Edexcel A Level Further Maths M5 · Mechanics 5 June 2009 question paper 6681

At time t = 0, a particle P of mass 3 kg is at rest at the point A with position vector (j – 3k) m. Two constant forces F₁ and F₂ then act on the particle P and it passes through the point B with position vector (8 i – 3 j + 5 k) m. Given that F₁ = (4i – 2j + 5k) N and F₂ = (8i – 4j + 7k) N and that F₁ and F₂ are the only two forces acting on P, find the velocity of P as it passes through B, giving your answer as a vector.

At time t seconds, the position vector of a particle P is r metres, where r satisfies the vector differential equation

When t = 0, P has position vector (i + j) m and velocity 2i m s^–1.

Find an expression for r in terms of t.

A spaceship is moving in a straight line in deep space and needs to increase its speed. This is done by ejecting fuel backwards from the spaceship at a constant speed c relative to the spaceship. When the speed of the spaceship is v, its mass is m.

Show that, while the spaceship is ejecting fuel,

The initial mass of the spaceship is m₀ and at time t the mass of the spaceship is given by m = m₀ (1 – kt), where k is a positive constant.

Find the acceleration of the spaceship at time t.

A uniform lamina of mass M is in the shape of a right-angled triangle OAB. The angle OAB is 90°, OA = a and AB = 2 a, as shown in Figure 1.

Prove, using integration, that the moment of inertia of the lamina OAB about the edge

(You may assume without proof that the moment of inertia of a uniform rod of mass m and length 2l about an axis through one end and perpendicular to the rod

The lamina OAB is free to rotate about a fixed smooth horizontal axis along the edge OA and hangs at rest with B vertically below A. The lamina is then given a horizontal impulse of magnitude J. The impulse is applied to the lamina at the point B, in a direction which is perpendicular to the plane of the lamina. Given that the lamina first comes to instantaneous rest after rotating through an angle of 120°,

find an expression for J, in terms of M, a and g.

Two forces F₁ = (2i + j) N and F₂ = (–2j – k) N act on a rigid body. The force F₁ acts at the point with position vector r₁ = (3i + j + k) m and the force F₂ acts at the point with position vector r₂ = (i – 2j) m. A third force F₃ acts on the body such that F₁, F₂ and F₃ are in equilibrium.

Find the magnitude of F₃.

Find a vector equation of the line of action of F₃. The force F₃ is replaced by a fourth force F₄, acting through the origin O, such that F₁, F₂ and F₄ are equivalent to a couple.

Find the magnitude of this couple.

A pendulum consists of a uniform rod AB, of length 4 a and mass 2 m, whose end A is rigidly attached to the centre O of a uniform square lamina PQRS, of mass 4 m and side a. The rod AB is perpendicular to the plane of the lamina. The pendulum is free to rotate about a fixed smooth horizontal axis L which passes through B. The axis L is perpendicular to AB and parallel to the edge PQ of the square.

Show that the moment of inertia of the pendulum about L is 75ma².

The pendulum is released from rest when BA makes an angle α with the downward vertical through B, where tan α = . When BA makes an angle θ with the downward vertical through B, the magnitude of the component, in the direction AB, of the force exerted by the axis L on the pendulum is X.

Find an expression for X in terms of m, g and θ.

Using the approximation sin,

find an estimate of the time for the pendulum to rotate through an angle α from its initial rest position.