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 7 m/s² until it stops. Find the braking distance s.

v₀ = 72 km/ha = −7 m/s²v = 0

s = m

2.KinematicsHard

A tram starts from rest and accelerates at 1.5 m/s² for 6 s, then runs at constant speed for 24 s, and finally brakes at 2.5 m/s² until it stops. Find the total travel time t.

a₁ = 1.5 m/s²t₁ = 6 st₂ = 24 sa₃ = −2.5 m/s²

t = s

3.Newton’s lawsMedium

A box slides across a level floor with an initial speed of 4.5 m/s and comes to rest after 4.1 m. Use g = 9.81 m/s². Find the coefficient of kinetic friction μ.

v₀ = 4.5 m/sd = 4.1 mg = 9.81 m/s²

μ =

4.Newton’s lawsHard

A 2.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 1.5 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 = 2.5 kgm₂ = 1.5 kgμ = 0.1g = 9.81 m/s²

a = m/s²

5.Projectile motionMedium

A ball is launched at 65° above horizontal from level ground. Ignore air resistance. Find the maximum height H the ball reaches.

v₀ = 38 m/sθ = 65°g = 9.81 m/s²

H = m

6.Projectile motionHard

A ball is launched at 35° above horizontal from the top of a 60 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.

v₀ = 17 m/sθ = 35°h = 60 mg = 9.81 m/s²

x = m

7.Circular motionMedium

In a fairground rotor, riders stand against the wall of a vertical cylinder of radius 3 m. When it spins fast enough the floor drops away and friction holds them up. The coefficient of static friction is 0.4. Use g = 9.81 m/s². Find the smallest rotation rate n in revolutions per minute.

r = 3 mμ_s = 0.4g = 9.81 m/s²

n = rpm

8.Circular motionHard

A curve of radius 70 m is banked at 20°, and the coefficient of static friction between tyres and road is 0.2. 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 = 70 mθ = 20°μ_s = 0.2g = 9.81 m/s²

v_max = m/s

9.Work & energyMedium

A toy launcher has a vertical spring of stiffness 2000 N/m. A 0.1 kg ball is placed on it and the spring is compressed by 3 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 = 2000 N/mx = 3 cmg = 9.81 m/s²

h = m

10.Work & energyHard

A 0.5 kg block is released from rest at height 4.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 3 m long (μ = 0.3) and then runs onto a spring (k = 4000 N/m) on a smooth surface. Find the maximum compression x of the spring.

m = 0.5 kgh = 4.5 mL = 3 mμ = 0.3k = 4000 N/mg = 9.81 m/s²

x = cm

11.Momentum & collisionsMedium

An 8 g bullet is fired horizontally into a 1.5 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 5 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.

m = 8 gM = 1.5 kgh = 5 cmg = 9.81 m/s²

v = m/s

12.Momentum & collisionsHard

A 0.45 kg ball hits a wall at 17 m/s and bounces straight back at 20 m/s. The contact lasts 50 ms. Find the magnitude of the impulse J the wall gives the ball.

m = 0.45 kgv_in = 17 m/sv_out = 20 m/sΔt = 50 ms

J = N·s

13.Springs & pendulumsMedium

How long must a simple pendulum be to have a period of 2.8 s? Use g = 9.81 m/s². Find L.

T = 2.8 sg = 9.81 m/s²

L = m

14.Springs & pendulumsHard

A 2.4 kg mass is attached to the lower end of an unstretched vertical spring with k = 120 N/m and released from rest. Use g = 9.81 m/s². Find the greatest speed v_max of the mass.

m = 2.4 kgk = 120 N/mg = 9.81 m/s²

v_max = m/s

15.Waves, sound & opticsMedium

A small source sends sound equally in all directions. The level is 65 dB at 1 m from it. Find the level β₂ at 8 m, ignoring absorption and reflections.

β₁ = 65 dBr₁ = 1 mr₂ = 8 m

β₂ = dB

16.Waves, sound & opticsHard

Light of wavelength 589 nm passes through a single slit 0.15 mm wide. Find the angle θ₂ of dark fringe number 2 from the straight-through direction.

