SI units and modelling assumptions

• B1 for the dimensions of the sledge are ignored (its mass acts at a single point), so, e.g., rotation and air resistance can be ignored

Worked solution: Modelling the sledge as a particle means ignoring its size and shape: all its mass is treated as being at one point, so effects such as rotation (and air resistance, which depends on shape) are ignored.

• B1 for the tension is the same at every point of the rope (or: the rope’s weight/mass is ignored)

Worked solution: A light rope has no mass, so its weight is ignored and the tension is the same all along the rope.

• B1 for in reality there is friction (resistance) between the sledge and the snow
• B1 for the model predicts an acceleration that is too large

Worked solution: Real snow exerts a frictional force opposing the motion. A model with no friction leaves out a force that opposes the pull, so it predicts a larger acceleration than the real one.

• B1 for 2.5 (m s⁻²) from \(30 = 12a\)

Worked solution: \(F = ma\): \(30 = 12a\), so \(a = 2.5\) m s⁻².

• B1 for the units are not consistent (km h⁻¹ and minutes are mixed; the answer is not in metres)
• M1 for 72 km h⁻¹ = 20 m s⁻¹ and 5 minutes = 300 s (or 72 × \(\frac{5}{60}\) = 6 km)
• A1 for 6000 m

Worked solution: The student multiplied a speed in km h⁻¹ by a time in minutes, so the units are inconsistent.
\(72\text{ km h}^{-1} = \frac{72 \times 1000}{3600} = 20\text{ m s}^{-1}\) and 5 minutes = 300 s, so the distance is \(20 \times 300 = 6000\) m.

• M1 for \(\frac{20}{40}\) (speed in m s⁻¹)
• A1 for 0.5 (m s⁻²)

Worked solution: \(a = \frac{v - u}{t} = \frac{0 - 20}{40} = -0.5\), so the deceleration is 0.5 m s⁻².

• M1 for the distance while decelerating: \(\tfrac{1}{2} \times 20 \times 40 = 400\) m (or \(s = \left(\frac{u + v}{2}\right)t\))
• M1 for \(\frac{\text{total distance}}{\text{total time}} = \frac{6000 + 400}{300 + 40}\) using their distances
• A1 for awrt 18.8 (m s⁻¹)

Worked solution: Distance while decelerating: \(s = \left(\frac{20 + 0}{2}\right) \times 40 = 400\) m.
\(\text{average speed} = \frac{6000 + 400}{300 + 40} = \frac{6400}{340} = 18.82\ldots = 18.8\text{ m s}^{-1}\) (3 s.f.)

• M1 for using \(v = u + at\) with \(u = 0, \ a = 9.8, \ t = 3\)
• A1 for 29.4 or 29 (m s⁻¹)

Worked solution: \(v = u + at = 0 + 9.8 \times 3 = 29.4\text{ m s}^{-1}\)

• M1 for using \(s = ut + \tfrac{1}{2}at^2\) (or \(s = \left(\frac{u + v}{2}\right)t\)) with \(u = 0, \ a = 9.8, \ t = 3\)
• A1 for 44.1 or 44 (m)

Worked solution: \(s = ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2} \times 9.8 \times 3^2 = 44.1\text{ m}\)

• B1 for air resistance increases with speed
• B1 for so the acceleration is less than \(g\) (and model A overestimates the speed / distance)

Worked solution: Model A ignores air resistance. In reality air resistance grows as the skydiver speeds up, so the resultant force, and hence the acceleration, decreases. Model A keeps the acceleration at \(g\), so its predictions become too large.

• M1 for \(k = \frac{\text{force}}{v^2}\) with N = kg m s⁻²
• A1 for kg m⁻¹

Worked solution: kg m s⁻² ÷ (m s⁻¹)² = kg m s⁻² ÷ m² s⁻² = kg m⁻¹

• M1 for the resultant force being zero at constant speed (air resistance balances the weight)
• A1 for a correct equation: \(k \times 50^2 = 80 \times 9.8\)
• A1 for 0.314 or 0.31

Worked solution: At constant speed the acceleration is zero, so the air resistance balances the weight:
\(k \times 50^2 = 80 \times 9.8 \Rightarrow k = \frac{784}{2500} = 0.3136 = 0.314\) (3 s.f.)

• B1 for \(k\) is constant, with a reason why it may vary, e.g. the skydiver’s position or shape changes / a parachute opens

Worked solution: Model B assumes \(k\) is constant, but the air resistance depends on the shape and orientation of the skydiver, which can change during the fall (for example when a parachute opens).