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

v₀ = 54 km/ha = −5.5 m/s²v = 0

s = m

2.KinematicsHard

A tram starts from rest and accelerates at 3 m/s² for 6 s, then runs at constant speed for 12 s, and finally brakes at 4 m/s² until it stops. Find the total distance s between the two stops.

a₁ = 3 m/s²t₁ = 6 st₂ = 12 sa₃ = −4 m/s²

s = m

3.Newton’s lawsMedium

A 50 kg box is pushed up a ramp at 40° to the horizontal at constant speed, by a force parallel to the ramp. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². Find the pushing force F.

m = 50 kgθ = 40°μ = 0.3g = 9.81 m/s²

F = N

4.Newton’s lawsHard

A 5.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 2.5 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 = 5.5 kgm₂ = 2.5 kgμ = 0.1g = 9.81 m/s²

T = N

5.Projectile motionMedium

A ball is kicked from level ground at 12 m/s and 65° above the horizontal towards a wall 7 m away. Ignore air resistance and use g = 9.81 m/s². At what height y does the ball pass the wall?

v₀ = 12 m/sθ = 65°x = 7 mg = 9.81 m/s²

y = m

6.Projectile motionHard

A ball is launched at 18 m/s and 40° above the horizontal from the top of a 20 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 = 18 m/sθ = 40°h = 20 mg = 9.81 m/s²

H = m

7.Circular motionMedium

A small ball on a string of length 0.4 m moves in a horizontal circle so that the string makes a constant angle of 50° with the vertical (a conical pendulum). Use g = 9.81 m/s². Find the speed v of the ball.

L = 0.4 mθ = 50°g = 9.81 m/s²

v = m/s

8.Circular motionHard

A ball on a string 1.2 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 = 1.2 mg = 9.81 m/s²

v_min = m/s

9.Work & energyMedium

A 4 kg block slides 7 m down a slope at 40° to the horizontal, with a coefficient of kinetic friction of 0.35. Find the work W_f done against friction.

m = 4 kgθ = 40°μ = 0.35d = 7 mg = 9.81 m/s²

W_f = J

10.Work & energyHard

A 10 kg block is released from rest at height 1 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 1.5 m long (μ = 0.15) 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 = 10 kgh = 1 mL = 1.5 mμ = 0.15k = 2500 N/mg = 9.81 m/s²

v = m/s

11.Momentum & collisionsMedium

A 2.5 kg glider moving at 8 m/s makes a perfectly elastic head-on collision with a 4 kg glider at rest on a frictionless track. Find the velocity v′₂ of the second glider after the collision.

m₁ = 2.5 kgm₂ = 4 kgv₁ = 8 m/sv₂ = 0

v′₂ = m/s

12.Momentum & collisionsHard

A 1000 kg car runs into a parked 1400 kg car. The wrecks lock together and skid 9.5 m before stopping. The coefficient of kinetic friction between tyres and road is 0.6. Use g = 9.81 m/s². Find the speed v₁ of the first car just before the collision.

m₁ = 1000 kgm₂ = 1400 kgd = 9.5 mμ = 0.6g = 9.81 m/s²

v₁ = m/s

13.Springs & pendulumsMedium

A simple pendulum has a length of 2.6 m. Use g = 9.81 m/s². Find its period T for small swings.

L = 2.6 mg = 9.81 m/s²

T = s

14.Springs & pendulumsHard

A 3.1 kg mass on a spring with k = 260 N/m is pulled 18 cm from equilibrium and released. Find its maximum acceleration a_max.

m = 3.1 kgk = 260 N/mA = 18 cm

a_max = m/s²

15.Waves, sound & opticsMedium

A pipe open at both ends must have its harmonic number 5 at 1429 Hz. The speed of sound is 343 m/s. Find the length L of the pipe.

v = 343 m/sn = 5f₅ = 1429 Hz

L = m

16.Waves, sound & opticsHard

A parking sensor at rest sends ultrasound of 35 kHz towards a car that approaches at 15 m/s. Find the frequency f″ of the echo received back at the sensor (speed of sound 343 m/s).

f = 35 kHzu = 15 m/sv = 343 m/s

f″ = Hz

3Solutions

1.Kinematics

Answer: s = 20.45 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/(2·5.5) = 20.45 m