λ = 589 nma = 0.15 mmm = 2

θ₂ = °

3Solutions

1.Kinematics

Answer: s = 28.57 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/(2·7) = 28.57 m

Answer

s = 28.57 m

2.Kinematics

Answer: t = 33.6 s

Top speed

v = a₁·t₁ = 1.5·6 = 9 m/s

Braking time

t₃ = v/a₃ = 9/2.5 = 3.6 s

Total time

t = 6 + 24 + 3.6 = 33.6 s

Answer

t = 33.6 s

3.Newton’s laws

Answer: μ = 0.2517

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) = 4.5²/(2·4.1) = 2.47 m/s²

Friction coefficient

μ·m·g = m·a → μ = a/g = 2.47/9.81 = 0.2517

Answer

μ = 0.2517

4.Newton’s laws

Answer: a = 3.066 m/s²

Acceleration of the system

a = (m₂·g − μ·m₁·g)/(m₁ + m₂) = (14.71 − 2.453)/4 = 3.066 m/s²

Answer

a = 3.066 m/s²

5.Projectile motion

Answer: H = 60.45 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

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

v_x = v₀·cos θ = 38·cos 65° = 16.06 m/s

v_y = v₀·sin θ = 38·sin 65° = 34.44 m/s

Highest point

At the top the vertical velocity is zero: v_y² = 2gH.

H = v_y²/(2g) = 34.44²/(2·9.81) = 60.45 m

Answer

H = 60.45 m

6.Projectile motion

Answer: x = 64.47 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

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

v_x = v₀·cos θ = 17·cos 35° = 13.93 m/s

v_y = v₀·sin θ = 17·sin 35° = 9.751 m/s

Time in the air

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

0 = 60 + 9.751·t − 4.905·t²

t = (v_y + √(v_y² + 2gh))/g = (9.751 + √(95.08 + 1177))/9.81 = 4.63 s

Horizontal distance

x = v_x·t = 13.93·4.63 = 64.47 m

Answer

x = 64.47 m

7.Circular motion

Answer: n = 27.3 rpm

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.

Forces on a rider

N = m·ω²·r

μ·N = m·g → μ·ω²·r = g

Angular speed

ω = √(g/(μ·r)) = √(9.81/(0.4·3)) = 2.859 rad/s

Rotation rate

n = ω·60/(2π) = 27.3 rpm

Answer

n = 27.3 rpm

8.Circular motion

Answer: v_max = 20.44 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 = √(70·9.81·0.53/0.8713) = 20.44 m/s

Answer

v_max = 20.44 m/s

(73.57 km/h)

9.Work & energy

Answer: h = 0.9174 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 = 3 cm = 0.03 m

½kx² = ½·2000·0.03² = 0.9 J

Height

m·g = 0.1·9.81 = 0.981 N

h = ½kx²/(mg) = 0.9/0.981 = 0.9174 m

Answer

h = 0.9174 m

10.Work & energy

Answer: x = 9.396 cm

Energy left after the rough strip

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

mgh − μmgL = 4.905·(4.5 − 0.3·3) = 17.66 J

Maximum compression

½kx² = 17.66 → x = √(2·17.66/4000) = 0.09396 m = 9.396 cm

Answer

x = 9.396 cm

11.Momentum & collisions

Answer: v = 186.7 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) = √(2·9.81·0.05) = 0.9905 m/s

Collision

m·v = (m + M)·V → v = (0.008 + 1.5)·0.9905/0.008 = 186.7 m/s

Answer

v = 186.7 m/s

12.Momentum & collisions

Answer: J = 16.65 N·s

Change in momentum

J = m·Δv = 0.45·(20 + 17) = 16.65 N·s

Answer

J = 16.65 N·s

13.Springs & pendulums

Answer: L = 1.948 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·(2.8/2π)² = 1.948 m

Answer

L = 1.948 m

14.Springs & pendulums

Answer: v_max = 1.387 m/s

Equilibrium and amplitude

x₀ = m·g/k = 2.4·9.81/120 = 0.1962 m = A

Greatest speed

The speed is largest at equilibrium.

ω = √(120/2.4) = 7.071 rad/s

v_max = A·ω = 0.1962·7.071 = 1.387 m/s

Answer

v_max = 1.387 m/s

15.Waves, sound & optics

Answer: β₂ = 46.94 dB

Waves: a periodic wave moves one wavelength in one period, so v = fλ with T = 1/f. On a stretched string v = √(T/μ), with T the tension and μ = m/L the mass per metre. Sound in air travels at about 343 m/s at 20 °C, light in vacuum at c = 3.00 × 10⁸ m/s. Loudness is measured on a logarithmic scale: β = 10·log₁₀(I/I₀) dB with I₀ = 10⁻¹² W/m², so 10 times the intensity adds 10 dB. A point source spreads its power P over a sphere: I = P/(4πr²).

Intensity ratio

I₂/I₁ = (1/8)² = 0.01563

New level

β₂ = 65 + 10·log₁₀(0.01563) = 46.94 dB

Answer

β₂ = 46.94 dB

Every doubling of the distance lowers the level by about 6 dB.

16.Waves, sound & optics

Answer: θ₂ = 0.45 °

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, …

Dark fringe condition

sin θ = mλ/a = 0.007853

θ = 0.45°

Answer

θ₂ = 0.45°

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