Edexcel A-Level Mathematics A2 Unit M2 Mechanics 2 Questions

Practise M2 mechanics across motion, centres of mass, work-energy, collisions and statics using equations, vectors, diagrams and modelling decisions.

Covered syllabus points

M2.1 - Kinematics of a particle moving in a straight line or plane M2.2 - Centres of mass M2.3 - Work and energy M2.4 - Collisions M2.5 - Statics of rigid bodies

[Maximum number: 5]

Find the exact value of k

Hemisphere / Cone

$\begin{aligned} \frac{2}{3}\times\frac{3}{8}r &= \frac{k}{3}\times\frac{1}{4}kr,\\ k^2&=3,\quad k=\sqrt{3}. \end{aligned}$
M1: Form a dimensionally correct moments equation with the correct value for $\bar{X}$ .
A1ft: Correct equation, follow through their masses and distances;
signs correct.

\begin{array}{lll} \text{Mass} & \frac{2}{3}\pi r^3 & \frac{1}{3}\pi k r^3\\ \text{Distance of centre of mass from centre of common plane} & \frac{3}{8}r & \frac{1}{4}kr \end{array}

[Maximum number: 6]

Three particles of masses 2 m, 3 m and 4 m are placed at the points with coordinates (-2,5),(2,-3) and (3 k, k) respectively, where k is a constant.
The centre of mass of the three particles is at the point $(\bar{x}, \bar{y})$ .

(a)

Show that $\bar{x}=\frac{2+12 k}{9}$

The centre of mass of the three particles lies at a point on the straight line with equation x+2 y=3

[ 2 ]

M( x axis)

M1

Need all terms. Dimensionally consistent. Condone if m missing throughout.

Accept as part of a vector equation

$\begin{aligned} 2 m \times-+\times+\times=\times(2) 3 m 24 m 3 k \quad 9 m \quad x x=\frac{2+12 k}{9} * \end{aligned}$

A1*

Obtain given result

2

(b)

Find the value of k.

[ 4 ]

M(y axis)
Need all terms. Dimensionally consistent. Might be seen as part of a vector equation in (a). It does not score any marks until referred to in part (b). Condone if m missing throughout.

$\begin{aligned} 2 m \times+\times-+\times=\times 53 m(3) 4 m k \quad 9 m \quad y \square-14+k \square \square y=\square \square \end{aligned}$
Correct unsimplified equation. Allow if m missing
throughout.

Form and solve equation in k
$(2+12 k++=2 \quad 8 k \quad 27)$
Use their y ānd $x \overline{+}^{-}=2 y^{-} \quad 3$

Dependent on the two preceding M marks

$k=\frac{23}{20}(1.15)$
Correct answer only

4

(6)

[Maximum number: 13]

A cyclist is travelling on a straight horizontal road and working at a constant rate of 500 W .

The total mass of the cyclist and her cycle is 80 kg .
The total resistance to the motion of the cyclist is modelled as a constant force of magnitude 60 N .

(a)

Using this model, find the acceleration of the cyclist at the instant when her speed is $6 \mathrm{~m} \mathrm{~s}^{-1}$

On the following day, the cyclist travels up a straight road from a point A to a point B.
The distance from A to B is 20 km.
Point A is 500 m above sea level and point B is 800 m above sea level.
The cyclist starts from rest at A.
At the instant she reaches B her speed is $8 \mathrm{~ms}^{-1}$
The total resistance to the motion of the cyclist from non-gravitational forces is modelled as a constant force of magnitude 60 N.

[ 4 ]

Use P=Fv:

$F=\frac{500}{6}$
Equation of motion:

F-60=80a $a=\frac{7}{24}\ \mathrm{m\,s^{-2}}$

(b)

Using this model, find the total work done by the cyclist in the journey from A to B.

Later on, the cyclist is travelling up a straight road which is inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha=\frac{1}{20}$

The cyclist is now working at a constant rate of P watts and has a constant speed of $7 \mathrm{~m} \mathrm{~s}^{-1}$

The total resistance to the motion of the cyclist from non-gravitational forces is again modelled as a constant force of magnitude 60 N.

[ 5 ]

$\text{Gain in KE}=\frac12\times80\times8^2=2560\ \mathrm J$ $\text{Gain in GPE}=80\times9.8\times300=235200\ \mathrm J$ $\text{Work against resistance}=20000\times60$ $\text{Total work}=40\times64+80\times9.8\times300+20000\times60=1437760\ \mathrm J$ $1440\ \mathrm{kJ}\quad\text{or}\quad 1400\ \mathrm{kJ}$

(c)

Using this model, find the value of P

[ 4 ]

Equation of motion
$\begin{aligned} F-60-80g\sin\alpha&=0\\ \frac{P}{7}-60-80g\left(\frac1{20}\right)&=0 \end{aligned}$ $P=694\ \text{W}\quad\text{or}\quad 690\ \text{W}$

[Maximum number: 14]

Particle P has mass 4 m and particle Q has mass 2 m.

The particles are moving in opposite directions along the same straight line on a smooth horizontal surface.

Particle P collides directly with particle Q.
Immediately before the collision, the speed of P is 2 u and the speed of Q is 3 u.
Immediately after the collision, the speed of P is x and the speed of Q is y.
The direction of motion of each particle is reversed as a result of the collision.
The total kinetic energy of P and Q after the collision is half of the total kinetic energy of P and Q before the collision.

(a)

Show that $y=\frac{8}{3} u$

The coefficient of restitution between P and Q is e.

[ 6 ]

Official answer figure

Equation for CLM
8mu-6mu=2my-4mx
Equation for kinetic energy
$2mx^2+my^2=\frac12(2m\cdot4u^2+m\cdot9u^2)$ $17u^2=2y^2+(y-u)^2 \Rightarrow 3y^2-2uy-16u^2=0$ $y=\frac83u$

A1*

(b)

Find the value of e.

After the collision, Q hits a smooth fixed vertical wall that is perpendicular to the direction of motion of Q.

Particle Q rebounds.
The coefficient of restitution between Q and the wall is f.
Given that there is no second collision between P and Q,

[ 3 ]

Use of impact law:

x+y=5eu $e=\frac{\frac12\left(\frac83u-u\right)+\frac83u}{5u}$ $e=\frac{7}{10}$

(c)

find the range of possible values of f.

Given that $f=\frac{1}{4}$

[ 3 ]

Velocity of Q after impact with the wall:

$f\cdot\frac83u$
No second collision if

$f\cdot\frac83u\le \frac56u$ $0<f\le\frac{5}{16}$

(d)

find, in terms of m and u, the magnitude of the impulse received by Q as a result of its impact with the wall.

[ 2 ]

Use

$I=\pm2m\left(y-\left(-\frac14y\right)\right)$ $|I|=\frac{20}{3}mu$

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Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)

Question (c)

Question 4

Question (a)

Question (b)

Question (c)

Question (d)