A uniform object is made by joining a solid cone of height 0.8 m and base radius 0.6 m and a cylinder. The cylinder has length 0.4 m and radius 0.5m. The cylinder has a cylindrical hole of length 0.4 m and radius x m drilled through it along the axis of symmetry. A plane face of the cylinder is attached to the base of the cone so that the object has an axis of symmetry perpendicular to its base and passing through the vertex of the cone. The object is placed with points on the base of the cone and the base of the cylinder in contact with a horizontal surface [Figure 5.1]. The object is on the point of toppling.
- (i)Show that the centre of mass of the object is 0.15 m from the base of the cone.[3]
- (ii)Find x. [The volume of a cone is ⅓πr²h.][4]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- [B1] (i) tanθ = (0.6−0.5)/0.4 (= 1/4)
- [M1] (i) tanθ = x/0.6
- [A1] (i) x = 0.15 m
- [M1] (ii) (π0.6² × 0.8/3) × (0.8/4) – [π( 0.5² − x²) × 0.4] × (0.4/2) = [π0.6² × 0.8/3 + π(0.5² − x²) × 0.4] x̄
- [M1] (ii) Attempts to solve the equation
- [A1] (ii) Note x² = 0.216
- [A1] (ii) x = 0.464
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →