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 m/s² along a runway. Find its speed v after 1800 m.

v₀ = 0a = 3 m/s²s = 1800 m

v = m/s

2.KinematicsHard

A model rocket is launched straight up from rest. Its engine gives it a constant upward acceleration of 26 m/s² for 2.5 s, after which the engine stops and the rocket moves freely. Ignore air resistance and use g = 9.81 m/s². Find the greatest height h_max the rocket reaches.

a = 26 m/s²t₁ = 2.5 sg = 9.81 m/s²

h_max = m

3.Newton’s lawsMedium

A 40 kg box is pushed up a ramp at 15° to the horizontal at constant speed, by a force parallel to the ramp. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². Find the pushing force F.

m = 40 kgθ = 15°μ = 0.1g = 9.81 m/s²

F = N

4.Newton’s lawsHard

A 9.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 4 kg mass. The coefficient of kinetic friction between block and table is 0.1. Use g = 9.81 m/s². Find the tension T in the string.

m₁ = 9.5 kgm₂ = 4 kgμ = 0.1g = 9.81 m/s²

T = N

5.Projectile motionMedium

A ball is launched from level ground at 70° above the horizontal and lands 95 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v₀.

θ = 70°R = 95 mg = 9.81 m/s²

v₀ = m/s

6.Projectile motionHard

A ball is launched at 20 m/s and 70° above the horizontal from the top of a 10 m cliff. Ignore air resistance and use g = 9.81 m/s². Find the greatest height H above the ground at the foot of the cliff.

v₀ = 20 m/sθ = 70°h = 10 mg = 9.81 m/s²

H = m

7.Circular motionMedium

A small ball on a string of length 0.7 m moves in a horizontal circle so that the string makes a constant angle of 15° with the vertical (a conical pendulum). Use g = 9.81 m/s². Find the time T for one revolution.

L = 0.7 mθ = 15°g = 9.81 m/s²

T = s

8.Circular motionHard

A 2 kg ball on a light string 0.8 m long is swung in a vertical circle. At the lowest point its speed is 10.5 m/s. Ignore air resistance and use g = 9.81 m/s². Find the tension T_top in the string at the highest point.

m = 2 kgr = 0.8 mv_bottom = 10.5 m/sg = 9.81 m/s²

T_top = N

9.Work & energyMedium

A 0.5 kg block slides 14 m down a slope at 25° to the horizontal, with a coefficient of kinetic friction of 0.1. Find the work W_f done against friction.

m = 0.5 kgθ = 25°μ = 0.1d = 14 mg = 9.81 m/s²

W_f = J

10.Work & energyHard

A 10 kg block is released from rest at height 1.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 4.5 m long (μ = 0.2) and then runs onto a spring (k = 500 N/m) on a smooth surface. Find the maximum compression x of the spring.

m = 10 kgh = 1.5 mL = 4.5 mμ = 0.2k = 500 N/mg = 9.81 m/s²

x = cm

11.Momentum & collisionsMedium

A 3 kg glider moving at 3 m/s makes a perfectly elastic head-on collision with a 2.5 kg glider at rest on a frictionless track. Find the velocity v′₂ of the second glider after the collision.

m₁ = 3 kgm₂ = 2.5 kgv₁ = 3 m/sv₂ = 0

v′₂ = m/s

12.Momentum & collisionsHard

A 10 g bullet is fired horizontally into a 3 kg block hanging from light strings 0.6 m long, and stays in it. The strings swing out to a largest angle of 40° from the vertical. Use g = 9.81 m/s². Find the speed v of the bullet before impact.

m = 10 gM = 3 kgL = 0.6 mθ = 40°g = 9.81 m/s²

v = m/s

13.Springs & pendulumsMedium

A simple pendulum has a length of 2.7 m. Use g = 9.81 m/s². Find its period T for small swings.

L = 2.7 mg = 9.81 m/s²

T = s

14.Springs & pendulumsHard

A 0.4 kg mass on a spring with k = 450 N/m is pulled 13 cm from equilibrium and released. Find its maximum speed v_max.

m = 0.4 kgk = 450 N/mA = 13 cm

v_max = m/s

15.Waves, sound & opticsMedium

A source of 500 Hz moving at 15 m/s and an observer moving at 15 m/s approach each other. The speed of sound is 343 m/s. Find the frequency f′ that the observer hears.

f = 500 Hzv = 343 m/sv_s = 15 m/sv_o = 15 m/s

f′ = Hz

16.Waves, sound & opticsHard

Light of wavelength 650 nm falls on two slits 0.5 mm apart. The screen is 3 m away. Find the spacing Δy between neighbouring bright fringes.