Answer

s = 20.45 m

2.Kinematics

Answer: s = 310.5 m

Top speed

v = a₁·t₁ = 3·6 = 18 m/s

Distance of each phase

s₁ = ½·3·6² = 54 m

s₂ = 18·12 = 216 m

s₃ = 18²/(2·4) = 40.5 m

Total distance

s = 54 + 216 + 40.5 = 310.5 m

Answer

s = 310.5 m

3.Newton’s laws

Answer: F = 428 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 = 315.3 + 112.7 = 428 N

Answer

F = 428 N

4.Newton’s laws

Answer: T = 18.55 N

Acceleration of the system

a = (m₂·g − μ·m₁·g)/(m₁ + m₂) = (24.53 − 5.396)/8 = 2.391 m/s²

Tension from the block on the table

T − μ·m₁·g = m₁·a

T = 5.5·2.391 + 5.396 = 18.55 N

Answer

T = 18.55 N

5.Projectile motion

Answer: y = 5.667 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.

Time to the wall

v_x = 12·cos 65° = 5.071 m/s

t = x/v_x = 7/5.071 = 1.38 s

Height at that time

y = 10.88·1.38 − ½·9.81·1.38² = 5.667 m

Answer

y = 5.667 m

The ball is already on its way down at the wall.

6.Projectile motion

Answer: H = 26.82 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 = 18·sin 40° = 11.57 m/s

Rise above the cliff

Δh = v_y²/(2g) = 11.57²/(2·9.81) = 6.823 m

Height above the ground

H = 20 + 6.823 = 26.82 m

Answer

H = 26.82 m

7.Circular motion

Answer: v = 1.893 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.

Radius of the circle

r = L·sin θ = 0.4·sin 50° = 0.3064 m

Speed

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

v = √(r·g·tan θ) = √(0.3064·9.81·tan 50°) = 1.893 m/s

Answer

v = 1.893 m/s

8.Circular motion

Answer: v_min = 3.431 m/s

Tension zero at the top

m·g = m·v²/r → v = √(g·r) = √(9.81·1.2) = 3.431 m/s

Answer

v_min = 3.431 m/s

9.Work & energy

Answer: W_f = 73.65 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 = 39.24·0.766 = 30.06 N

F = μN = 0.35·30.06 = 10.52 N

Work against friction

W_f = F·d = 10.52·7 = 73.65 J

Answer

W_f = 73.65 J

10.Work & energy

Answer: v = 3.899 m/s

Energy left after the rough strip

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

mgh − μmgL = 98.1·(1 − 0.15·1.5) = 76.03 J

Speed at the spring

½mv² = mg(h − μL) → v = √(2g(h − μL)) = √(2·9.81·0.775) = 3.899 m/s

Answer

v = 3.899 m/s

11.Momentum & collisions

Answer: v′₂ = 6.154 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 = (2.5 − 4)/(2.5 + 4)·8 = −1.846 m/s

v′₂ = 2·2.5/(2.5 + 4)·8 = 6.154 m/s

Answer

v′₂ = 6.154 m/s

12.Momentum & collisions

Answer: v₁ = 25.38 m/s

Speed after the collision

V = √(2·μ·g·d) = √(2·0.6·9.81·9.5) = 10.58 m/s

Momentum in the collision

m₁·v₁ = (m₁ + m₂)·V → v₁ = 2400·10.58/1000 = 25.38 m/s

Answer

v₁ = 25.38 m/s

(91.37 km/h)

13.Springs & pendulums

Answer: T = 3.235 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·√(2.6/9.81) = 3.235 s

Answer

T = 3.235 s

14.Springs & pendulums

Answer: a_max = 15.1 m/s²

Angular frequency

ω = √(260/3.1) = 9.158 rad/s

Maximum acceleration

a_max = A·ω² = 0.18·9.158² = 15.1 m/s²

Answer

a_max = 15.1 m/s²

15.Waves, sound & optics

Answer: L = 0.6001 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 = 343/1429 = 0.24 m

Length

L = nλ/2 = 5·0.24/2 = 0.6001 m

Answer

L = 0.6001 m

16.Waves, sound & optics

Answer: f″ = 38201 Hz

Frequency at the car

f′ = 35000·(343 + 15)/343 = 36531 Hz

Frequency of the echo

f″ = f′·343/(343 − 15) = 38201 Hz

Answer

f″ = 38201 Hz

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