Mechanics for Cambridge International AS & A Level Physics

Reading cue. Read the arrow as the whole of the initial velocity, drawn along one direction only, and the 45 m as the vertical drop — not as the length of the dashed path.

1. aState State the vertical component of the stone's velocity at the instant it leaves the cliff edge, and give the feature of Fig. 2.1 that tells you this.

   recall2 marks

   Check answer · 2 marks

   1. vertical component of the initial velocity is zero
   2. because the velocity arrow at the cliff edge is drawn horizontal

2. bCalculate Calculate the speed of the stone at the instant it reaches the ground.

   routine3 marks

   Check answer · 3 marks

   1. time to fall 45 m: t = √(2 × 45 / 9.81) = 3.03 s
   2. vertical component on landing: v = 9.81 × 3.03 = 29.7 m s⁻¹
   3. resultant speed = √(12² + 29.7²) = 32 m s⁻¹

3. cDetermine Determine the angle between the dashed path and the ground at the point where the stone lands.

   demanding3 marks

   Check answer · 3 marks

   1. uses horizontal component 12 m s⁻¹ with vertical component 29.7 m s⁻¹
   2. tan θ = 29.7 / 12
   3. θ = 68° to the horizontal

4. dShow (that) Taking the cliff edge as origin, with x measured horizontally and y measured vertically downwards, show that the dashed curve in Fig. 2.1 is a parabola, and state the numerical constant in your equation.

   top of the paper4 marks

   Show a hint

   Write x and y separately in terms of t, then eliminate t.

   Check answer · 4 marks

   1. x = 12t, so t = x/12
   2. y = ½ × 9.81 × t²
   3. substitution gives y = 9.81x² / (2 × 12²), which is of the form y ∝ x², a parabola
   4. y = 0.034x², with x and y in metres

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