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: 9]

Figure 5

A thin smooth hollow spherical shell has centre O and radius r. Part of the shell is removed to form a bowl with a plane circular rim. The bowl is fixed with the circular rim uppermost and horizontal. The point A is the lowest point of the bowl, as shown in Figure 5.

The point B is on the rim of the bowl, with OB at an angle $\theta$ to the upward vertical, where $\tan \theta=\frac{12}{5}$
A small ball is placed in the bowl at A. The ball is projected from A with horizontal speed u and moves in the vertical plane AOB. The ball stays in contact with the bowl until it reaches B.

At the instant when the ball reaches B, the speed of the ball is v.
By modelling the ball as a particle and ignoring air resistance,

(a)

use the principle of conservation of mechanical energy to show that

v^{2}=u^{2}-\frac{36}{13} g r

[ 3 ]

Use conservation of energy equation with 2KE terms and 1EPE term. Note there are rearrangements. Dimensionally correct, terms of the correct structure,
condone sign errors. EPE of the form $\frac{1}{2} k x^{2}$
For an unsimplified equation with at most one error
For a correct unsimplified equation

A1*

Given answer obtained from complete and correct working. Must include a line
of working before reaching the given answer.

(b)

The point C is such that BC is a diameter of the rim of the bowl.

Given that $u^{2}=4 g r$

use the model to show that, after leaving the inner surface of the bowl at B, the ball falls back into the bowl before reaching C.

[ 6 ]

BC=2r\sin\theta=\frac{24r}{13}

v^2=u^2-\frac{36gr}{13}=\frac{16gr}{13}

t=\frac{2v\sin\theta}{g}=\frac{24}{13}\sqrt{\frac{r}{g}}

x=v\cos\theta\, t

x=\frac{1920r}{2197}

A1*

$\frac{1920r}{2197}<\frac{24r}{13}=BC$ , so the ball falls back into the bowl before reaching C.

[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 4

Question (a)

Question (b)

Question (c)

Question (d)