Edexcel A Level Further Maths Paper 4C: Further Mechanics 2 June 2025 question paper 9FM0/4C

A uniform rod is cut into six pieces. The pieces are used to form the framework ABCDE shown in Figure 1.

• All the pieces of the framework lie in the same plane.

• AB = BC = CD = 2a

• Point E is the midpoint of AC

• Angle ABC = angle BCD = 90°

The distance of the centre of mass of the framework from AB is d

The mass of the framework is M. A particle of mass kM is attached to the framework at B

The distance of the centre of mass of the loaded framework from AB is a

Find the value of k

The shaded region shown in Figure 2 is bounded by the curve with equation y = , the line with equation x = 2, the line with equation x = 4, and the x-axis. This region is rotated through 360° about the x-axis to form a solid.

This solid is used to model a uniform solid pedestal of height 2 m, upper radius 0.5 m and base radius 0.25 m.

Given that the volume of the pedestal is

show that the centre of mass of the pedestal is (4 – 4 ln 2) m from its base.

The pedestal is placed on a rough plane that is inclined at an angle α° to the horizontal. The plane base of the pedestal is in contact with the inclined plane, as shown in Figure 3. The inclined plane is sufficiently rough to prevent the pedestal from sliding.

Given that the pedestal is on the point of toppling,

find the value of α

A particle P of mass 0.2 kg is attached to one end of a light inextensible string of length 0.6 m. The other end of the string is attached to the fixed point A. A second light inextensible string of length 0.6 m has one end attached to P and the other end attached to the fixed point B. The point B is vertically below A such that AB = 0.8 m, as shown in Figure 4.

The particle moves in a horizontal circle with constant angular speed ω radians per second, with both strings taut.

Given that the tension in the string from P to A is twice the tension in the string from P to B, find the value of ω.

The uniform rectangular lamina ACDF shown in Figure 5 has AC = FD = 15a and AF = CD = 3 a. The point B on AC is such that AB = 6 a. The point E on FD is such that ED = 6 a.

The rectangular lamina is folded along BE to form the folded lamina shown in Figure 6. The folded lamina has AB perpendicular to BC and the two sections, ABEF and BEDC, of the lamina lie in the same plane.

Show that the distance of the centre of mass of the folded lamina from EF is a.

Explain why the centre of mass of the folded lamina lies on the perpendicular bisector of BE.

The folded lamina is freely suspended from F and hangs in equilibrium.

Find the size of the angle between FE and the downward vertical.

A car of mass 1000 kg is moving along a straight horizontal road. The engine of the car produces a constant driving force of 800 N. At time t seconds, t 0, the speed of the car is vm s^–1 and the resistance to the motion of the car has magnitude 2v²N.

The time taken for the speed of the car to increase from 5 m s^–1 to 15 m s^–1 is T seconds.

When t = 0, v = 2

Show that pe^qt where p and q are rational numbers to be found.

Hence explain why v 20 for all values of t 0

A uniform semicircular lamina has radius r and the midpoint of the diameter of the lamina is at O. The distance of the centre of mass of the lamina from O is d.

Use algebraic integration to show that d =

The uniform plane template T, shown shaded in Figure 7, is formed from a rectangle and a semicircular ring. The rectangle has AD = BC = 2a and AB = DC = 8a. The midpoint of CD is O. The semicircular ring has centre O, outer radius OD = 4a and inner radius OE = 2a. The template is modelled as a uniform lamina.

The template is freely suspended from A. The weight of T is W newtons. A horizontal force is applied to T at B so that T is held in equilibrium with AB vertical. The force acts in the same vertical plane as T and has magnitude λW newtons.

Find the value of λ

A particle P of mass 0.4 kg is moving in a straight line with simple harmonic motion. The maximum speed of P is 5 m s^–1 and the maximum magnitude of acceleration of P is 12.5 m s^–2 The time taken for one complete oscillation is T seconds.

Find the time within each complete oscillation for which P is more than 0.5 m from the centre of the oscillation.

Find the kinetic energy of P at the instant when P is 0.2 m from the centre of the oscillation.

In this question use g = 10 m s^–2

A particle P of mass 0.4 kg is attached to one end of a light inextensible string of length 1.5 m. The other end of the string is attached to a fixed point O. The particle is at rest vertically below O with the string taut.

The particle is then projected horizontally with speed 6 m s^–1

When the string has turned through an angle θ, the string is still taut and the speed of P is vm s^–1, as shown in Figure 8.