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 60 km/h brakes with a constant deceleration of 4.5 m/s² until it stops. Find the braking distance s.

v₀ = 60 km/ha = −4.5 m/s²v = 0

s = m

2.KinematicsHard

A model rocket is launched straight up from rest. Its engine gives it a constant upward acceleration of 22 m/s² for 1.5 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 = 22 m/s²t₁ = 1.5 sg = 9.81 m/s²

h_max = m

3.Newton’s lawsMedium

A 20 kg block slides down a slope at 45° to the horizontal. The coefficient of kinetic friction is 0.25. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.

m = 20 kgθ = 45°μ = 0.25g = 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 45 N pulls the front 3 kg block; the 5 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 = 5 kgm₂ = 3 kgF = 45 Nμ = 0.15g = 9.81 m/s²

T = N

5.Projectile motionMedium

A ball is launched from level ground at 32 m/s and must land 35 m away. Ignore air resistance and use g = 9.81 m/s². Find the smaller of the two possible launch angles θ.

v₀ = 32 m/sR = 35 mg = 9.81 m/s²

θ = °

6.Projectile motionHard

A ball is launched at 26 m/s and 30° above the horizontal from the top of a 50 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 = 26 m/sθ = 30°h = 50 mg = 9.81 m/s²

H = m

7.Circular motionMedium

A car takes a flat curve of radius 75 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 = 75 mμ_s = 0.5g = 9.81 m/s²

v_max = m/s

8.Circular motionHard

A 1.9 kg ball on a string 0.9 m long is swung in a vertical circle. At the lowest point it moves at 5 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.

m = 1.9 kgr = 0.9 mv = 5 m/sg = 9.81 m/s²

T = N

9.Work & energyMedium

A block slides from rest down a slope at 40° to the horizontal. The coefficient of kinetic friction is 0.35. Find its speed v after it has slid 3 m along the slope.

θ = 40°μ = 0.35d = 3 mg = 9.81 m/s²

v = m/s

10.Work & energyHard

A 10 kg block starts from rest at the bottom of a 20° slope and is pulled up the slope by a constant force of 95 N parallel to the slope. The coefficient of kinetic friction is 0.15. Find the speed v of the block after it has moved 6 m up the slope.

m = 10 kgF = 95 Nθ = 20°μ = 0.15d = 6 mg = 9.81 m/s²

v = m/s

11.Momentum & collisionsMedium

A 10 g bullet is fired horizontally into a 4.5 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 23 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.

m = 10 gM = 4.5 kgh = 23 cmg = 9.81 m/s²

v = m/s

12.Momentum & collisionsHard

A 0.6 kg ball hits a wall at 12 m/s and bounces straight back at 20 m/s. The contact lasts 50 ms. Find the average force F_avg of the wall on the ball.

m = 0.6 kgv_in = 12 m/sv_out = 20 m/sΔt = 50 ms

F_avg = N

13.Springs & pendulumsMedium

How long must a simple pendulum be to have a period of 3 s? Use g = 9.81 m/s². Find L.

T = 3 sg = 9.81 m/s²

L = m

14.Springs & pendulumsHard

A 1.7 kg mass on a spring with k = 430 N/m is pulled 7 cm from equilibrium and released. Find the total energy E of the oscillation.

m = 1.7 kgk = 430 N/mA = 7 cm

E = J

15.Waves, sound & opticsMedium

An object stands 8 cm in front of a thin converging lens with a focal length of 5 cm. Find the image distance d_i. Use the sign convention 1/f = 1/d_o + 1/d_i (f < 0 for a diverging lens, d_i < 0 for a virtual image).

d_o = 8 cmf = 5 cm

d_i = cm

16.Waves, sound & opticsHard

Light of wavelength 600 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 = 600 nma = 0.15 mmL = 1.5 m

w = mm

3Solutions

1.Kinematics

Answer: s = 30.86 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 = 60/3.6 = 16.67 m/s

Braking distance

0 = v₀² − 2·a·s → s = v₀²/(2a) = 16.67²/(2·4.5) = 30.86 m

Answer

s = 30.86 m

2.Kinematics

Answer: h_max = 80.25 m

Powered phase

v₁ = 22·1.5 = 33 m/s

h₁ = ½·22·1.5² = 24.75 m

Free flight up

At the top the velocity is zero.

