant drawn, the drag force F is 3.4 × 10² N. Calculate the weight of the skydiver and her acceleration at that instant.

   routine3 marks

   Check answer · 3 marks

   1. W = 85 × 9.81 = 8.3 × 10² N
   2. resultant force = 834 − 340 = 4.9 × 10² N downwards
   3. a = 494 / 85 = 5.8 m s⁻² downwards

3. cDetermine The skydiver falls 250 m from rest and is then moving at 50 m s⁻¹. Determine the average drag force acting on her over that fall.

   demanding4 marks

   Check answer · 4 marks

   1. loss of gravitational potential energy = 85 × 9.81 × 250 = 2.08 × 10⁵ J
   2. gain in kinetic energy = ½ × 85 × 50² = 1.06 × 10⁵ J
   3. work done against drag = 2.08 × 10⁵ − 1.06 × 10⁵ = 1.02 × 10⁵ J
   4. average drag force = 1.02 × 10⁵ / 250 = 4.1 × 10² N

4. dExplain The skydiver is falling at constant velocity when she turns into a head-down dive, presenting a much smaller area to the airflow. Explain how the two arrows of Fig. 7.1 change, and describe her subsequent motion.

   top of the paper4 marks

   Check answer · 4 marks

   1. the drag is reduced, so F becomes shorter while W is unchanged
   2. there is now a resultant downward force, so she accelerates again
   3. as her speed rises the drag increases until F is once more equal to W
   4. she then falls at constant velocity at a higher terminal speed than before