A-Level Further Mathematics: Further Mechanics
Exam Board Note
Further Mechanics is an optional module in all major UK specifications. Edexcel offers Further Mechanics 1 (FM1/8FM0:26) and FM2. AQA has Mechanics options covering collisions, SHM, and circular motion. OCR A includes this material in Mechanics A (Y534). CIE covers mechanics topics in Further Mathematics Papers 4 and further options. Dimensional analysis is found primarily on Edexcel and CIE. Check your specification for the exact combination of topics required.
Further Mechanics extends the A-Level Mechanics with more sophisticated models of motion and forces. The topics here link directly to physics, engineering, and applied mathematics at degree level. Pay particular attention to SHM, which appears frequently in examination contexts, and to the coefficient of restitution, which introduces non-trivial collision problems.
1. Momentum, Impulse, and Conservation
Definitions: Momentum and Impulse
The momentum of a particle of mass $m$ moving with velocity $\mathbf{v}$ is: $$\mathbf{p} = m\mathbf{v}$$ The SI unit is kg m s$^{-1}$ (or N s).
The impulse $\mathbf{J}$ of a force $\mathbf{F}$ acting over time interval $[t_1, t_2]$ is: $$\mathbf{J} = \int_{t_1}^{t_2}\mathbf{F}\,dt$$ For a constant force: $\mathbf{J} = \mathbf{F}(t_2 - t_1) = \mathbf{F}\Delta t$.
Impulse-Momentum Theorem: $\mathbf{J} = \Delta\mathbf{p} = m\mathbf{v} - m\mathbf{u}$
Law of Conservation of Linear Momentum
For a system of particles with no external forces (or when external forces cancel), the total momentum is conserved: $$\sum m_i\mathbf{u}_i = \sum m_i\mathbf{v}_i$$ where $\mathbf{u}_i$ and $\mathbf{v}_i$ are the velocities before and after the collision.
Worked Example 1.1 — Conservation of Momentum
Particle $A$ (mass 3 kg, velocity $5$ m s$^{-1}$) collides with stationary particle $B$ (mass 2 kg). After the collision, $A$ moves at $1$ m s$^{-1}$ in the original direction. Find $B$'s velocity after the collision and the impulse exerted by $A$ on $B$.
Step 1: Conservation of momentum (taking right as positive): $$3(5) + 2(0) = 3(1) + 2v_B$$ $$15 = 3 + 2v_B \implies v_B = 6 \text{ m s}^{-1}$$
Step 2: Impulse on B $$J = m_B(v_B - u_B) = 2(6 - 0) = 12 \text{ N s}$$ By Newton's 3rd law, the impulse exerted by $B$ on $A$ is $-12$ N s (in the negative direction).
2. Coefficient of Restitution and Collisions
Newton's Law of Restitution
When two particles collide directly, the coefficient of restitution $e$ (where $0 \leq e \leq 1$) satisfies: $$e = \frac{\text{speed of separation along line of impact}}{\text{speed of approach along line of impact}} = \frac{v_B - v_A}{u_A - u_B}$$ (taking positive direction from $A$ to $B$, with $u_A > u_B$ for a collision to occur).
- $e = 1$: perfectly elastic (kinetic energy is conserved).
- $e = 0$: perfectly inelastic (particles coalesce, maximum KE loss).
- $0 < e < 1$: partially elastic (some KE is lost, converted to heat/sound).
Worked Example 2.1 — Collision with Coefficient of Restitution
Sphere $A$ (mass 4 kg) moves at 8 m s$^{-1}$ and collides directly with sphere $B$ (mass 6 kg) moving at 2 m s$^{-1}$ in the same direction. The coefficient of restitution is $e = 0.5$. Find the velocities after impact and the loss in kinetic energy.
Step 1: Set up equations Let $v_A$ and $v_B$ be velocities after collision.
Conservation of momentum: $$4(8) + 6(2) = 4v_A + 6v_B \implies 44 = 4v_A + 6v_B \quad (1)$$
Newton's law of restitution: $$v_B - v_A = e(u_A - u_B) = 0.5(8-2) = 3 \quad (2)$$
Step 2: Solve From (2): $v_B = v_A + 3$. Substitute into (1): $$44 = 4v_A + 6(v_A + 3) = 10v_A + 18 \implies v_A = 2.6 \text{ m s}^{-1}$$ $$v_B = 2.6 + 3 = 5.6 \text{ m s}^{-1}$$
Step 3: KE loss $$\text{KE before} = \tfrac{1}{2}(4)(8^2) + \tfrac{1}{2}(6)(2^2) = 128 + 12 = 140 \text{ J}$$ $$\text{KE after} = \tfrac{1}{2}(4)(2.6^2) + \tfrac{1}{2}(6)(5.6^2) = 13.52 + 94.08 = 107.60 \text{ J}$$ $$\text{Loss} = 140 - 107.6 = 32.4 \text{ J}$$
Oblique Collisions with a Smooth Wall
When a ball strikes a smooth wall at angle $\alpha$ to the normal:
- The component of velocity parallel to the wall is unchanged (wall is smooth, no friction).
- The component of velocity perpendicular to the wall is reversed and reduced by factor $e$.
3. Work, Energy, and Power
Work-Energy Theorem
The net work done on a particle equals the change in its kinetic energy: $$W_{\text{net}} = \Delta KE = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2$$
The work done by a force $\mathbf{F}$ along a path from $A$ to $B$: $$W = \int_A^B \mathbf{F}\cdot d\mathbf{r}$$ For a constant force at angle $\theta$ to displacement $s$: $W = Fs\cos\theta$.
The principle of conservation of energy: in the absence of non-conservative forces (friction, air resistance), the total mechanical energy $KE + PE$ is constant.
Power: The rate of doing work. $P = \mathbf{F}\cdot\mathbf{v}$. For a vehicle engine: $$P = Fv \quad \Rightarrow \quad F = \frac{P}{v}$$ The maximum speed $v_{\max}$ on a given gradient occurs when the driving force equals the resistance (zero acceleration), so $P = F_{\text{resistance}}\,v_{\max}$.
4. Elastic Strings and Hooke's Law
Hooke's Law for Elastic Strings and Springs
For an elastic string or spring with natural length $l_0$ and modulus of elasticity $\lambda$, the tension when stretched by extension $x$ is: $$T = \frac{\lambda x}{l_0}$$ Alternatively, using the spring constant $k = \dfrac{\lambda}{l_0}$: $T = kx$ (Hooke's Law).
Note: an elastic string can only pull (tension only, no compression); an elastic spring can both push and pull.
Elastic potential energy (EPE) stored in the string: $$E_{PE} = \frac{\lambda x^2}{2l_0} = \frac{kx^2}{2}$$
Worked Example 4.1 — Energy Method with Elastic String
A particle of mass 2 kg is attached to one end of an elastic string of natural length 1 m and modulus of elasticity 40 N. The other end is fixed to a point $O$ on a smooth horizontal table. The particle is pulled to a distance 1.5 m from $O$ and released from rest. Find its speed when the string returns to its natural length.
Step 1 Initially, the extension is $x = 1.5 - 1.0 = 0.5$ m. Initial EPE: $$E_0 = \frac{40 \times 0.5^2}{2 \times 1} = \frac{40 \times 0.25}{2} = 5 \text{ J}$$
Step 2 When the string is at its natural length, EPE = 0. The table is smooth (no friction). By conservation of energy: $$E_0 = KE = \tfrac{1}{2}mv^2$$ $$5 = \tfrac{1}{2}(2)v^2 = v^2 \implies v = \sqrt{5} \approx 2.24 \text{ m s}^{-1}$$
Worked Example 4.2 — Vertical Elastic String
A particle of mass 0.5 kg hangs at rest on an elastic string of natural length 0.8 m and modulus of elasticity 20 N. Find (a) the extension at rest, (b) the amplitude when the particle is pulled down a further 0.2 m and released.
Part (a) At equilibrium, tension $=$ weight: $$\frac{20x}{0.8} = 0.5 \times 9.8 \implies 25x = 4.9 \implies x = 0.196 \text{ m}$$
Part (b) Take the equilibrium position as origin. Let $e = 0.196$ m (equilibrium extension), $a = 0.2$ m (initial displacement below equilibrium).
By energy conservation between the lowest point (displacement $a$ below equilibrium) and the highest point (displacement $A$ above equilibrium):
Total extension at lowest: $e + a = 0.396$ m. EPE at lowest point: $\dfrac{20(0.396)^2}{2(0.8)} = 1.960$ J.
At highest point, check whether string is slack. If $A < e$, string still taut. Using energy conservation (measuring PE from equilibrium position): $$\frac{1}{2}mv^2 + \frac{\lambda(e-A)^2}{2l_0} - mg A = \frac{\lambda(e+a)^2}{2l_0} + mg a$$ At the top ($v = 0$): this yields $A = a = 0.2$ m. The amplitude is 0.2 m.
5. Circular Motion
A particle moving in a circle of radius $r$ with speed $v$ (or angular speed $\omega$) has a centripetal acceleration directed towards the centre: $$a = \frac{v^2}{r} = r\omega^2$$ where $v = r\omega$. By Newton's second law, the centripetal force needed to maintain circular motion is: $$F = ma = \frac{mv^2}{r} = mr\omega^2$$ This force is provided by a real force (tension, normal reaction, friction, or a component thereof) directed towards the centre.
Angular quantities:
- Angular velocity: $\omega = \dfrac{d\theta}{dt}$, measured in rad s$^{-1}$.
- Period: $T = \dfrac{2\pi}{\omega}$, measured in seconds.
- Frequency: $f = \dfrac{1}{T} = \dfrac{\omega}{2\pi}$, measured in Hz.
Worked Example 5.1 — Conical Pendulum
A particle of mass 0.3 kg is attached to a string of length 0.5 m and moves in a horizontal circle. The string makes angle $\theta$ with the vertical. Find the tension $T$ and angular speed $\omega$ if $\theta = 30°$.
Step 1 Resolve vertically: $T\cos\theta = mg$ $$T\cos 30° = 0.3 \times 9.8 \implies T = \frac{2.94}{\cos 30°} = \frac{2.94}{\sqrt{3}/2} = \frac{5.88}{\sqrt{3}} \approx 3.39 \text{ N}$$
Step 2 The radius of the circle: $r = l\sin\theta = 0.5\sin 30° = 0.25$ m.
Step 3 Resolve horizontally (centripetal direction): $T\sin\theta = mr\omega^2$ $$T\sin 30° = 0.3(0.25)\omega^2$$ $$3.39 \times 0.5 = 0.075\omega^2 \implies \omega^2 = \frac{1.695}{0.075} = 22.6 \implies \omega \approx 4.75 \text{ rad s}^{-1}$$
6. Circular Motion in a Vertical Plane
In a vertical circle, the speed of the particle varies as it gains or loses height. Energy conservation gives: $$\tfrac{1}{2}mv^2 + mgh = \text{constant}$$ where $h$ is the height above some reference level.
Particle on a string: For the particle to complete the circle, the tension must be $\geq 0$ at the top. At the top of the circle (radius $r$, speed $v_{\text{top}}$): $$T + mg = \frac{mv_{\text{top}}^2}{r}$$ For $T \geq 0$: $v_{\text{top}}^2 \geq gr$, so $v_{\text{top}} \geq \sqrt{gr}$.
Particle on the outside of a sphere: The particle leaves the surface when the normal reaction $N = 0$: $$mg\cos\theta - N = \frac{mv^2}{r} \implies N = mg\cos\theta - \frac{mv^2}{r}$$ Setting $N = 0$ and using energy conservation gives the angle at which the particle leaves the sphere.
Worked Example 6.1 — Minimum Speed at Top of Loop
A particle on a string of length 0.6 m moves in a vertical circle. Find (a) the minimum speed at the top to maintain circular motion, and (b) the minimum speed at the bottom to achieve this.
Part (a) Minimum speed at top: $v_{\text{top}} = \sqrt{gr} = \sqrt{9.8 \times 0.6} = \sqrt{5.88} \approx 2.42$ m s$^{-1}$.
Part (b) Using energy conservation between bottom and top (height difference $= 2r = 1.2$ m): $$\tfrac{1}{2}mv_{\text{bot}}^2 = \tfrac{1}{2}mv_{\text{top}}^2 + mg(2r)$$ $$v_{\text{bot}}^2 = v_{\text{top}}^2 + 4gr = gr + 4gr = 5gr = 5(9.8)(0.6) = 29.4$$ $$v_{\text{bot}} = \sqrt{29.4} \approx 5.42 \text{ m s}^{-1}$$
7. Simple Harmonic Motion
Definition: Simple Harmonic Motion
A particle exhibits simple harmonic motion (SHM) if its acceleration is directed towards a fixed point (the equilibrium position) and is proportional to its displacement from that point: $$\ddot{x} = -\omega^2 x$$ where $\omega > 0$ is the angular frequency. The general solution is: $$x(t) = A\cos(\omega t + \phi)$$ where $A$ is the amplitude and $\phi$ is the initial phase angle.
Key SHM Formulae
$$x = A\cos(\omega t + \phi) \qquad v = -A\omega\sin(\omega t + \phi)$$
$$v^2 = \omega^2(A^2 - x^2) \quad \text{(velocity-displacement relation)}$$
$$\text{Period: } T = \frac{2\pi}{\omega} \qquad \text{Frequency: } f = \frac{\omega}{2\pi}$$
Maximum speed: $v_{\max} = A\omega$ (at $x = 0$)
Maximum acceleration: $|\ddot{x}|_{\max} = A\omega^2$ (at $x = \pm A$)
FMECH·1 — SHM displacement $x = A\cos(\omega t)$ (blue) and velocity $v = -A\omega\sin(\omega t)$ (green, dashed) plotted against time, with $A = 3$, $\omega = 1$ (so period $T = 2\pi$). The velocity is zero at the amplitude positions and maximum in magnitude at $x = 0$.
Worked Example 7.1 — SHM: Mass on a Spring
A particle of mass 0.5 kg is attached to a spring of stiffness $k = 18$ N m$^{-1}$. It is displaced 0.2 m from equilibrium and released from rest. Find (a) the angular frequency and period, (b) the velocity when $x = 0.1$ m, (c) the maximum acceleration.
Part (a) From $F = -kx = m\ddot{x}$: $$\omega^2 = \frac{k}{m} = \frac{18}{0.5} = 36 \implies \omega = 6 \text{ rad s}^{-1}$$ $$T = \frac{2\pi}{6} = \frac{\pi}{3} \approx 1.05 \text{ s}$$
Part (b) Amplitude $A = 0.2$ m. Use $v^2 = \omega^2(A^2 - x^2)$: $$v^2 = 36(0.04 - 0.01) = 36(0.03) = 1.08$$ $$v = \sqrt{1.08} = 0.6\sqrt{3} \approx 1.04 \text{ m s}^{-1}$$
Part (c) $|\ddot{x}|_{\max} = A\omega^2 = 0.2 \times 36 = 7.2$ m s$^{-2}$
Worked Example 7.2 — SHM: Time to Reach a Given Position
A particle undergoing SHM has amplitude 5 cm and period 4 s. Starting from the equilibrium position moving in the positive direction, find the first time the particle reaches $x = 3$ cm.
Step 1 $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{4} = \dfrac{\pi}{2}$ rad s$^{-1}$. Starting from $x = 0$ moving positively: $x = A\sin(\omega t) = 5\sin\!\left(\dfrac{\pi t}{2}\right)$ cm.
Step 2 Set $x = 3$: $$5\sin\!\left(\frac{\pi t}{2}\right) = 3 \implies \sin\!\left(\frac{\pi t}{2}\right) = 0.6$$ $$\frac{\pi t}{2} = \arcsin(0.6) \approx 0.6435 \text{ rad}$$ $$t \approx \frac{2 \times 0.6435}{\pi} \approx 0.410 \text{ s}$$
8. The Simple Pendulum and Oscillation Problems
A simple pendulum consists of a particle of mass $m$ on a string of length $l$, swinging in a vertical plane. For small oscillations (angle $\theta \ll 1$ radian), the restoring force is approximately $-mg\theta$ (using $\sin\theta \approx \theta$), giving SHM:
$$\ddot{\theta} = -\frac{g}{l}\theta$$Period of a Simple Pendulum (small oscillations)
$$T = 2\pi\sqrt{\frac{l}{g}}$$
This result is independent of the mass of the bob and the amplitude (for small oscillations). It is used to determine $g$ experimentally.
