Edexcel A-Level Mathematics A2 Unit M3 Mechanics 3 Questions

Practise Mechanics 3 topics involving variable acceleration, elastic strings, circular motion, energy methods and rigid body statics.

Covered syllabus points

M3.1 - Further kinematics M3.2 - Elastic strings and springs M3.3 - Further dynamics M3.4 - Motion in a circle M3.5 - Statics of rigid bodies

[Maximum number: 10]

In this question you must show all stages in your working.

Solutions relying entirely on calculator technology are not acceptable.

A particle P is moving along the x-axis.
At time t seconds, where $0 \leqslant t \leqslant \frac{2}{3}, P$ is x metres from the origin O and is moving with velocity $v \mathrm{~m} \mathrm{~s}^{-1}$ in the positive x direction where

v=(2 x+1)^{\frac{3}{2}}

When t=0, P passes through O.

(a)

Find the value of x when the acceleration of P is $243 \mathrm{~m} \mathrm{~s}^{-2}$

[ 4 ]

$a=v \frac{\mathrm{~d} v}{\mathrm{~d} x}$ $=\frac{3}{2}(2 x+1)^{\frac{1}{2}} \times 2 \times(2 x+1)^{\frac{3}{2}}=3(2 x+1)^{2}$ $3(2 x+1)^{2}=243$
x=4
(4)

(b)

Find v in terms of t.

[ 6 ]

$(2x+1)^{3/2}=\frac{\mathrm{d}x}{\mathrm{d}t}$
or $a=3v^{4/3}=\frac{\mathrm{d}v}{\mathrm{d}t}$ $\mathrm{d}t=(2x+1)^{-3/2}\,\mathrm{d}x$ $t=-(2x+1)^{-1/2}+C$ $t=0,\ x=0\Rightarrow C=1$
and obtain an equation in v and t only.
$v=\frac{1}{(1-t)^3}$

[Maximum number: 9]

Figure 1

A light elastic spring has natural length l and modulus of elasticity $\lambda$ .
One end of the spring is attached to a point A on a smooth plane.
The plane is inclined at angle $\theta$ to the horizontal, where $\tan\theta=\frac5{12}$ .
A particle P of mass m is attached to the other end of the spring.
Initially P is held at the point B on the plane, where AB is a line of greatest slope of the plane.
The point B is lower than A and AB=2l, as shown in Figure 1.
The particle is released from rest at B and first comes to instantaneous rest at the point C on AB, where AC=0.7l.

(a)

Use the principle of conservation of mechanical energy to show that

$\lambda=\frac{100}{91}mg.$

[ 5 ]

Change in GPE
$=mg\cdot1.3l\sin\theta=0.5lmg$ $\mathrm{EPE}=\frac{\lambda l^2}{2l}\quad\text{or}\quad \frac{\lambda(0.3l)^2}{2l}$
Energy equation from B to A:
$\frac{\lambda l^2}{2l}-\frac{\lambda(0.3l)^2}{2l}=0.5lmg$ $\lambda=mg\frac1{1-0.09}=\frac{100}{91}mg$
A1*

(b)

Find the acceleration of P when it is released from rest at B.

[ 4 ]

Equation of motion:
$T-mg\sin\theta=ma$ $\frac{\lambda l}{l}-mg\sin\theta=ma$ $\frac{100}{91}mg-\frac5{13}mg=ma$ $a=\frac57g$

[Maximum number: 8]

A particle P of mass $m \mathrm{~kg}$ is initially held at rest at the point O on a smooth inclined plane. The plane is inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha=\frac{2}{5}$

The particle is released from rest and slides down the plane against a force which acts towards O. The force has magnitude $\frac{1}{3} m x^{2} \mathrm{~N}$ , where x metres is the distance of P from O.

(a)

Find the speed of P when x=2

The particle first comes to instantaneous rest at the point A.

[ 6 ]

Using the equation of motion along the plane:
$mv\frac{dv}{dx}=mg\sin\alpha-\frac13mx^2$ $\frac12v^2=xg\sin\alpha-\frac19x^3$

At x=2:
$\frac12v^2=2g\sin\alpha-\frac89$

Since $\sin\alpha=\frac25$ ,
$v=3.7\ \text{or}\ 3.73\text{ m s}^{-1}$

M1 A1 DM1 A1

Alternative energy method:
$mg\sin\alpha\,x=\int\frac13mx^2\,dx+\frac12mv^2$

Marking guidance for this part:
M1 for attempting an equation of motion parallel to the plane.
A1 for correct integration.
DM1 for substituting x=2 into the expression for $v^2$ .
A1 for 3.7 or $3.73\text{ m s}^{-1}$ .

(b)

Find the distance OA.

Figure for Question (b) — Edexcel A-Level Mathematics A2

[ 2 ]

$v=0\Rightarrow x^2=9g\sin\alpha=9g\times\frac25,\qquad x\ne0$ $x=5.939\ldots$ $OA=5.9\text{ or }5.94\text{ m}$
Marking guidance for this part:
M1 for using v=0 in their expression for v and obtaining a value of x.
A1 for 5.9 or $5.94\text{ m}$ .

[Maximum number: 5]

Use algebraic integration to show that the centre of mass of a uniform semicircular disc of radius r and centre O is at a distance $\frac{4 r}{3 \pi}$ from the diameter through O [You may assume, without proof, that the area of a circle of radius r is $\pi r^{2}$ ]

A uniform lamina L is in the shape of a semicircle with centre B and diameter A C=8 a. The semicircle with diameter AB is removed from L and attached to the straight edge BC to form the template T, shown shaded in Figure 4.

Figure 4

The distance of the centre of mass of T from AC is d.

\bar{x}=\frac{\pi\int_0^r x(r^2-x^2)\,dx}{\frac{2}{3}\pi r^3}

=\frac{\pi\left[\frac{r^2x^2}{2}-\frac{x^4}{4}\right]_0^r}{\frac{2}{3}\pi r^3}

=\frac{\pi\cdot\frac{r^4}{4}}{\frac{2}{3}\pi r^3}

A1*

\bar{x}=\frac{4r}{3\pi}

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)

Question 4