CAIE A-Level Mathematics AS 4.4 Newtons Laws of Motion Questions

Practise applying Newton's second law to particles, vehicles and connected systems with weight, reaction, tension, thrust, resistance and friction on horizontal or inclined paths.

Exam points

draw and resolve the forces on each body before writing ΣF = ma along its motion
use a shared acceleration and appropriate tension relations for connected particles
solve a whole-system equation for acceleration, then an individual equation for tension or thrust

Covered syllabus points

4.4.1—Forces 4.4.2—Mass and weight 4.4.3—Constant acceleration 4.4.4—Connected-particle models

[Maximum number: 1]

A car has mass 1400 kg . When the speed of the car is $\mathrm{vms}^{-1}$ the magnitude of the resistance to motion is $k v^{2} \mathrm{~N}$ where k is a constant.

The car moves at a constant speed of $24 \mathrm{~m} \mathrm{~s}^{-1}$ up a hill inclined at an angle of $\alpha$ to the horizontal where $\sin \alpha=0.12$ . At this speed the magnitude of the resistance to motion is 480 N .

Find the value of k.

$\left[k \times 24^{2}=480 \Rightarrow\right] k=\frac{5}{6}$
OE, e.g. $\frac{480}{576}, 0.833$ or better.

[Maximum number: 8]

A straight slope of length 60 m is inclined at an angle of $12^{\circ}$ to the horizontal. A bobsled starts at the top of the slope with a speed of $5 \mathrm{~ms}^{-1}$ . The bobsled slides directly down the slope.

(a)

It is given that there is no resistance to the bobsled's motion.

Find its speed when it reaches the bottom of the slope.

[ 3 ]

$a=2.08 \quad[2.07911 \ldots]$
From $m g \sin 12=m a$ .

Marking guidance:

Allow exact (e.g. $a=g \sin 12$ ).

$v^{2}=5^{2}+2 a \times 60$
For use of $v^{2}=u^{2}+2 a s$ with u=5 and s=60.
Allow sign errors but a must be either $g \sin 12$ or $g \cos 12$ only.

Speed $=16.6 \mathrm{~ms}^{-1}[16.567861 \ldots]$
AWRT 16.6

Alternative Method for Q5(a)

For attempt at work energy equation
3 terms, dimensionally correct.
Allow sign errors;
allow sin/cos mix on PE term - condone m missing
from all terms. Must be a weight component.

$\frac{1}{2} m v^{2}=\frac{1}{2} m \times 5^{2}+m g \times 60 \sin 12$
Correct equation.
(for reference: $60 \sin 12=12.4747 \ldots$ )

Speed $=16.6 \mathrm{~ms}^{-1}[16.567861 \ldots]$
AWRT 16.6

(b)

It is given instead that the coefficient of friction between the bobsled and the slope is 0.03 .

Find the time that it takes for the bobsled to reach the bottom of the slope.

[ 5 ]

$R=m g \cos 12$
Resolving correctly perpendicular to the plane.

$m g \sin 12-F=m a$

*M1

Use of Newton's second law, correct number of terms;
allow sign errors;

Marking guidance:

allow $\sin / \cos$ mix (must be a weight component).

For use of F=0.03 R to get equation in a (and m ) only
Where R is a component of weight only (dimensionally correct but allow $\sin / \cos \operatorname{mix})$ .
$[\Rightarrow a=1.79[1.78567 \ldots .]]$ $60=5 t+\frac{1}{2} a t^{2}$ and solve for t
Dependent on previous two M marks.

For use of $s=u t+\frac{1}{2} a t^{2}$ with s=60, u=5 and their a or other complete method to find positive value(s) of t.

Time $=5.86 \mathrm{~s}[5.86260 \ldots]$
AWRT 5.86 (from using a=1.79 or better).

AWRT 5.85 (from using a=1.8 ).

AWRT 5.87 (from correct working).

Alternative method for Question 5(b): Using energy

$R=m g \cos 12$
Resolving correctly perpendicular to the plane.