Worked Example 8.1 — Simple Pendulum
A pendulum clock beats seconds (period $T = 2$ s). Find the length of the pendulum (take $g = 9.81$ m s$^{-2}$). If the pendulum is taken to the Moon where $g = 1.62$ m s$^{-2}$, what is the new period?
On Earth $$T = 2\pi\sqrt{\frac{l}{g}} \implies 2 = 2\pi\sqrt{\frac{l}{9.81}} \implies \sqrt{\frac{l}{9.81}} = \frac{1}{\pi} \implies l = \frac{9.81}{\pi^2} \approx 0.994 \text{ m}$$
On the Moon $$T_{\text{Moon}} = 2\pi\sqrt{\frac{0.994}{1.62}} = 2\pi\sqrt{0.6136} \approx 2\pi(0.7832) \approx 4.92 \text{ s}$$ The pendulum runs much slower on the Moon, as expected.
Particle Attached to Elastic String: SHM Connection
When a particle of mass $m$ hangs on a vertical elastic string (spring constant $k$) and is displaced from its equilibrium position by $x$, the net restoring force is $-kx$ and the motion is SHM with $\omega^2 = k/m = \lambda/(ml_0)$. The particle only performs SHM while the string remains taut (i.e., while $x > -(e_0)$ where $e_0$ is the equilibrium extension).
9. Dimensional Analysis
Every physical quantity can be expressed in terms of the fundamental dimensions: mass $[\text{M}]$, length $[\text{L}]$, time $[\text{T}]$, and sometimes temperature, current, etc. Dimensional analysis uses this to:
- Check equations are dimensionally consistent ("dimensional checking").
- Derive possible forms of physical laws (the Buckingham $\Pi$ theorem).
- Find unknown indices in power-law relationships.
Dimensions of Common Quantities
| Quantity | Symbol | Dimensions | SI Unit |
|---|---|---|---|
| Mass | $m$ | M | kg |
| Length | $l$ | L | m |
| Time | $t$ | T | s |
| Velocity | $v$ | LT$^{-1}$ | m s$^{-1}$ |
| Acceleration | $a$ | LT$^{-2}$ | m s$^{-2}$ |
| Force | $F$ | MLT$^{-2}$ | N = kg m s$^{-2}$ |
| Energy / Work | $E$ | ML$^2$T$^{-2}$ | J = kg m$^2$ s$^{-2}$ |
| Power | $P$ | ML$^2$T$^{-3}$ | W = kg m$^2$ s$^{-3}$ |
| Momentum | $p$ | MLT$^{-1}$ | kg m s$^{-1}$ |
| Pressure | $P$ | ML$^{-1}$T$^{-2}$ | Pa = kg m$^{-1}$ s$^{-2}$ |
| Density | $\rho$ | ML$^{-3}$ | kg m$^{-3}$ |
Worked Example 9.1 — Deriving a Formula by Dimensional Analysis
The period $T$ of a simple pendulum is believed to depend on the string length $l$, the gravitational acceleration $g$, and the bob mass $m$. Use dimensional analysis to find the form of $T$.
Step 1 Assume $T = k\,l^a g^b m^c$ for some dimensionless constant $k$ and unknown indices $a, b, c$.
Step 2 Write dimensions on both sides: $$[\text{T}] = [\text{L}]^a [\text{LT}^{-2}]^b [\text{M}]^c = \text{L}^{a+b}\,\text{T}^{-2b}\,\text{M}^c$$
Step 3 Equate exponents of M, L, T:
- M: $0 = c \implies c = 0$ (mass does not affect period)
- L: $0 = a + b$
- T: $1 = -2b \implies b = -\frac{1}{2}$
Result $T = k\,l^{1/2}\,g^{-1/2} = k\sqrt{\dfrac{l}{g}}$. The actual formula (from physics) confirms $k = 2\pi$.
Worked Example 9.2 — Checking Dimensional Consistency
The escape velocity from a planet is given by $v = \sqrt{\dfrac{2GM}{R}}$, where $G$ is the gravitational constant, $M$ is the planet's mass, and $R$ is its radius. Show this is dimensionally consistent, given that $[G] = \text{M}^{-1}\text{L}^3\text{T}^{-2}$.