Δh = v₁²/(2g) = 33²/(2·9.81) = 55.5 m

h_max = 24.75 + 55.5 = 80.25 m

Answer

h_max = 80.25 m

3.Newton’s laws

Answer: a = 5.203 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 45° − 0.25·cos 45°) = 9.81·(0.7071 − 0.1768) = 5.203 m/s²

Answer

a = 5.203 m/s²

4.Newton’s laws

Answer: T = 28.13 N

Whole system

a = (F − μ·(m₁ + m₂)·g)/(m₁ + m₂) = (45 − 11.77)/8 = 4.154 m/s²

Rear block alone

T = m₁·(a + μ·g) = 5·(4.154 + 1.472) = 28.13 N

Answer

T = 28.13 N

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

5.Projectile motion

Answer: θ = 9.795 °

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 = 9.81·35/32² = 0.3353

Angle

2θ = 19.59° → θ = 9.795°

Answer

θ = 9.795°

The other solution is 90° − θ = 80.2°.

6.Projectile motion

Answer: H = 58.61 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 = 26·sin 30° = 13 m/s

Rise above the cliff

Δh = v_y²/(2g) = 13²/(2·9.81) = 8.614 m

Height above the ground

H = 50 + 8.614 = 58.61 m

Answer

H = 58.61 m

7.Circular motion

Answer: v_max = 19.18 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) = √(0.5·9.81·75) = 19.18 m/s

Answer

v_max = 19.18 m/s (69.05 km/h)

8.Circular motion

Answer: T = 71.42 N

Forces at the lowest point

Towards the centre (up) is positive.

T − m·g = m·v²/r

T = 1.9·9.81 + 1.9·5²/0.9 = 18.64 + 52.78 = 71.42 N

Answer

T = 71.42 N

9.Work & energy

Answer: v = 4.696 m/s

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).

Energy balance along the slope

The block drops a height d·sin θ while friction removes μmg·cos θ·d. The mass cancels.

½v² = g·d·(sin θ − μ·cos θ)

v = √(2·9.81·3·(0.6428 − 0.35·0.766)) = 4.696 m/s

Answer

v = 4.696 m/s

10.Work & energy

Answer: v = 7.559 m/s

Forces along the slope

m·g = 10·9.81 = 98.1 N

mg·sin θ = 33.55 N

μ·mg·cos θ = 13.83 N

Work–energy

½mv² = (95 − 47.38)·6 = 285.7 J

v = √(2·285.7/10) = 7.559 m/s

Answer

v = 7.559 m/s

11.Momentum & collisions

Answer: v = 958.1 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) = √(2·9.81·0.23) = 2.124 m/s

Collision

m·v = (m + M)·V → v = (0.01 + 4.5)·2.124/0.01 = 958.1 m/s

Answer

v = 958.1 m/s

12.Momentum & collisions

Answer: F_avg = 384 N

Change in momentum

J = m·Δv = 0.6·(20 + 12) = 19.2 N·s

Average force

F_avg = J/Δt = 19.2/0.05 = 384 N

Answer

F_avg = 384 N

13.Springs & pendulums

Answer: L = 2.236 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·(3/2π)² = 2.236 m

Answer

L = 2.236 m

14.Springs & pendulums

Answer: E = 1.054 J

Angular frequency

ω = √(430/1.7) = 15.9 rad/s

Energy

E = ½·k·A² = ½·430·0.07² = 1.054 J

Answer

E = 1.054 J

15.Waves, sound & optics

Answer: d_i = 13.33 cm

Geometric optics: at a boundary, n₁·sin θ₁ = n₂·sin θ₂ (Snell's law), with angles measured from the normal and n = c/v. Going into a slower medium the ray bends towards the normal. From a denser medium (n₁ > n₂) beyond the critical angle θ_c = arcsin(n₂/n₁) all light is reflected: total internal reflection. Thin lenses and spherical mirrors obey 1/f = 1/d_o + 1/d_i with the magnification m = −d_i/d_o. Sign convention: f > 0 for a converging lens and a concave mirror (f = R/2), f < 0 for a diverging lens and a convex mirror; d_i > 0 for a real image, d_i < 0 for a virtual one, and m < 0 for an inverted image.

Thin-lens equation

1/d_i = 1/f − 1/d_o = 1/5 − 1/8

d_i = 13.33 cm

Ray diagram

real, on the far side of the lens

Answer

d_i = 13.33 cm

16.Waves, sound & optics

Answer: w = 12 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 = 600 × 10⁻⁹·1.5/(0.15 × 10⁻³) = 6 mm

Width of the central maximum

w = 2y₁ = 12 mm

Answer

w = 12 mm

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