Practice Paper 1 (Edexcel IAL Maths)

A rectangular framework is made from a uniform wire and is illustrated in the diagram below.

Write down the coordinates of the centre of mass of the framework.

The framework has a total mass of 45 kg. Particles of masses 25 kg, 50 kg, 30 kg are loaded onto the framework at the midpoint of the sides $AB,BC,CD$ respectively. Find the coordinates of the centre of mass of the loaded framework.

Two particles $P$ and $Q$ are moving in a straight line on a smooth horizontal plane with velocities $u_{1}$ and $u_{2}$ respectively. The two particles collide directly. Immediately after the collision, the velocities of $P$ and $Q$ are $v_{1}$ and $v_{2}$ respectively.

The coefficient of restitution between $P$ and $Q$ is given by

$e=\frac{\mathrm{Speed} \mathrm{of} \mathrm{separation}}{\mathrm{Speed} \mathrm{of} \mathrm{approach}}=\frac{v_{2}-v_{1}}{u_{1}-u_{2}}.$

Find the coefficient of restitution for each of the scenarios:

i) Immediately before the collision, $P$ and $Q$ are moving in the same direction with speeds 5 m s^-1 and 2 m s^-1 respectively. Immediately after the collision, $P$ and $Q$ are moving in the same direction with speeds 3 m s^-1 and 4 m s^-1 respectively.

ii) Immediately before the collision, $P$ and $Q$ are moving in opposite directions with speeds 5 m s^-1 and 2 m s^-1 respectively. Immediately after the collision, $P$ and $Q$ are moving in opposite directions with speeds 3 m s^-1 and 4 m s^-1 respectively.

iii) Immediately before the collision, $P$ and $Q$ are moving in opposite directions with speeds 5 m s^-1 and 2 m s^-1 respectively. Immediately after the collision, $P$ and $Q$ move with speeds 3 m s^-1 and 4 m s^-1 respectively. The direction of motion of $P$ is unchanged but the direction of motion of $Q$ is reversed.

A particle is projected with an initial speed of 24 ms^-1 at an angle of 30° above the horizontal. Find the horizontal and vertical components of the initial velocity, writing your answer in the form $(u_{x}i+u_{y}j)ms^{-1}$.

A go kart manufacturer is testing out a new model on a straight horizontal road.

Starting from rest, the velocity of the go kart is modelled by the equation

$v = \frac{1}{10} t ( 36 - t ) t \text{greater} \text{or} \text{equal} \text{than} 0$

where $v$ m s⁻¹ is the velocity at time $t$ seconds.

Find the maximum velocity of the go kart and the time at which this occurs. Justify that this is a maximum.

The go kart does not move backwards at any point during the test. Find the time it takes to complete the test.

Once an aircraft reaches its cruising height (at time $t=0$ seconds) its acceleration is modelled by

$a = ( ( 30 - 2 \mathrm{t} ) i + ( 4 \mathrm{t} - 3 ) j ) \mathrm{m} \mathrm{s}^{- 2} \mathrm{t} \text{greater} \text{or} \text{equal} \text{than} 0$

Given that the velocity of the aircraft at $t=0$ is $v = ( 200 i + 150 j ) \mathrm{m} \mathrm{s}^{- 1}$, find the velocity of the aircraft in terms of $t$.

Find the speed of the aircraft at time $t=4$ seconds, giving your answer in kilometres per hour to three significant figures.

A non-uniform barge pole $AB$ , of length 4.5 m and mass $m$ kg, is placed with $A$ resting on the floor at an angle of $α$ to the horizontal and the point $C$, 0.5 m from $B$ resting against the corner of a rough barge, as shown in the diagram below. The pole is at the point of slipping when the coefficient of friction between the pole and the ground is 0.2, and the pole and the barge is $μ$, and $\mathrm{tan}α=0.75$.

Later, the same pole is placed with $A$ on the ground and $B$ against a smooth wall. The pole is now on the point of slipping when the coefficient of friction between the pole and the ground is $\frac{5}{18}$ . Show that the angle between the pole and the ground now can be given as:

$\tan ^{- 1} ( \frac{20}{19 - 17 \\mu } )$

A car of mass 850 kg is moving up a straight road, inclined at an angle $θ$ to the horizontal, along the line of greatest slope, where $\mathrm{sin}θ=\frac{1}{50}$. The engine of the car is working at a constant rate of 12.1 kW and the car is moving with a constant speed of $v$ m s^-1. The resistance to motion on the car from non-gravitational forces is $RN$ and acts parallel to the slope.

Justifying your answer, show that

$R=\frac{12100}{v}-17g.$

The car later moves down the same hill along the line of greatest slope. It travels at the same speed as it did when it travelled up the hill. The resistance to motion on the car from non-gravitational forces has increased to $( R + 300 ) \mathrm{N}$ and the engine is now working at a constant rate of $P$ kW.

Show that

$R=\frac{1000P}{v}+17g-300$.

Given that $P$ =11.4, find the values of $v$ and $R$.

A lawnmower of mass 9 kg is pushed up a hill, along the line of greatest slope, inclined at 15° to the horizontal by a force of 80 N acting at an angle of 60° to the slope as shown in the diagram below. The frictional force acting on the lawnmower is 15 N and acts in the direction opposing the motion of the lawnmower parallel to the slope.

The lawnmower is pushed 8 metres up the hill.

Calculate:

i) the work done against gravity,

ii) the work done against friction,

iii) the work done by the force of 80 N pushing the lawnmower.

Given that the lawnmower was initially at rest at the bottom of the hill, find the speed of the lawnmower in the instant that it has been pushed 8 metres up the hill.

For a uniform sector of a circle with radius $r$ and centre angle $2α$ the position of the centre of mass lies along the axis of symmetry a distance

$\frac{2r\mathrm{sin}α}{3α}$

from the centre of the circle.

A uniform sector of a circle, centre $C$, has its centre of mass located at the point $G$. Given the radius of the sector is $6π$ cm and the centre angle is $\frac{π}{3}$ radians, find the distance $CG$.

Find the exact coordinates of the centre of mass for the uniform sector of a circle shown in the diagram below.