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 54 km/h brakes with a constant deceleration of 3.5 m/s² until it stops. Find the braking distance s.

v₀ = 54 km/ha = −3.5 m/s²v = 0

s = m

2.KinematicsHard

A ball is thrown straight up at 13 m/s from the edge of a roof 20 m above the ground. Ignore air resistance and use g = 9.81 m/s². How long after the throw does it hit the ground? Find t.

v₀ = 13 m/sh₀ = 20 mg = 9.81 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 5.2 m. Use g = 9.81 m/s². Find the coefficient of kinetic friction μ.

v₀ = 4.5 m/sd = 5.2 mg = 9.81 m/s²

μ =

4.Newton’s lawsHard

A 6.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 3.5 kg mass. The table is frictionless. Use g = 9.81 m/s². Find the acceleration a of the system.

m₁ = 6.5 kgm₂ = 3.5 kgμ = 0g = 9.81 m/s²

a = m/s²

5.Projectile motionMedium

A ball is launched from level ground at 20° above the horizontal and lands 65 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v₀.

θ = 20°R = 65 mg = 9.81 m/s²

v₀ = m/s

6.Projectile motionHard

A ball is launched at 22 m/s and 45° above the horizontal from the top of a 40 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 = 22 m/sθ = 45°h = 40 mg = 9.81 m/s²

H = m

7.Circular motionMedium

A car takes a flat curve of radius 20 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 = 20 mμ_s = 0.6g = 9.81 m/s²

v_max = m/s

8.Circular motionHard

A ball on a string 1 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 mg = 9.81 m/s²

v_min = m/s

9.Work & energyMedium

A car travelling at 40 km/h on a level road brakes hard and skids to a stop with its wheels locked. The coefficient of kinetic friction between the tyres and the road is 0.5. Find the stopping distance d.

v = 40 km/hμ = 0.5g = 9.81 m/s²

d = 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 4 m long (μ = 0.2) and then runs onto a spring (k = 800 N/m) on a smooth surface. Find the speed v of the block when it reaches the spring.

m = 1 kgh = 1.5 mL = 4 mμ = 0.2k = 800 N/mg = 9.81 m/s²

v = m/s

11.Momentum & collisionsMedium

A 20 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 15 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.

m = 20 gM = 1.5 kgh = 15 cmg = 9.81 m/s²

v = m/s

12.Momentum & collisionsHard

A 0.4 kg ball hits a wall at 23 m/s and bounces straight back at 6 m/s. The contact lasts 10 ms. Find the average force F_avg of the wall on the ball.

m = 0.4 kgv_in = 23 m/sv_out = 6 m/sΔt = 10 ms

F_avg = N

13.Springs & pendulumsMedium

How long must a simple pendulum be to have a period of 2 s? Use g = 9.81 m/s². Find L.

T = 2 sg = 9.81 m/s²

L = m

14.Springs & pendulumsHard

A 1.4 kg mass on a spring with k = 220 N/m is pulled 6 cm from equilibrium and released. Find its maximum speed v_max.

m = 1.4 kgk = 220 N/mA = 6 cm

v_max = m/s

15.Waves, sound & opticsMedium

An object 1 cm tall stands 10 cm in front of a thin diverging lens with a focal length of 10 cm. Find the image height h_i (negative for an inverted image).

h_o = 1 cmd_o = 10 cmf = −10 cm

h_i = cm

16.Waves, sound & opticsHard

Light of wavelength 480 nm falls on two slits 0.2 mm apart. The screen is 1.5 m away. Find the distance y₃ from the central bright fringe to bright fringe number 3.

λ = 480 nmd = 0.2 mmL = 1.5 mm = 3

y₃ = mm

3Solutions

1.Kinematics

Answer: s = 32.14 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 = 54/3.6 = 15 m/s

Braking distance

0 = v₀² − 2·a·s → s = v₀²/(2a) = 15²/(2·3.5) = 32.14 m

Answer

s = 32.14 m

2.Kinematics

Answer: t = 3.74 s

Equation of motion

Up is positive and the ground is h₀ below the starting point.