$\frac{1}{2} m v^{2}-\frac{1}{2} m \times 5^{2}=60 \times m g \sin 12-60 \times F$

(*M1)

Use of work-energy principle, correctly number of relevant terms;
allow sign errors;
allow sin/cos mix on PE term (must be a weight component).

For use of F=0.03 R to get equation in v (and m ) only
Where R is a component of weight only (dimensionally correct but allow sin/cos mix)
$[\Rightarrow v=15.5[15.46870 \ldots .]] .$ $60=\frac{1}{2}(5+v) t$ and solve for t
Dependent on previous two M marks.

Use of $s=\frac{1}{2}(u+v) t$ with s=60, u=5 and their v or other complete method to find positive value(s) of t.

Time $=5.86 \mathrm{~s}[5.86260 \ldots]$

[Maximum number: 9]

A van of mass 4500 kg is towing a trailer of mass 750 kg down a straight hill inclined at an angle of $\theta$ to the horizontal where $\sin \theta=0.05$ . The van and the trailer are connected by a light rigid tow-bar which is parallel to the road. There are constant resistance forces of 2500 N on the van and 300 N on the trailer.

(a)

It is given that the tension in the tow-bar is 450 N .

Find the acceleration of the trailer and the driving force of the van's engine.

On another occasion, the van and trailer ascend a straight hill inclined at an angle of $\alpha$ to the horizontal where $\sin \alpha=0.09$ . The driving force of the van's engine is now 9100 N , and the speed of the van at the bottom of the hill is $20 \mathrm{~ms}^{-1}$ . The resistances to motion are unchanged.

[ 4 ]

Use of Newton's second law for van or trailer or system

Note: Trailer has 4 terms

Van has 5 terms

System has 7 terms (or 5 if counting van and trailer as one body)

M1*

Must have correct number of relevant terms. Allow sign errors.

Marking guidance:

Allow $\sin / \cos$ mix. Allow g missing.

Masses must be correct for their equation(s).

Forces must have components (or not) as required.

Must have either 0.05 or $\sin 2.86 \ldots$ or $\sin 2.9$ , not just $\sin \theta$ .

Trailer: $450+750 g \times 0.05-300=750 a$ [525=750 a]

Van: $D+4500 g \times 0.05-2500-450=4500 a$ [D-700=4500 a]

System:

$D+4500 g \times 0.05+750 g \times 0.05-2500-300=(4500+750) a$ [D-175=5250 a]
For any two correct equations.

For attempt to solve for a or D
Must get to ' a= ' or ' D= '. Must have correct number of relevant
terms in the equation(s) which they are using to find a or D . g
must be present. Allow sign errors. Allow sin/cos mix. If no working shown to solve their equations, then their answers should be correct for their equations.

$a=0.7 \mathrm{~ms}^{-2}$ and $D=3850 \mathrm{~N}$

(b)

Find the acceleration of the van and the tension in the tow-bar.

[ 5 ]

Use of Newton's second law for van or trailer or system

Note: Trailer has 4 terms

Van has 5 terms

System has 7 terms (or 5 if counting van and trailer as one body)

M1*

Must have correct number of relevant terms Allow sign errors. Allow sin/cos mix. Allow g missing.

Masses must be appropriate for their equation(s). Forces must have components (or not) as required. Must have either 0.09 or $\sin 5.16 \ldots$ or $\sin 5.2$ not just $\sin \theta$ .

Trailer: $T-300-750 g \times 0.09=750 a$ [T-975=750 a]
Van: $9100-2500-4500 g \times 0.09-T=4500 a$ [2550-T=4500 a]
System: $9100-2500-300-(4500+750) g \times 0.09=(4500+750) a$ [1575=5250 a]
A1 for one correct equation, second A1 for another correct equation.

If using Van and Trailer equations, must be using the same T for both to get the second A1.

For attempt to solve for a or T
Must get to ' a= ' or ' T= '. Must have correct number of relevant
terms in the equation(s) which they are using to find a or T . g
must be present. Allow sign errors. Allow sin/cos mix.
If no working shown to solve their equations, then their answers should be correct for their equations.

$T=1200 \mathrm{~N}$ and $a=0.3 \mathrm{~ms}^{-2}$

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Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)