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 72 km/h brakes with a constant deceleration of 6 m/s² until it stops. Find the braking distance s.

v₀ = 72 km/ha = −6 m/s²v = 0

s = m

2.KinematicsHard

A ball is thrown straight up at 18 m/s from the edge of a roof 10 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height h_max above the ground.

v₀ = 18 m/sh₀ = 10 mg = 9.81 m/s²

h_max = m

3.Newton’s lawsMedium

A 16 kg block slides down a frictionless slope at 35° to the horizontal. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.

m = 16 kgθ = 35°μ = 0g = 9.81 m/s²

a = m/s²

4.Newton’s lawsHard

A 5.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 2.5 kg mass. The coefficient of kinetic friction between block and table is 0.2. Use g = 9.81 m/s². Find the acceleration a of the system.

m₁ = 5.5 kgm₂ = 2.5 kgμ = 0.2g = 9.81 m/s²

a = m/s²

5.Projectile motionMedium

A ball is launched at 20° above horizontal from level ground. Ignore air resistance. Find the time of flight t until the ball hits the ground.

v₀ = 32 m/sθ = 20°g = 9.81 m/s²

t = s

6.Projectile motionHard

A ball is thrown from ground level at 38 m/s and 40° above the horizontal. It lands on the flat roof of a building 7 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 = 38 m/sθ = 40°h_roof = 7 mg = 9.81 m/s²

x = m

7.Circular motionMedium

A car takes a flat curve of radius 110 m. The coefficient of static friction between tyres and road is 0.6. Use g = 9.81 m/s². Find the highest speed v_max at which the car can take the curve without skidding.

r = 110 mμ_s = 0.6g = 9.81 m/s²

v_max = m/s

8.Circular motionHard

A curve of radius 260 m is banked at 12°, and the coefficient of static friction between tyres and road is 0.3. Use g = 9.81 m/s². Find the highest speed v_max at which a car can take the curve without sliding up and out.

r = 260 mθ = 12°μ_s = 0.3g = 9.81 m/s²

v_max = m/s

9.Work & energyMedium

A toy launcher has a vertical spring of stiffness 800 N/m. A 0.1 kg ball is placed on it and the spring is compressed by 8 cm, then released. Ignoring air resistance and the spring’s mass, find the height h the ball rises, measured from its position when the spring was compressed.

m = 0.1 kgk = 800 N/mx = 8 cmg = 9.81 m/s²

h = m

10.Work & energyHard

A 1 kg block is released from rest at height 1.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 1 m long (μ = 0.15) and then runs onto a spring (k = 500 N/m) on a smooth surface. Find the speed v of the block when it reaches the spring.

m = 1 kgh = 1.5 mL = 1 mμ = 0.15k = 500 N/mg = 9.81 m/s²

v = m/s

11.Momentum & collisionsMedium

A 17 kg projectile moving horizontally at 38 m/s explodes into two fragments that move along the same line. The 8 kg fragment continues forwards at 77 m/s. Find the velocity v′₂ of the 9 kg fragment (positive in the original direction).

M = 17 kgu = 38 m/sm₁ = 8 kgv′₁ = 77 m/sm₂ = 9 kg

v′₂ = m/s

12.Momentum & collisionsHard

A 1100 kg car runs into a parked 1100 kg car. The wrecks lock together and skid 10.5 m before stopping. The coefficient of kinetic friction between tyres and road is 0.7. Use g = 9.81 m/s². Find the speed v₁ of the first car just before the collision.

m₁ = 1100 kgm₂ = 1100 kgd = 10.5 mμ = 0.7g = 9.81 m/s²

v₁ = m/s

13.Springs & pendulumsMedium

How long must a simple pendulum be to have a period of 0.9 s? Use g = 9.81 m/s². Find L.

T = 0.9 sg = 9.81 m/s²

L = m

14.Springs & pendulumsHard

A 2.8 kg mass on a spring with k = 180 N/m is pulled 3 cm from equilibrium and released. Find the total energy E of the oscillation.

m = 2.8 kgk = 180 N/mA = 3 cm

E = J

15.Waves, sound & opticsMedium

A source of 440 Hz moving at 10 m/s and an observer moving at 5 m/s approach each other. The speed of sound is 343 m/s. Find the frequency f′ that the observer hears.

f = 440 Hzv = 343 m/sv_s = 10 m/sv_o = 5 m/s

f′ = Hz

16.Waves, sound & opticsHard

A diffraction grating has 100 lines per millimetre. Light of wavelength 589 nm falls on it at right angles. What is the highest order m_max of bright maximum that can be seen?