−20 = 13·t − ½·9.81·t²

4.905·t² − 13·t − 20 = 0

Solve the quadratic

Only the positive root is physical.

t = (13 + √(13² + 2·9.81·20))/9.81 = 3.74 s

Answer

t = 3.74 s

3.Newton’s laws

Answer: μ = 0.1985

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/(2·5.2) = 1.947 m/s²

Friction coefficient

μ·m·g = m·a → μ = a/g = 1.947/9.81 = 0.1985

Answer

μ = 0.1985

4.Newton’s laws

Answer: a = 3.433 m/s²

Acceleration of the system

a = (m₂·g − μ·m₁·g)/(m₁ + m₂) = (34.34 − 0)/10 = 3.433 m/s²

Answer

a = 3.433 m/s²

5.Projectile motion

Answer: v₀ = 31.5 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) = √(9.81·65/sin 40°) = 31.5 m/s

Answer

v₀ = 31.5 m/s

6.Projectile motion

Answer: H = 52.33 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 = 22·sin 45° = 15.56 m/s

Rise above the cliff

Δh = v_y²/(2g) = 15.56²/(2·9.81) = 12.33 m

Height above the ground

H = 40 + 12.33 = 52.33 m

Answer

H = 52.33 m

7.Circular motion

Answer: v_max = 10.85 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) = √(0.6·9.81·20) = 10.85 m/s

Answer

v_max = 10.85 m/s (39.06 km/h)

8.Circular motion

Answer: v_min = 3.132 m/s

Tension zero at the top

m·g = m·v²/r → v = √(g·r) = √(9.81·1) = 3.132 m/s

Answer

v_min = 3.132 m/s

9.Work & energy

Answer: d = 12.58 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).

Speed in base units

v = 40/3.6 = 11.11 m/s

Energy balance

½mv² = μmg·d → d = v²/(2μg) = 11.11²/(2·0.5·9.81) = 12.58 m

Answer

d = 12.58 m

Doubling the speed makes the stopping distance four times longer.

10.Work & energy

Answer: v = 3.706 m/s

Energy left after the rough strip

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

mgh − μmgL = 9.81·(1.5 − 0.2·4) = 6.867 J

Speed at the spring

½mv² = mg(h − μL) → v = √(2g(h − μL)) = √(2·9.81·0.7) = 3.706 m/s

Answer

v = 3.706 m/s

11.Momentum & collisions

Answer: v = 130.4 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) = √(2·9.81·0.15) = 1.716 m/s

Collision

m·v = (m + M)·V → v = (0.02 + 1.5)·1.716/0.02 = 130.4 m/s

Answer

v = 130.4 m/s

12.Momentum & collisions

Answer: F_avg = 1160 N

Change in momentum

J = m·Δv = 0.4·(6 + 23) = 11.6 N·s

Average force

F_avg = J/Δt = 11.6/0.01 = 1160 N

Answer

F_avg = 1160 N

13.Springs & pendulums

Answer: L = 0.994 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·(2/2π)² = 0.994 m

Answer

L = 0.994 m

14.Springs & pendulums

Answer: v_max = 0.7521 m/s

Angular frequency

ω = √(220/1.4) = 12.54 rad/s

Maximum speed

v_max = A·ω = 0.06·12.54 = 0.7521 m/s

Answer

v_max = 0.7521 m/s

15.Waves, sound & optics

Answer: h_i = 0.5 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/(−10) − 1/10

d_i = −5 cm

Magnification and height

m = −(−5)/10 = 0.5

h_i = 0.5·1 = 0.5 cm

Answer

h_i = 0.5 cm

16.Waves, sound & optics

Answer: y₃ = 10.8 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 = 480 × 10⁻⁹·1.5/(0.2 × 10⁻³) = 3.6 mm

Fringe number 3

y₃ = 3·3.6 = 10.8 mm

Answer

y₃ = 10.8 mm

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