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 plane starts from rest and accelerates uniformly at 3.75 m/s² along a runway. Find its speed v after 1800 m.

v₀ = 0a = 3.75 m/s²s = 1800 m

v = m/s

2.KinematicsHard

Ball A is dropped from rest from a height of 15 m. At the same instant ball B is thrown straight up from the ground directly below at 27 m/s. Ignore air resistance and use g = 9.81 m/s². When do the balls meet? Find t.

H = 15 mv_0B = 27 m/sg = 9.81 m/s²

t = s

3.Newton’s lawsMedium

A box slides across a level floor with an initial speed of 7 m/s and comes to rest after 25 m. Use g = 9.81 m/s². Find the coefficient of kinetic friction μ.

v₀ = 7 m/sd = 25 mg = 9.81 m/s²

μ =

4.Newton’s lawsHard

Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 75 N pulls the front 6 kg block; the 2 kg block is towed behind. The coefficient of kinetic friction is 0.15 for both blocks. Use g = 9.81 m/s². Find the acceleration a of the blocks.

m₁ = 2 kgm₂ = 6 kgF = 75 Nμ = 0.15g = 9.81 m/s²

a = m/s²

5.Projectile motionMedium

A ball is kicked from level ground at 40 m/s and 30° above the horizontal towards a wall 99 m away. Ignore air resistance and use g = 9.81 m/s². At what height y does the ball pass the wall?

v₀ = 40 m/sθ = 30°x = 99 mg = 9.81 m/s²

y = m

6.Projectile motionHard

A ball is launched at 55° above horizontal from the top of a 10 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.

v₀ = 29 m/sθ = 55°h = 10 mg = 9.81 m/s²

t = s

7.Circular motionMedium

A car takes a flat curve of radius 75 m. The coefficient of static friction between tyres and road is 0.4. Use g = 9.81 m/s². Find the highest speed v_max at which the car can take the curve without skidding.

r = 75 mμ_s = 0.4g = 9.81 m/s²

v_max = m/s

8.Circular motionHard

A 1.7 kg ball on a string 1 m long is swung in a vertical circle. At the lowest point it moves at 4.5 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.

m = 1.7 kgr = 1 mv = 4.5 m/sg = 9.81 m/s²

T = N

9.Work & energyMedium

A 10 kg block slides 16 m down a slope at 20° to the horizontal, with a coefficient of kinetic friction of 0.15. Find the work W_f done against friction.

m = 10 kgθ = 20°μ = 0.15d = 16 mg = 9.81 m/s²

W_f = J

10.Work & energyHard

A 2 kg block starts from rest at the bottom of a 35° slope and is pulled up the slope by a constant force of 25 N parallel to the slope. The coefficient of kinetic friction is 0.25. Find the speed v of the block after it has moved 2 m up the slope.

m = 2 kgF = 25 Nθ = 35°μ = 0.25d = 2 mg = 9.81 m/s²

v = m/s

11.Momentum & collisionsMedium

A 5 g bullet is fired horizontally into a 3.5 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 25 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.

m = 5 gM = 3.5 kgh = 25 cmg = 9.81 m/s²

v = m/s

12.Momentum & collisionsHard

A 5 g bullet is fired horizontally into a 1 kg block hanging from light strings 1.9 m long, and stays in it. The strings swing out to a largest angle of 15° from the vertical. Use g = 9.81 m/s². Find the speed v of the bullet before impact.

m = 5 gM = 1 kgL = 1.9 mθ = 15°g = 9.81 m/s²

v = m/s

13.Springs & pendulumsMedium

A 2.7 kg mass on a spring with k = 150 N/m is pulled 14 cm in the positive direction and released from rest at t = 0. Find its displacement x from equilibrium at t = 1.2 s (with sign).

m = 2.7 kgk = 150 N/mA = 14 cmt = 1.2 s

x = cm

14.Springs & pendulumsHard

An 8 g bullet moving at 250 m/s hits and stays in a 2.5 kg block resting on a frictionless surface. The block is attached to a horizontal spring with k = 1700 N/m. Find the amplitude A of the resulting oscillation in cm.

m = 8 gv = 250 m/sM = 2.5 kgk = 1700 N/m

A = cm

15.Waves, sound & opticsMedium

A pipe open at both ends is 0.6 m long. The speed of sound is 343 m/s. Find the frequency f₄ of its harmonic number 4.

v = 343 m/sn = 4L = 0.6 m

f₄ = Hz

16.Waves, sound & opticsHard

Light of wavelength 633 nm passes through a single slit 0.15 mm wide. Find the width w of the central bright maximum on a screen 1.5 m away (between the first dark fringes on either side).

λ = 633 nma = 0.15 mmL = 1.5 m

w = mm

3Solutions

1.Kinematics

Answer: v = 116.2 m/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.

Speed after the distance

v = √(2·a·s) = √(2·3.75·1800) = 116.2 m/s

Answer

v = 116.2 m/s

(418.3 km/h)

2.Kinematics

Answer: t = 0.5556 s

Meeting time

Up is positive, the ground is y = 0.

15 − ½g·t² = 27·t − ½g·t² → t = 15/27 = 0.5556 s

Answer

t = 0.5556 s

3.Newton’s laws

Answer: μ = 0.0999

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.

Deceleration

a = v₀²/(2d) = 7²/(2·25) = 0.98 m/s²

Friction coefficient

μ·m·g = m·a → μ = a/g = 0.98/9.81 = 0.0999

Answer

μ = 0.0999

4.Newton’s laws

Answer: a = 7.904 m/s²

Whole system

a = (F − μ·(m₁ + m₂)·g)/(m₁ + m₂) = (75 − 11.77)/8 = 7.904 m/s²

Answer

a = 7.904 m/s²

5.Projectile motion

Answer: y = 17.1 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.

Time to the wall

v_x = 40·cos 30° = 34.64 m/s

t = x/v_x = 99/34.64 = 2.858 s

Height at that time

y = 20·2.858 − ½·9.81·2.858² = 17.1 m

Answer

y = 17.1 m

The ball is already on its way down at the wall.

6.Projectile motion

Answer: t = 5.233 s

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

Horizontal velocity stays constant; vertical motion has a constant acceleration g = 9.81 m/s² downward.

v_x = v₀·cos θ = 29·cos 55° = 16.63 m/s

v_y = v₀·sin θ = 29·sin 55° = 23.76 m/s

Time in the air

Solve y(t) = h + v_y·t − ½gt² = 0 for the positive root.

0 = 10 + 23.76·t − 4.905·t²

t = (v_y + √(v_y² + 2gh))/g = (23.76 + √(564.3 + 196.2))/9.81 = 5.233 s

Answer

t = 5.233 s

7.Circular motion

Answer: v_max = 17.16 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.

Friction provides the centripetal force

m·v²/r = μ·m·g

v = √(μ·g·r) = √(0.4·9.81·75) = 17.16 m/s

Answer

v_max = 17.16 m/s (61.76 km/h)

8.Circular motion

Answer: T = 51.1 N

Forces at the lowest point

Towards the centre (up) is positive.

T − m·g = m·v²/r

T = 1.7·9.81 + 1.7·4.5²/1 = 16.68 + 34.42 = 51.1 N

Answer

T = 51.1 N

9.Work & energy

Answer: W_f = 221.2 J

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).

Friction force

N = mg·cos θ = 98.1·0.9397 = 92.18 N

F = μN = 0.15·92.18 = 13.83 N

Work against friction

W_f = F·d = 13.83·16 = 221.2 J

Answer

W_f = 221.2 J

10.Work & energy

Answer: v = 4.411 m/s

Forces along the slope

m·g = 2·9.81 = 19.62 N

mg·sin θ = 11.25 N

μ·mg·cos θ = 4.018 N

Work–energy

½mv² = (25 − 15.27)·2 = 19.46 J

v = √(2·19.46/2) = 4.411 m/s

Answer

v = 4.411 m/s

11.Momentum & collisions

Answer: v = 1553 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) = √(2·9.81·0.25) = 2.215 m/s

Collision

m·v = (m + M)·V → v = (0.005 + 3.5)·2.215/0.005 = 1553 m/s

Answer

v = 1553 m/s

12.Momentum & collisions

Answer: v = 226.5 m/s

Rise of the block

h = L·(1 − cos θ) = 1.9·(1 − cos 15°) = 0.06474 m

Speed after impact

V = √(2·9.81·0.06474) = 1.127 m/s

Collision

v = (m + M)·V/m = 1.005·1.127/0.005 = 226.5 m/s

Answer

v = 226.5 m/s

13.Springs & pendulums

Answer: x = −12.41 cm

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.

Angular frequency

ω = √(150/2.7) = 7.454 rad/s

ω·t = 7.454·1.2 = 8.944 rad

Displacement

x = A·cos(ω·t) = 14·cos(8.944) = −12.41 cm

Answer

x = −12.41 cm

14.Springs & pendulums

Answer: A = 3.063 cm

Speed after impact

V = m·v/(m + M) = 0.008·250/2.508 = 0.7974 m/s

Angular frequency

ω = √(k/(m + M)) = √(1700/2.508) = 26.04 rad/s

Amplitude

A = V/ω = 0.7974/26.04 = 0.03063 m = 3.063 cm

Answer

A = 3.063 cm

15.Waves, sound & optics

Answer: f₄ = 1143 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 = 2L/n = 2·0.6/4 = 0.3 m

Frequency

f = v/λ = 343/0.3 = 1143 Hz

Answer

f₄ = 1143 Hz

16.Waves, sound & optics

Answer: w = 12.66 mm

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, …

First dark fringe

y₁ = λL/a = 633 × 10⁻⁹·1.5/(0.15 × 10⁻³) = 6.33 mm

Width of the central maximum

w = 2y₁ = 12.66 mm

Answer

w = 12.66 mm

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