N = 100 mm⁻¹λ = 589 nm

m_max =

3Solutions

1.Kinematics

Answer: s = 33.33 m

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 = 72/3.6 = 20 m/s

Braking distance

0 = v₀² − 2·a·s → s = v₀²/(2a) = 20²/(2·6) = 33.33 m

Answer

s = 33.33 m

2.Kinematics

Answer: h_max = 26.51 m

Rise above the roof

0 = v₀² − 2·g·Δh → Δh = 18²/(2·9.81) = 16.51 m

Height above the ground

h_max = 10 + 16.51 = 26.51 m

Answer

h_max = 26.51 m

3.Newton’s laws

Answer: a = 5.627 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·cos 35°) = 9.81·(0.5736 − 0) = 5.627 m/s²

Answer

a = 5.627 m/s²

4.Newton’s laws

Answer: a = 1.717 m/s²

Acceleration of the system

a = (m₂·g − μ·m₁·g)/(m₁ + m₂) = (24.53 − 10.79)/8 = 1.717 m/s²

Answer

a = 1.717 m/s²

5.Projectile motion

Answer: t = 2.231 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 = 32·cos 20° = 30.07 m/s

v_y = v₀·sin θ = 32·sin 20° = 10.94 m/s

Time in the air

Starting and landing at the same height: t = 2v_y/g.

t = 2·10.94/9.81 = 2.231 s

Answer

t = 2.231 s

6.Projectile motion

Answer: x = 136.1 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.

Split the launch velocity

v_x = 29.11 m/s, v_y = 24.43 m/s

Time to land on the roof

Larger root: the ball is descending.

7 = 24.43·t − 4.905·t²

t = (24.43 + √(24.43² − 2·9.81·7))/9.81 = 4.675 s

Horizontal distance

x = v_x·t = 29.11·4.675 = 136.1 m

Answer

x = 136.1 m

7.Circular motion

Answer: v_max = 25.45 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) = √(0.6·9.81·110) = 25.45 m/s

Answer

v_max = 25.45 m/s (91.6 km/h)

8.Circular motion

Answer: v_max = 37.37 m/s

Force balances

Friction points down the slope.

N·(sin θ + μ·cos θ) = m·v²/r

N·(cos θ − μ·sin θ) = m·g

Speed

v = √(260·9.81·0.5014/0.9158) = 37.37 m/s

Answer

v_max = 37.37 m/s

(134.5 km/h)

9.Work & energy

Answer: h = 2.61 m

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).

Spring energy

x = 8 cm = 0.08 m

½kx² = ½·800·0.08² = 2.56 J

Height

m·g = 0.1·9.81 = 0.981 N

h = ½kx²/(mg) = 2.56/0.981 = 2.61 m

Answer

h = 2.61 m

10.Work & energy

Answer: v = 5.147 m/s

Energy left after the rough strip

The ramp is smooth, so only the strip removes energy.

mgh − μmgL = 9.81·(1.5 − 0.15·1) = 13.24 J

Speed at the spring

½mv² = mg(h − μL) → v = √(2g(h − μL)) = √(2·9.81·1.35) = 5.147 m/s

Answer

v = 5.147 m/s

11.Momentum & collisions

Answer: v′₂ = 3.333 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.

Momentum before

p = M·u = 17·38 = 646 kg·m/s

Momentum after

646 = 8·77 + 9·v′₂

v′₂ = (646 − 616)/9 = 3.333 m/s

Answer

v′₂ = 3.333 m/s

12.Momentum & collisions

Answer: v₁ = 24.02 m/s

Speed after the collision

V = √(2·μ·g·d) = √(2·0.7·9.81·10.5) = 12.01 m/s

Momentum in the collision

m₁·v₁ = (m₁ + m₂)·V → v₁ = 2200·12.01/1100 = 24.02 m/s

Answer

v₁ = 24.02 m/s

(86.46 km/h)

13.Springs & pendulums

Answer: L = 0.2013 m

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.

Solve for the length

L = g·(T/2π)² = 9.81·(0.9/2π)² = 0.2013 m

Answer

L = 0.2013 m

14.Springs & pendulums

Answer: E = 0.081 J

Angular frequency

ω = √(180/2.8) = 8.018 rad/s

Energy

E = ½·k·A² = ½·180·0.03² = 0.081 J

Answer

E = 0.081 J

15.Waves, sound & optics

Answer: f′ = 459.8 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|.

Doppler formula

f′ = f·(v + v_o)/(v − v_s) = 440·(343 + 5)/(343 − 10)

f′ = 459.8 Hz

Answer

f′ = 459.8 Hz

16.Waves, sound & optics

Answer: m_max = 16

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 = 1/100 mm = 10 µm

Highest order

m ≤ d/λ = 10 µm/0.589 µm = 16.98 → m_max = 16

Answer

m_max = 16

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