λ = 650 nmd = 0.5 mmL = 3 m

Δy = mm

3Solutions

1.Kinematics

Answer: v = 103.9 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) = √(2·3·1800) = 103.9 m/s

Answer

v = 103.9 m/s

(374.1 km/h)

2.Kinematics

Answer: h_max = 296.6 m

Powered phase

v₁ = 26·2.5 = 65 m/s

h₁ = ½·26·2.5² = 81.25 m

Free flight up

At the top the velocity is zero.

Δh = v₁²/(2g) = 65²/(2·9.81) = 215.3 m

h_max = 81.25 + 215.3 = 296.6 m

Answer

h_max = 296.6 m

3.Newton’s laws

Answer: F = 139.5 N

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

Constant speed: the forces balance.

F = m·g·sin θ + μ·m·g·cos θ

Pushing force

F = 101.6 + 37.9 = 139.5 N

Answer

F = 139.5 N

4.Newton’s laws

Answer: T = 30.37 N

Acceleration of the system

a = (m₂·g − μ·m₁·g)/(m₁ + m₂) = (39.24 − 9.32)/13.5 = 2.216 m/s²

Tension from the block on the table

T − μ·m₁·g = m₁·a

T = 9.5·2.216 + 9.32 = 30.37 N

Answer

T = 30.37 N

5.Projectile motion

Answer: v₀ = 38.08 m/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.

Solve the range formula for v₀

v₀ = √(g·R/sin 2θ) = √(9.81·95/sin 140°) = 38.08 m/s

Answer

v₀ = 38.08 m/s

6.Projectile motion

Answer: H = 28 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.

Vertical launch speed

v_y = 20·sin 70° = 18.79 m/s

Rise above the cliff

Δh = v_y²/(2g) = 18.79²/(2·9.81) = 18 m

Height above the ground

H = 10 + 18 = 28 m

Answer

H = 28 m

7.Circular motion

Answer: T = 1.65 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.

Radius of the circle

r = L·sin θ = 0.7·sin 15° = 0.1812 m

Speed

Tension and gravity together give the inward force: tan θ = v²/(r·g).

v = √(r·g·tan θ) = √(0.1812·9.81·tan 15°) = 0.6901 m/s

Period

T = 2π·r/v = 2π·0.1812/0.6901 = 1.65 s

Answer

T = 1.65 s

8.Circular motion

Answer: T_top = 177.5 N

Speed at the top

Kinetic energy is traded for potential energy over a height 2r.

v_top² = 10.5² − 4·9.81·0.8 = 78.86 m²/s²

Forces at the top

Tension and gravity both point down, towards the centre.

T + m·g = m·v_top²/r

T = 2·78.86/0.8 − 2·9.81 = 177.5 N

Answer

T_top = 177.5 N

9.Work & energy

Answer: W_f = 6.224 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 = 4.905·0.9063 = 4.445 N

F = μN = 0.1·4.445 = 0.4445 N

Work against friction

W_f = F·d = 0.4445·14 = 6.224 J

Answer

W_f = 6.224 J

10.Work & energy

Answer: x = 48.52 cm

Energy left after the rough strip

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

mgh − μmgL = 98.1·(1.5 − 0.2·4.5) = 58.86 J

Maximum compression

½kx² = 58.86 → x = √(2·58.86/500) = 0.4852 m = 48.52 cm

Answer

x = 48.52 cm

11.Momentum & collisions

Answer: v′₂ = 3.273 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.

Elastic collision with a target at rest

Solving momentum and energy conservation together gives these two results.

v′₁ = (3 − 2.5)/(3 + 2.5)·3 = 0.2727 m/s

v′₂ = 2·3/(3 + 2.5)·3 = 3.273 m/s

Answer

v′₂ = 3.273 m/s

12.Momentum & collisions

Answer: v = 499.5 m/s

Rise of the block

h = L·(1 − cos θ) = 0.6·(1 − cos 40°) = 0.1404 m

Speed after impact

V = √(2·9.81·0.1404) = 1.66 m/s

Collision

v = (m + M)·V/m = 3.01·1.66/0.01 = 499.5 m/s

Answer

v = 499.5 m/s

13.Springs & pendulums

Answer: T = 3.296 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·√(2.7/9.81) = 3.296 s

Answer

T = 3.296 s

14.Springs & pendulums

Answer: v_max = 4.36 m/s

Angular frequency

ω = √(450/0.4) = 33.54 rad/s

Maximum speed

v_max = A·ω = 0.13·33.54 = 4.36 m/s

Answer

v_max = 4.36 m/s

15.Waves, sound & optics

Answer: f′ = 545.7 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) = 500·(343 + 15)/(343 − 15)

f′ = 545.7 Hz

Answer

f′ = 545.7 Hz

16.Waves, sound & optics

Answer: Δy = 3.9 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, …

Fringe spacing

Δy = λL/d = 650 × 10⁻⁹·3/(0.5 × 10⁻³) = 3.9 mm

Answer

Δy = 3.9 mm

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