Physics 1: Mechanics: cram pack

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

New problems Back to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 1.5 m/s² along a runway. Find its speed v after 1100 m.

v₀ = 0a = 1.5 m/s²s = 1100 m

v = m/s

2.KinematicsHard

A stone is dropped from rest into a well. The splash is heard 4.8 s after the stone is released. The speed of sound is 343 m/s; ignore air resistance and use g = 9.81 m/s². Find the depth d of the well.

t_total = 4.8 sv_sound = 343 m/sg = 9.81 m/s²

d = m

3.Newton’s lawsMedium

A 12 kg block slides down a slope at 30° to the horizontal. The coefficient of kinetic friction is 0.2. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.

m = 12 kgθ = 30°μ = 0.2g = 9.81 m/s²

a = m/s²

4.Newton’s lawsHard

Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 95 N pulls the front 7 kg block; the 6 kg block is towed behind. The coefficient of kinetic friction is 0.15 for both blocks. Use g = 9.81 m/s². Find the tension T in the connecting string.

m₁ = 6 kgm₂ = 7 kgF = 95 Nμ = 0.15g = 9.81 m/s²

T = N

5.Projectile motionMedium

A ball is launched from level ground at 34 m/s and must land 75 m away. Ignore air resistance and use g = 9.81 m/s². Find the smaller of the two possible launch angles θ.

v₀ = 34 m/sR = 75 mg = 9.81 m/s²

θ = °

6.Projectile motionHard

A ball is launched at 27 m/s and 60° above the horizontal from the top of a 5 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 = 27 m/sθ = 60°h = 5 mg = 9.81 m/s²

H = m

7.Circular motionMedium

A car takes a flat curve of radius 85 m. The coefficient of static friction between tyres and road is 0.5. Use g = 9.81 m/s². Find the highest speed v_max at which the car can take the curve without skidding.

r = 85 mμ_s = 0.5g = 9.81 m/s²

v_max = m/s

8.Circular motionHard

A ball on a string 1.4 m long is swung in a vertical circle. Use g = 9.81 m/s². What is the smallest speed v_min at the top for which the string just stays tight?

r = 1.4 mg = 9.81 m/s²

v_min = m/s

9.Work & energyMedium

A 2 kg ball hangs on a light string of length 0.8 m. It is pulled aside until the string makes 30° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.

m = 2 kgL = 0.8 mθ = 30°g = 9.81 m/s²

T = N

10.Work & energyHard

A 1 kg block is released from rest at height 1 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 1 m long (μ = 0.2) and then runs onto a spring (k = 2500 N/m) on a smooth surface. Find the speed v of the block when it reaches the spring.

m = 1 kgh = 1 mL = 1 mμ = 0.2k = 2500 N/mg = 9.81 m/s²

v = m/s

11.Momentum & collisionsMedium

A 10 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 19 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.

m = 10 gM = 3.5 kgh = 19 cmg = 9.81 m/s²

v = m/s

12.Momentum & collisionsHard

At a crossing a 1100 kg car travelling east at 13 m/s collides with a 1800 kg car travelling north at 17 m/s. The cars lock together. Find the speed v′ of the wreck just after the collision.

m₁ = 1100 kgv₁ = 13 m/s (E)m₂ = 1800 kgv₂ = 17 m/s (N)

v′ = m/s

13.Springs & pendulumsMedium

A 1.8 kg mass on a spring with k = 440 N/m oscillates on a frictionless surface with an amplitude of 20 cm. Find its speed v when it is 13 cm from equilibrium.

m = 1.8 kgk = 440 N/mA = 20 cmx = 13 cm

v = m/s

14.Springs & pendulumsHard

A 3.8 kg mass on a spring with k = 200 N/m is pulled 7 cm from equilibrium and released. Find its maximum acceleration a_max.

m = 3.8 kgk = 200 N/mA = 7 cm

a_max = m/s²

15.Waves, sound & opticsMedium

A pipe closed at one end must have its fundamental (first harmonic) at 132 Hz. The speed of sound is 343 m/s. Find the length L of the pipe.

v = 343 m/sn = 1f₁ = 132 Hz

L = m

16.Waves, sound & opticsHard

Light of wavelength 633 nm passes through a single slit 0.05 mm wide. Find the angle θ₃ of dark fringe number 3 from the straight-through direction.