Check Dimensions of the right-hand side: $$\left[\sqrt{\frac{GM}{R}}\right] = \sqrt{\frac{\text{M}^{-1}\text{L}^3\text{T}^{-2} \cdot \text{M}}{\text{L}}} = \sqrt{\frac{\text{L}^3\text{T}^{-2}}{\text{L}}} = \sqrt{\text{L}^2\text{T}^{-2}} = \text{L}\,\text{T}^{-1}$$ This is the dimension of velocity (LT$^{-1}$), confirming the formula is dimensionally consistent.
10. Practice Problems
Problem 1 — Impulse
A ball of mass 0.4 kg moving at 12 m s$^{-1}$ is hit by a bat and returns in the opposite direction at 20 m s$^{-1}$. Find the magnitude of the impulse exerted by the bat on the ball. If the contact time is 0.05 s, find the average force exerted.
Show Solution
Taking the original direction as positive: $u = 12$ m s$^{-1}$, $v = -20$ m s$^{-1}$.
Impulse $= m(v - u) = 0.4(-20 - 12) = 0.4 \times (-32) = -12.8$ N s.
Magnitude of impulse $= 12.8$ N s.
Average force $= \dfrac{12.8}{0.05} = 256$ N (in the direction of the return).
Problem 2 — Newton's Law of Restitution
A ball of mass 0.5 kg falls vertically and hits a smooth horizontal floor at 8 m s$^{-1}$. The coefficient of restitution between the ball and the floor is 0.6. Find the speed after the first bounce and the height to which it rises.
Show Solution
Speed after bounce $= e \times u = 0.6 \times 8 = 4.8$ m s$^{-1}$ (upward).
Using $v^2 = u^2 - 2gh$ with $v = 0$ at maximum height: $$h = \frac{(4.8)^2}{2 \times 9.8} = \frac{23.04}{19.6} \approx 1.18 \text{ m}$$
Alternatively, $h = e^2 H_0$ where $H_0 = \dfrac{8^2}{2g} \approx 3.27$ m was the original drop height: $h = (0.6)^2 \times 3.27 \approx 1.18$ m ✓.
Problem 3 — Elastic String Energy
An elastic string of natural length 2 m and modulus of elasticity $\lambda = 60$ N has one end fixed to a ceiling. A particle of mass 3 kg is attached to the other end. Find the extension at equilibrium and the total energy stored in the system (EPE + GPE change from natural length position).
Show Solution
At equilibrium, tension = weight: $\dfrac{60x}{2} = 3 \times 9.8 \implies 30x = 29.4 \implies x = 0.98$ m.
EPE at equilibrium: $\dfrac{60 \times 0.98^2}{2 \times 2} = \dfrac{60 \times 0.9604}{4} = 14.41$ J.
Loss in GPE from natural length position to equilibrium: $mgh = 3 \times 9.8 \times 0.98 = 28.81$ J.
At equilibrium, KE = 0. Energy balance: $\text{GPE lost} = \text{EPE gained} + \text{KE}$... but wait, we want the equilibrium position, not the lowest point. The equilibrium is a static position where the particle would hang at rest if placed gently. The energy analysis for this static position gives EPE $= 14.41$ J.
Problem 4 — Circular Motion
A car of mass 1200 kg travels over a hill that is approximately a circular arc of radius 80 m. Find the maximum speed for the car to remain in contact with the road at the top of the hill.
Show Solution
At the top of the hill, applying Newton's 2nd law toward the centre (downward): $$mg - N = \frac{mv^2}{r}$$ The car leaves the road when $N = 0$: $$mg = \frac{mv^2_{\max}}{r} \implies v_{\max}^2 = gr = 9.8 \times 80 = 784$$ $$v_{\max} = 28 \text{ m s}^{-1} \approx 101 \text{ km h}^{-1}$$
Problem 5 — Vertical Circular Motion
A particle of mass 0.2 kg is on the inside of a smooth circular loop of radius 0.8 m in a vertical plane. Find the minimum speed at the bottom of the loop for the particle to maintain contact throughout.
