Physics 1: Mechanics: cram pack

Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.

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1Formulas and methods

Kinematics

Velocity: v = v₀ + a·t
Displacement: s = v₀·t + ½·a·t²
Without time: v² = v₀² + 2·a·s
Units: 1 m/s = 3.6 km/h

Method

List what you know: v₀, v, a, t and s, with units.
Pick the equation that contains your three knowns and the unknown.
Choose one positive direction and give opposing quantities a minus sign.
Solve and check that the answer is physically sensible.

Newton’s laws

Second law: ΣF = m·a
Weight: W = m·g
Kinetic friction: F_f = μ·N
On a slope: along: m·g·sin θ, normal: m·g·cos θ

Method

Draw a free-body diagram with every force on the body.
Choose a positive direction along the motion.
Split forces into components along and perpendicular to the motion (on a slope: m·g·sin θ along, m·g·cos θ into the slope).
Apply ΣF = m·a along the motion; friction is μ·N against the motion.

Projectile motion

Components: v_x = v₀cos θ, v_y = v₀sin θ
Position: x = v_xt, y = h + v_yt − ½gt²
Maximum height: H = h + v_y²/(2g)
Range (level ground): R = v₀²·sin 2θ/g

Method

Split the launch velocity into components: v_x = v₀·cos θ and v_y = v₀·sin θ.
The horizontal motion has constant speed; the vertical motion has acceleration −g.
Find the time of flight from the vertical motion.
Use that time in the horizontal motion to find the range.

Circular motion

Centripetal acceleration: a = v²/r = ω²·r
Centripetal force: F = m·v²/r
Angular speed: ω = v/r = 2π/T
Flat curve, friction: v_max = √(μ·g·r)

Work & energy

Kinetic energy: E_k = ½mv²
Potential energy: E_p = mgh
Energy balance: E_k1 + E_p1 + W = E_k2 + E_p2 + losses
Work of friction: W = μ·N·d
Power: P = W/t

Momentum & collisions

Momentum: p = m·v
Conservation: Σm·v (before) = Σm·v (after)
Perfectly inelastic: v′ = (m₁v₁ + m₂v₂)/(m₁ + m₂)
Impulse: J = F·Δt = m·Δv

Method

Choose a positive direction and give each velocity its sign.
Write the total momentum before: Σm·v.
For a perfectly inelastic collision the bodies move together after; for an elastic one kinetic energy is also conserved.
Set momentum before equal to momentum after and solve.

Springs & pendulums

Mass on a spring: T = 2π·√(m/k)
Pendulum: T = 2π·√(L/g)
Angular frequency: ω = √(k/m) = 2π/T
Maximum speed and acceleration: v_max = A·ω, a_max = A·ω²
Energy: E = ½·k·A²

Waves, sound & optics

Wave speed: v = fλ, T = 1/f
String: v = √(T/μ), μ = m/L
Sound level: β = 10·log₁₀(I/I₀), I₀ = 10⁻¹² W/m²
Point source: I = P/(4πr²)
Standing waves: string or open pipe f_n = n·v/(2L); closed pipe f_n = n·v/(4L), n odd
Doppler effect: f′ = f·(v ± v_o)/(v ∓ v_s)
Refractive index: n = c/v, λ = λ₀/n
Snell’s law: n₁sin θ₁ = n₂sin θ₂
Critical angle: sin θ_c = n₂/n₁
Optical fibre: NA = sin θ_a = √(n₁² − n₂²)
Lenses and mirrors: 1/f = 1/d_o + 1/d_i, m = −d_i/d_o, mirror f = R/2
Double slit: d·sin θ = mλ, y_m = mλL/d
Grating: d = 1/N, d·sin θ = mλ, m_max ≤ d/λ
Single slit: a·sin θ = mλ (dark), w = 2λL/a

Method

Identify the kind of problem: wave speed, sound level, standing wave, Doppler, refraction, lens or mirror, or interference.
Write the relation first: v = fλ, β = 10·log(I/I₀), fₙ = n·v/(2L) or n·v/(4L), f′ = f(v ± v_o)/(v ∓ v_s), n₁sin θ₁ = n₂sin θ₂, 1/f = 1/d₀ + 1/dᵢ, d·sin θ = mλ.
Convert units: nm, mm and cm to m, and check whether angles are measured from the normal.
Check signs and plausibility: a virtual image has dᵢ < 0, an approaching source sounds higher, a closed pipe has only odd harmonics.

2Practice problems

1.KinematicsMedium

A car driving at 80 km/h brakes with a constant deceleration of 4.5 m/s² until it stops. How long does braking take? Find t.

v₀ = 80 km/ha = −4.5 m/s²v = 0

t = s

2.KinematicsHard

A tram starts from rest and accelerates at 1.5 m/s² for 11 s, then runs at constant speed for 20 s, and finally brakes at 4.5 m/s² until it stops. Find the total distance s between the two stops.

a₁ = 1.5 m/s²t₁ = 11 st₂ = 20 sa₃ = −4.5 m/s²

s = m

3.Newton’s lawsMedium

A 3 kg block slides down a slope at 35° to the horizontal. The coefficient of kinetic friction is 0.25. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.

m = 3 kgθ = 35°μ = 0.25g = 9.81 m/s²

a = m/s²

4.Newton’s lawsHard

An 8.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 5.5 kg mass. The coefficient of kinetic friction between block and table is 0.1. Use g = 9.81 m/s². Find the acceleration a of the system.

m₁ = 8.5 kgm₂ = 5.5 kgμ = 0.1g = 9.81 m/s²

a = m/s²

5.Projectile motionMedium

A ball is launched from level ground at 30 m/s and must land 40 m away. Ignore air resistance and use g = 9.81 m/s². Find the smaller of the two possible launch angles θ.

v₀ = 30 m/sR = 40 mg = 9.81 m/s²

θ = °

6.Projectile motionHard

A ball is thrown from ground level at 34 m/s and 50° above the horizontal. It lands on the flat roof of a building 2 m high, on its way down. Ignore air resistance and use g = 9.81 m/s². Find the horizontal distance x from the launch point to where it lands.

v₀ = 34 m/sθ = 50°h_roof = 2 mg = 9.81 m/s²

x = m

7.Circular motionMedium

A curve of radius 310 m is banked at 5°. At what speed v can a car take it with no friction needed at all? Use g = 9.81 m/s².

r = 310 mθ = 5°g = 9.81 m/s²

v = m/s

8.Circular motionHard

A curve of radius 130 m is banked at 20°, and the coefficient of static friction between tyres and road is 0.3. Use g = 9.81 m/s². Find the lowest speed v_min at which a car can take the curve without sliding down the bank.

r = 130 mθ = 20°μ_s = 0.3g = 9.81 m/s²

v_min = m/s

9.Work & energyMedium

A block slides from rest down a slope at 25° to the horizontal. The coefficient of kinetic friction is 0.35. Find its speed v after it has slid 14 m along the slope.

θ = 25°μ = 0.35d = 14 mg = 9.81 m/s²

v = m/s

10.Work & energyHard

A spring of stiffness 3000 N/m at the bottom of a 35° slope is compressed by 12 cm against a 2 kg block and released. The coefficient of kinetic friction between block and slope is 0.15. How far d up the slope does the block travel from the release point before it stops? (Ignore the short distance over which the spring pushes, i.e. take friction and gravity over the whole distance d.)

m = 2 kgk = 3000 N/mx = 12 cmθ = 35°μ = 0.15g = 9.81 m/s²

d = m

11.Momentum & collisionsMedium

A 20 g bullet is fired horizontally into a 4 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 21 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.

m = 20 gM = 4 kgh = 21 cmg = 9.81 m/s²

v = m/s

12.Momentum & collisionsHard

A 0.1 kg ball hits a wall at 10 m/s and bounces straight back at 10 m/s. The contact lasts 10 ms. Find the average force F_avg of the wall on the ball.

m = 0.1 kgv_in = 10 m/sv_out = 10 m/sΔt = 10 ms

F_avg = N

13.Springs & pendulumsMedium

A simple pendulum has a length of 0.8 m. Use g = 9.81 m/s². Find its period T for small swings.

L = 0.8 mg = 9.81 m/s²

T = s

14.Springs & pendulumsHard

A 2.5 kg mass on a spring with k = 380 N/m is pulled 20 cm from equilibrium and released. Find its maximum speed v_max.

m = 2.5 kgk = 380 N/mA = 20 cm

v_max = m/s

15.Waves, sound & opticsMedium

A pipe closed at one end is 0.75 m long. The speed of sound is 343 m/s. Find the frequency f₁ of its fundamental (first harmonic).

v = 343 m/sn = 1L = 0.75 m

f₁ = Hz

16.Waves, sound & opticsHard

A diffraction grating has 800 lines per millimetre. Light of wavelength 550 nm falls on it at right angles. What is the highest order m_max of bright maximum that can be seen?

N = 800 mm⁻¹λ = 550 nm

m_max =

3Solutions

1.Kinematics

Answer: t = 4.938 s

Motion with constant acceleration: v = v₀ + a·t, s = v₀·t + ½·a·t² and v² = v₀² + 2·a·s. Pick the equation that contains the three things you know and the one you want, and keep the signs consistent with one positive direction.

Units

v₀ = 80/3.6 = 22.22 m/s

Braking time

0 = v₀ − a·t → t = v₀/a = 22.22/4.5 = 4.938 s

Answer

t = 4.938 s

2.Kinematics

Answer: s = 451 m

Top speed

v = a₁·t₁ = 1.5·11 = 16.5 m/s

Distance of each phase

s₁ = ½·1.5·11² = 90.75 m

s₂ = 16.5·20 = 330 m

s₃ = 16.5²/(2·4.5) = 30.25 m

Total distance

s = 90.75 + 330 + 30.25 = 451 m

Answer

s = 451 m

3.Newton’s laws

Answer: a = 3.618 m/s²

Newton’s second law: the net force on a body equals its mass times its acceleration, ΣF = m·a. Draw a free-body diagram, choose a positive direction along the motion, and add every force in that direction: applied forces, friction μ·N against the motion, and the components of gravity.

Forces along the slope

Down the slope is positive.

m·a = m·g·sin θ − μ·m·g·cos θ

Acceleration

a = 9.81·(sin 35° − 0.25·cos 35°) = 9.81·(0.5736 − 0.2048) = 3.618 m/s²

Answer

a = 3.618 m/s²

4.Newton’s laws

Answer: a = 3.258 m/s²

Acceleration of the system

a = (m₂·g − μ·m₁·g)/(m₁ + m₂) = (53.96 − 8.339)/14 = 3.258 m/s²

Answer

a = 3.258 m/s²

5.Projectile motion

Answer: θ = 12.92 °

Projectile motion: with no air resistance, x = v_xt and y = h + v_yt − ½gt². The two directions are independent except that they share the same time.

Range formula

R = v₀²·sin 2θ/g → sin 2θ = 9.81·40/30² = 0.436

Angle

2θ = 25.85° → θ = 12.92°

Answer

θ = 12.92°

The other solution is 90° − θ = 77.08°.

6.Projectile motion

Answer: x = 114.3 m

Projectile motion: with no air resistance, x = v_xt and y = h + v_yt − ½gt². The two directions are independent except that they share the same time.

Split the launch velocity

v_x = 21.85 m/s, v_y = 26.05 m/s

Time to land on the roof

Larger root: the ball is descending.

2 = 26.05·t − 4.905·t²

t = (26.05 + √(26.05² − 2·9.81·2))/9.81 = 5.232 s

Horizontal distance

x = v_x·t = 21.85·5.232 = 114.3 m

Answer

x = 114.3 m

7.Circular motion

Answer: v = 16.31 m/s

Uniform circular motion: an object moving at speed v on a circle of radius r accelerates towards the centre with a = v²/r, so it needs a net inward force F = m·v²/r. That force is supplied by something real: friction, tension, gravity or a normal force.

Force balance

N·cos θ = m·g, N·sin θ = m·v²/r → tan θ = v²/(r·g)

Speed

v = √(r·g·tan θ) = √(310·9.81·tan 5°) = 16.31 m/s

Answer

v = 16.31 m/s

(58.72 km/h)

8.Circular motion

Answer: v_min = 8.576 m/s

Force balances

Friction points up the slope.

N·(sin θ − μ·cos θ) = m·v²/r

N·(cos θ + μ·sin θ) = m·g

Speed

v = √(130·9.81·0.06011/1.042) = 8.576 m/s

Answer

v_min = 8.576 m/s

(30.87 km/h)

9.Work & energy

Answer: v = 5.381 m/s

Work–energy principle: the change in kinetic energy equals the work done by all forces. Writing gravity and springs as potential energy (mgh and ½kx²) turns this into a balance: initial energy = final energy + energy lost to friction (μN·d).

Energy balance along the slope

The block drops a height d·sin θ while friction removes μmg·cos θ·d. The mass cancels.

½v² = g·d·(sin θ − μ·cos θ)

v = √(2·9.81·14·(0.4226 − 0.35·0.9063)) = 5.381 m/s

Answer

v = 5.381 m/s

10.Work & energy

Answer: d = 1.581 m

Spring energy

x = 12 cm = 0.12 m

½kx² = ½·3000·0.12² = 21.6 J

Energy balance along the slope

m·g = 2·9.81 = 19.62 N

½kx² = mg·d·(sin θ + μ·cos θ)

d = 21.6/(19.62·(0.5736 + 0.15·0.8192)) = 1.581 m

Answer

d = 1.581 m

11.Momentum & collisions

Answer: v = 408 m/s

Conservation of momentum: when no external force acts during a collision, the total momentum Σm·v before equals the total after. Velocities have a direction, so pick a positive direction and give velocities the other way a minus sign. In an elastic collision kinetic energy is conserved too; in a perfectly inelastic one the bodies stick together.

Swing after impact

V = √(2·g·h) = √(2·9.81·0.21) = 2.03 m/s

Collision

m·v = (m + M)·V → v = (0.02 + 4)·2.03/0.02 = 408 m/s

Answer

v = 408 m/s

12.Momentum & collisions

Answer: F_avg = 200 N

Change in momentum

J = m·Δv = 0.1·(10 + 10) = 2 N·s

Average force

F_avg = J/Δt = 2/0.01 = 200 N

Answer

F_avg = 200 N

13.Springs & pendulums

Answer: T = 1.794 s

Simple harmonic motion: a mass m on a spring of stiffness k oscillates with angular frequency ω = √(k/m) and period T = 2π/ω = 2π·√(m/k). A simple pendulum of length L has T = 2π·√(L/g). With amplitude A, the largest speed is A·ω, the largest acceleration A·ω² and the energy ½·k·A².

Pendulum period

T = 2π·√(0.8/9.81) = 1.794 s

Answer

T = 1.794 s

14.Springs & pendulums

Answer: v_max = 2.466 m/s

Angular frequency

ω = √(380/2.5) = 12.33 rad/s

Maximum speed

v_max = A·ω = 0.2·12.33 = 2.466 m/s

Answer

v_max = 2.466 m/s

15.Waves, sound & optics

Answer: f₁ = 114.3 Hz

Standing waves and the Doppler effect: a string fixed at both ends, or a pipe open at both ends, fits whole half-wavelengths: L = nλ/2, f_n = n·v/(2L), n = 1, 2, 3, … A pipe closed at one end has a node at the closed end and an antinode at the open end: L = nλ/4 with odd n only, f_n = n·v/(4L). When source or observer move along the line between them, f′ = f·(v ± v_o)/(v ∓ v_s): use the upper signs for motion towards each other. Two waves of nearly equal frequency beat at |f₁ − f₂|.

Wavelength of this mode

λ = 4L/n = 4·0.75/1 = 3 m

Frequency

f = v/λ = 343/3 = 114.3 Hz

Answer

f₁ = 114.3 Hz

16.Waves, sound & optics

Answer: m_max = 2

Interference and diffraction: light from two slits a distance d apart reaches a point on the screen with a path difference d·sin θ. Bright fringes appear where d·sin θ = mλ (m = 0, ±1, ±2, …). For a screen at distance L ≫ d, the m-th bright fringe is at y = mλL/d and the fringe spacing is Δy = λL/d. A grating with N lines per millimetre has d = 1/N and sharp maxima at d·sin θ = mλ; since sin θ ≤ 1, the highest order is the largest whole number m ≤ d/λ. A single slit of width a has dark fringes at a·sin θ = mλ, m = ±1, ±2, …

Line spacing

d = 1/800 mm = 1.25 µm

Highest order

m ≤ d/λ = 1.25 µm/0.55 µm = 2.273 → m_max = 2

Answer

m_max = 2

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