CAIE A-Level Further Math A2 3 Further Mechanics Questions

Practise Further Mechanics through projectiles, equilibrium, circular motion, Hooke's law, variable forces and momentum, checking modelling steps against mark schemes.

Covered syllabus points

3.1 Motion of a projectile 3.2 Equilibrium of a rigid body 3.3 Circular motion 3.4 Hooke's law 3.5 Linear motion under a variable force 3.6 Momentum

[Maximum number: 11]

A particle P of mass m is attached to one end of a light inextensible rod of length 3 a. An identical particle Q is attached to the other end of the rod. The rod is smoothly pivoted at a point O on the rod, where O Q=x. The system, of rod and particles, rotates about O in a vertical plane.

At an instant when the rod is vertical, with P above Q, the particle P is moving horizontally with speed u. When the rod has turned through an angle of $60^{\circ}$ from the vertical, the speed of P is $2 \sqrt{a g}$ , and the tensions in the two parts of the rod, OP and OQ, have equal magnitudes.

(a)

Show that the speed of Q when the rod has turned through an angle of $60^{\circ}$ from the vertical is

\frac{2 x}{3 a-x} \sqrt{a g} .

[ 2 ]

Angular speeds of P and Q are equal,
so $\frac{\nu_{Q}}{x}=\frac{\nu_{P}}{3 a-x}$ $v_{Q}=\frac{2 x \sqrt{a g}}{3 a-x}$
Shown convincingly: angular speeds equal stated.
AG

2

(b)

Find x in terms of a.

[ 5 ]

For $P: T+m g \cos 60^{\circ}=\frac{m \times 4 a g}{3 a-x}$
For $Q: T-m g \cos 60^{\circ}=\frac{m v_{Q}^{2}}{x}$
Eliminate $T:-m g \cos 60^{\circ}+\frac{m \cdot 4 a g}{3 a-x}=m g \cos 60^{\circ}+\frac{m v_{Q}^{2}}{x}$ $\begin{aligned} \frac{m \times 4 a g}{3 a-x}=1+\frac{m x 4 a g}{(3 a-x)^{2}} 4 a(3 a-x)=(3 a-x)^{2}+4 a x, x^{2}+2 a x-3 a^{2}=0 \end{aligned}$
Solve to find x.
Obtain 3-term quadratic equation.

$(x-a)(x+3 a)=0, \quad x=a$
5

(c)

Find u in terms of a and g.

Additional page

If you use the following page to complete the answer to any question, the question number must be clearly shown.

[ 4 ]

KEs correct.

Energy changes from initial position:

Gain in KE of $P: \frac{1}{2} m\left(4 a g-u^{2}\right)$

Loss in KE of $Q: \frac{1}{2} m\left(\left(\frac{u}{2}\right)^{2}-v_{Q}^{2}\right)$

Loss in GPE of $P=m g(3 a-x)\left(1-\cos 60^{\circ}\right)(=m g a)$

Gain in GPE of $Q=m g x\left(1-\cos 60^{\circ}\right)\left(=\frac{1}{2} m g a\right)$

B1FT

GPEs correct.

$\frac{1}{2} m\left(4 a g-u^{2}\right)-\frac{1}{2} m\left(\left(\frac{u}{2}\right)^{2}-v_{Q}^{2}\right)=-m g x\left(1-\cos 60^{\circ}\right)+m g(3 a-x)\left(1-\cos 60^{\circ}\right)$
Energy equation.

Simplify: $4 a g-\frac{5}{4} u^{2}+a g=a g$ $u^{2}=\frac{16}{5} a g, \quad u=\frac{4}{5} \sqrt{5 a g}$
AEF

4

[Maximum number: 5]

One end of a light elastic string, of natural length a and modulus of elasticity 3 m g, is attached to a fixed point O. The other end of the string is attached to a particle P of mass m. The string hangs with P vertically below O. The particle P is pulled vertically downwards so that the extension of the string is 2 a. The particle P is then released from rest.

(a)

Find the speed of P when it is at a distance $\frac{3}{4} a$ below O.