Show Solution
Minimum condition at top: $N = 0$, so $mg = \dfrac{mv_{\text{top}}^2}{r}$, giving $v_{\text{top}}^2 = gr = 9.8 \times 0.8 = 7.84$ m$^2$ s$^{-2}$.
Energy conservation (height from bottom to top $= 2r = 1.6$ m): $$\frac{1}{2}mv_{\text{bot}}^2 = \frac{1}{2}mv_{\text{top}}^2 + mg(2r)$$ $$v_{\text{bot}}^2 = 7.84 + 2(9.8)(1.6) = 7.84 + 31.36 = 39.2$$ $$v_{\text{bot}} = \sqrt{39.2} \approx 6.26 \text{ m s}^{-1}$$
Problem 6 — SHM
A particle undergoes SHM with amplitude 4 cm and frequency 2 Hz. At $t = 0$ the particle is at $x = 4$ cm. Find (a) the equation of motion $x(t)$, (b) the speed when $x = 2$ cm, (c) the first time the particle reaches $x = -4$ cm.
Show Solution
(a) $\omega = 2\pi f = 4\pi$ rad s$^{-1}$. Starting at $x = A = 4$ cm at $t = 0$: $x(t) = 4\cos(4\pi t)$ cm.
(b) $v^2 = \omega^2(A^2 - x^2) = (4\pi)^2(16 - 4) = 16\pi^2 \times 12 = 192\pi^2$ cm$^2$ s$^{-2}$. $v = 8\pi\sqrt{3} \approx 43.5$ cm s$^{-1}$.
(c) $x = -4$ cm $= -A$ (amplitude on the other side). This is the first time the particle reaches the other extreme. $4\cos(4\pi t) = -4 \implies \cos(4\pi t) = -1 \implies 4\pi t = \pi \implies t = \frac{1}{4}$ s $= T/2$. This makes sense: it takes half a period to go from one extreme to the other.
Problem 7 — Simple Pendulum
A simple pendulum of length 1.5 m makes small oscillations. (a) Find the period. (b) How many complete oscillations does it make in 1 minute? (c) If the string is shortened to 0.6 m, by what percentage does the period change?
Show Solution
(a) $T = 2\pi\sqrt{\dfrac{l}{g}} = 2\pi\sqrt{\dfrac{1.5}{9.8}} = 2\pi\sqrt{0.1531} \approx 2\pi(0.3912) \approx 2.457$ s.
(b) Oscillations in 60 s $= \dfrac{60}{2.457} \approx 24.4$, so approximately $\mathbf{24}$ complete oscillations.
(c) $T' = 2\pi\sqrt{\dfrac{0.6}{9.8}} \approx 2\pi(0.2474) \approx 1.554$ s. Percentage change $= \dfrac{1.554 - 2.457}{2.457} \times 100 \approx -36.7\%$. The period decreases by approximately 36.7%.
Note: $T \propto \sqrt{l}$, so $\dfrac{T'}{T} = \sqrt{\dfrac{0.6}{1.5}} = \sqrt{0.4} \approx 0.632$, confirming a decrease of about 36.8%.
Problem 8 — Dimensional Analysis
The drag force $F$ on a sphere of radius $r$ moving through a fluid at speed $v$ depends on $r$, $v$, and the fluid's dynamic viscosity $\eta$ (dimensions: ML$^{-1}$T$^{-1}$). Show by dimensional analysis that $F = k\eta rv$ for some dimensionless constant $k$. (Stokes' Law gives $k = 6\pi$.)
Show Solution
Assume $F = k\,r^a\,v^b\,\eta^c$. Dimensions: $$[\text{MLT}^{-2}] = [\text{L}]^a [\text{LT}^{-1}]^b [\text{ML}^{-1}\text{T}^{-1}]^c = \text{M}^c\,\text{L}^{a+b-c}\,\text{T}^{-b-c}$$
Equating exponents: M: $1 = c$ T: $-2 = -b - c \implies b = 1$ L: $1 = a + b - c = a + 1 - 1 = a \implies a = 1$
So $F = k\,r^1\,v^1\,\eta^1 = k\eta r v$, confirming the Stokes' law form.