λ = 633 nma = 0.05 mmm = 3

θ₃ = °

3Solutions

1.Kinematics

Answer: v = 57.45 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) = √(2·1.5·1100) = 57.45 m/s

Answer

v = 57.45 m/s

(206.8 km/h)

2.Kinematics

Answer: d = 99.73 m

Two parts of the time

½·9.81·t₁² = 343·(4.8 − t₁)

4.905·t₁² + 343·t₁ − 1646 = 0

Fall time

Only the positive root is physical.

t₁ = (−343 + √(343² + 4·4.905·1646))/(2·4.905) = 4.509 s

Depth

d = ½·9.81·4.509² = 99.73 m

Answer

d = 99.73 m

3.Newton’s laws

Answer: a = 3.206 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 30° − 0.2·cos 30°) = 9.81·(0.5 − 0.1732) = 3.206 m/s²

Answer

a = 3.206 m/s²

4.Newton’s laws

Answer: T = 43.85 N

Whole system

a = (F − μ·(m₁ + m₂)·g)/(m₁ + m₂) = (95 − 19.13)/13 = 5.836 m/s²

Rear block alone

T = m₁·(a + μ·g) = 6·(5.836 + 1.472) = 43.85 N

Answer

T = 43.85 N

With the same μ on both blocks, friction drops out: T = F·m₁/(m₁ + m₂).

5.Projectile motion

Answer: θ = 19.76 °

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 = 9.81·75/34² = 0.6365

Angle

2θ = 39.53° → θ = 19.76°

Answer

θ = 19.76°

The other solution is 90° − θ = 70.24°.

6.Projectile motion

Answer: H = 32.87 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 = 27·sin 60° = 23.38 m/s

Rise above the cliff

Δh = v_y²/(2g) = 23.38²/(2·9.81) = 27.87 m

Height above the ground

H = 5 + 27.87 = 32.87 m

Answer

H = 32.87 m

7.Circular motion

Answer: v_max = 20.42 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) = √(0.5·9.81·85) = 20.42 m/s

Answer

v_max = 20.42 m/s (73.51 km/h)

8.Circular motion

Answer: v_min = 3.706 m/s

Tension zero at the top

m·g = m·v²/r → v = √(g·r) = √(9.81·1.4) = 3.706 m/s

Answer

v_min = 3.706 m/s

9.Work & energy

Answer: T = 24.88 N

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). On a circular path the net force towards the centre must equal mv²/r.

Speed at the bottom

v² = 2gL(1 − cos θ) = 2·9.81·0.8·(1 − cos 30°) = 2.103 m²/s²

Forces at the bottom

m·g = 2·9.81 = 19.62 N

T = mg + mv²/L = 19.62 + 2·2.103/0.8 = 24.88 N

Answer

T = 24.88 N

Check: T = mg(3 − 2cos θ) = 24.88 N.

10.Work & energy

Answer: v = 3.962 m/s

Energy left after the rough strip

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

mgh − μmgL = 9.81·(1 − 0.2·1) = 7.848 J

Speed at the spring

½mv² = mg(h − μL) → v = √(2g(h − μL)) = √(2·9.81·0.8) = 3.962 m/s

Answer

v = 3.962 m/s

11.Momentum & collisions

Answer: v = 677.7 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) = √(2·9.81·0.19) = 1.931 m/s

Collision

m·v = (m + M)·V → v = (0.01 + 3.5)·1.931/0.01 = 677.7 m/s

Answer

v = 677.7 m/s

12.Momentum & collisions

Answer: v′ = 11.65 m/s

Momentum components

p_x = 1100·13 = 14300 kg·m/s

p_y = 1800·17 = 30600 kg·m/s

Speed of the wreck

v′ = √(p_x² + p_y²)/(m₁ + m₂) = 33776/2900 = 11.65 m/s

Answer

v′ = 11.65 m/s

13.Springs & pendulums

Answer: v = 2.376 m/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.

Angular frequency

ω = √(440/1.8) = 15.63 rad/s

Speed at x

v = ω·√(A² − x²) = 15.63·√(0.2² − 0.13²) = 2.376 m/s

Answer

v = 2.376 m/s

14.Springs & pendulums

Answer: a_max = 3.684 m/s²

Angular frequency

ω = √(200/3.8) = 7.255 rad/s

Maximum acceleration

a_max = A·ω² = 0.07·7.255² = 3.684 m/s²

Answer

a_max = 3.684 m/s²

15.Waves, sound & optics

Answer: L = 0.6496 m

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

λ = v/f = 343/132 = 2.598 m

Length

L = nλ/4 = 1·2.598/4 = 0.6496 m

Answer

L = 0.6496 m

16.Waves, sound & optics

Answer: θ₃ = 2.177 °

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 = 0.03798

θ = 2.177°

Answer

θ₃ = 2.177°

ExamProblems · examproblems.com/courses/physics-1-mechanics