CAIE A-Level Mathematics AS 4 Mechanics Questions

Practise Mechanics through force diagrams, equilibrium, kinematics, Newton's laws, momentum, work, energy and power using connected-particle and inclined-plane models.

Covered syllabus points

4.1 Forces and equilibrium 4.2 Kinematics of motion in a straight line 4.3 Momentum 4.4 Newton's laws of motion 4.5 Energy, work and power

[Maximum number: 6]

A particle of mass 12 kg is stationary on a rough plane inclined at an angle of $25^{\circ}$ to the horizontal. A force of magnitude $P \mathrm{~N}$ acting parallel to a line of greatest slope of the plane is used to prevent the particle sliding down the plane. The coefficient of friction between the particle and the plane is 0.35 .

(a)

Draw a sketch showing the forces acting on the particle.

[ 1 ]

The force diagram should show the weight vertically downward, the normal reaction perpendicular to the plane, the force P parallel to the line of greatest slope, and the friction force F along the plane opposing the tendency to slide. The angle of inclination is $25^\circ$ .

Figure for Answer to Question (a) — CAIE A-Level Mathematics AS

(b)

Find the least possible value of P.

[ 5 ]

Resolving parallel to the plane, with the friction opposing the tendency to slide down,
$P+F=12g\sin25^\circ$ .
The normal reaction is
$R=12g\cos25^\circ$ ,
and the limiting friction is F=0.35R. Using $g=10\text{ m s}^{-2}$ ,
$F=0.35(120\cos25^\circ)=38.1\text{ N}$ .
Therefore,
$P=120\sin25^\circ-38.1=12.6\text{ N}$ .

[Maximum number: 5]

Figure for Question 2 — CAIE A-Level Mathematics AS

The displacement of a particle moving in a straight line is s metres at time t seconds after leaving a fixed point O. The particle starts from rest and passes through points P, Q and R, at times t=5, t=10 and t=15 respectively, and returns to O at time t=20. The distances O P, O Q and O R are 50 m , 150 m and 200 m respectively.

The diagram shows a displacement-time graph which models the motion of the particle from t=0 to t=20. The graph consists of two curved segments A B and C D and two straight line segments B C and D E.

(a)

Find the speed of the particle between t=5 and t=10.

[ 1 ]

Speed $=20 \mathrm{~ms}^{-1}$

(b)

Find the acceleration of the particle between t=0 and t=5, given that it is constant.

[ 2 ]

$20=0+a \times 5$
Use of v=u+a t OE

$a=4 \mathrm{~ms}^{-2}$

(c)

Find the average speed of the particle during its motion.

[ 2 ]

$\frac{50+100+50+200}{20}$
Use of $\frac{\text { total distance }}{\text { total time }}$ OE

Average speed $=20 \mathrm{~ms}^{-1}$

[Maximum number: 7]

A particle P of mass 0.2 kg lies at rest on a rough horizontal plane. A horizontal force of 1.2 N is applied to P.

(a)

Given that P is in limiting equilibrium, find the coefficient of friction between P and the plane.

[ 3 ]

$R=0.2 \mathrm{~g}$ $1.2=\mu \times 0.2 g$
Resolve horizontally and using
$F=\mu R$ to get an equation in $\mu ; 2$
relevant terms.

$\mu=0.6$
ое

(b)

Given instead that the coefficient of friction between P and the plane is 0.3 , find the distance travelled by P in the third second of its motion.

[ 4 ]

$1.2-0.3 \times 0.2 g=0.2 a$

*M1

Resolve horizontally using
Newton's Second Law;
3 relevant
terms;
allow sign errors;
R=0.2 g
only.

a=3
0.6=0.2 a only seen, allow with
BOD, but if 0.6 as friction being used as resultant force, this is
$s_{3}=0+\frac{1}{2} \times 3 \times 3^{2}[=13.5] s_{2}=0+\frac{1}{2} \times 3 \times 2^{2}[=6]$
For use of $s=u t+\frac{1}{2} a t^{2}$ (or a
complete method) to find a distance
at least once with u=0 and their
positive a and t=2 or t=3.

Distance $=13.5-6=7.5 \mathrm{~m}$
WWW

[Maximum number: 9]

A car of mass 1400 kg is moving on a straight road against a constant force of 1250 N resisting the motion.

(a)

The car moves along a horizontal section of the road at a constant speed of $36 \mathrm{~m} \mathrm{~s}^{-1}$ .

[ 7 ]

(i)

Calculate the work done against the resisting force during the first 8 seconds.

[ 2 ]

[WD = $1250 \times 36 \times 8$ ]
For using Work Done = Force × Distance.
$\mathrm{WD}=360000 \mathrm{~J}$
or 360 kJ

(ii)

Calculate, in kW , the power developed by the engine of the car.

[ 2 ]

Power $=1250 \times 36$ or $\mathrm{P}=\frac{360000}{8}[=45000 \mathrm{~J}]$
B1 FT
FT Work Done from $\frac{\mathbf{a}(\mathbf{i})}{8}$ .
$=45 \mathrm{~kW}$
B1

(iii)

Given that this power is suddenly increased by 12 kW , find the instantaneous acceleration of the car.

[ 3 ]

$\mathrm{DF}=\frac{57000}{36}[=1583.3 .$ .
Use changed Power in $P=\mathrm{DF} \times v$ .
$\frac{57000}{36}-1250=1400 a$
For using Newton's 2nd law applied to the car.
$a=0.238 \mathrm{~ms}^{-2}$

(b)

The car now travels at a constant speed of $32 \mathrm{~m} \mathrm{~s}^{-1}$ up a section of the road inclined at $\theta^{\circ}$ to the horizontal, with the engine working at 64 kW .

Find the value of $\theta$ .

[ 2 ]

$\frac{64000}{32}=1250+1400 g \sin \theta$
For using DF= resistance + component of the weight of
the car.
$\theta=3.1[3.0708 \ldots]$

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (b)

Question (c)

Question 3

Question (a)

Question (b)

Question 4

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (b)