[ 3 ]

$\frac{3 m g}{2 a}(2 a)^{2}$
Correct EPE term seen

$\frac{1}{2} m v^{2}+m g \times\left(3 a-\frac{3}{4} a\right)=\frac{3 m g}{2 a}(2 a)^{2}$
Dimensionally correct energy equation. Must have
one KE, one EPE term and at least one GPE.
Allow sign errors.

$v=\sqrt{\frac{15}{2} a g} \quad[2.74 \sqrt{a g}]$
AEF

3

(b)

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

[ 2 ]

T-m g=m A and $T=\frac{3 m g}{a} \times 2 a$
N2L and Hooke's law

Acceleration =5 g [upwards]
Allow $\pm 50$ or $\pm 5 g$

2

[Maximum number: 6]

A particle P of mass 0.5 kg moves in a straight line. At time $t \mathrm{~s}$ the velocity of P is $v \mathrm{~ms}^{-1}$ and its displacement from a fixed point O on the line is $x \mathrm{~m}$ . The only forces acting on P are a force of magnitude $\frac{150}{(x+1)^{2}} \mathrm{~N}$ in the direction of increasing displacement and a resistive force of magnitude $\frac{450}{(x+1)^{3}} \mathrm{~N}$ . When t=0, x=0 and v=20.

Find v in terms of x, giving your answer in the form $v=\frac{A x+B}{(x+1)}$ , where A and B are constants to be
determined. determined.

$0.5 v \frac{\mathrm{~d} v}{\mathrm{~d} x}=\frac{150}{(x+1)^{2}}-\frac{450}{(x+1)^{3}}$
Allow sign errors.

Integrate: $0.5 v^{2}=-\frac{300}{x+1}+\frac{450}{(x+1)^{2}}+A$
Correct powers, allow sign errors.

x=0, v=20;
A=50
Use initial condition.

Rearrange: $v^{2}=\frac{100\left(x^{2}-4 x+4\right)}{(x+1)^{2}}$
AEF

$v^{2}=\frac{100(x-2)^{2}}{(x+1)^{2}} \text { so } v= \pm \frac{10(x-2)}{(x+1)}$

From initial condition, sign must be negative, $v=\frac{20-10 x}{x+1}$
Signs dealt with convincingly.

6

[Maximum number: 7]

(a)

Show that $v=\frac{1}{2} u(4 \cos \theta-1)$ .

[ 1 ]

$\begin{aligned} \text { PCLM: } m v=-m u \cos 60^{\circ}+2 m u \cos \theta v=-\frac{1}{2} u+2 u \cos \theta=v=\frac{1}{2} u(4 \cos \theta-1) \end{aligned}$
First line must be seen.

AG

1

(b)

Find the value of $\cos \theta$ .

[ 4 ]

KE of $A=\frac{1}{2} m\left(v^{2}+\left(u \sin 60^{\circ}\right)^{2}\right)$
KE of $A=2 \times \mathrm{KE}$ of B, so
$\frac{1}{2} m\left(v^{2}+\left(u \sin 60^{\circ}\right)^{2}\right)=2 \times \frac{1}{2} \times 2 m(u \sin \theta)^{2}$ $\begin{aligned} \left(\frac{1}{2} u(4 \cos \theta-1)\right)^{2}+\frac{3}{4} u^{2}=4 u^{2}(\sin \theta)^{2} 8 \cos ^{2} \theta-2 \cos \theta-3=0 \end{aligned}$
Use result of (a) and rearrange.
Obtain 3-term quadratic.

$\cos \theta=\frac{3}{4}, \quad-\frac{1}{2}$ but angle is acute, so $\cos \theta=\frac{3}{4}$
4

(c)

Find the value of e.

[ 2 ]

NEL: $v=e\left(u \cos 60^{\circ}+u \cos \theta\right)$ $\begin{aligned} v=u, \quad u=e\left(\frac{1}{2} u+\frac{3}{4} u\right) e=\frac{4}{5} \end{aligned}$
2

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question (a)

Question (b)

Question 3

Question 4

Question (a)

Question (b)

Question (c)