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

  1. List what you know: v₀, v, a, t and s, with units.
  2. Pick the equation that contains your three knowns and the unknown.
  3. Choose one positive direction and give opposing quantities a minus sign.
  4. Solve and check that the answer is physically sensible.

Newton’s laws

Second law
ΣF = m·a
Weight
W = m·g
Kinetic friction
Ff = μ·N
On a slope
along: m·g·sin θ, normal: m·g·cos θ

Method

  1. Draw a free-body diagram with every force on the body.
  2. Choose a positive direction along the motion.
  3. 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).
  4. Apply ΣF = m·a along the motion; friction is μ·N against the motion.

Projectile motion

Components
vx = v0cos θ, vy = v0sin θ
Position
x = vxt, y = h + vyt − ½gt²
Maximum height
H = h + vy²/(2g)
Range (level ground)
R = v0²·sin 2θ/g

Method

  1. Split the launch velocity into components: vx = v₀·cos θ and vy = v₀·sin θ.
  2. The horizontal motion has constant speed; the vertical motion has acceleration −g.
  3. Find the time of flight from the vertical motion.
  4. 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
vmax = √(μ·g·r)

Work & energy

Kinetic energy
Ek = ½mv²
Potential energy
Ep = mgh
Energy balance
Ek1 + Ep1 + W = Ek2 + Ep2 + 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

  1. Choose a positive direction and give each velocity its sign.
  2. Write the total momentum before: Σm·v.
  3. For a perfectly inelastic collision the bodies move together after; for an elastic one kinetic energy is also conserved.
  4. 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
vmax = A·ω, amax = 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 fn = n·v/(2L); closed pipe fn = n·v/(4L), n odd
Doppler effect
f′ = f·(v ± vo)/(v ∓ vs)
Refractive index
n = c/v, λ = λ0/n
Snell’s law
n1sin θ1 = n2sin θ2
Critical angle
sin θc = n2/n1
Optical fibre
NA = sin θa = √(n1² − n2²)
Lenses and mirrors
1/f = 1/do + 1/di, m = −di/do, mirror f = R/2
Double slit
d·sin θ = mλ, ym = mλL/d
Grating
d = 1/N, d·sin θ = mλ, mmax ≤ d/λ
Single slit
a·sin θ = mλ (dark), w = 2λL/a

Method

  1. Identify the kind of problem: wave speed, sound level, standing wave, Doppler, refraction, lens or mirror, or interference.
  2. Write the relation first: v = fλ, β = 10·log(I/I₀), fₙ = n·v/(2L) or n·v/(4L), f′ = f(v ± vo)/(v ∓ vs), n₁sin θ₁ = n₂sin θ₂, 1/f = 1/d₀ + 1/dᵢ, d·sin θ = mλ.
  3. Convert units: nm, mm and cm to m, and check whether angles are measured from the normal.
  4. 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
policea = 2 m/s²car22 m/s

A car speeding at a constant 22 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 2 m/s². How far from its starting point does the police car catch up? Find d.

vcar = 22 m/sapolice = 2 m/s²v0,police = 0
d = m
2.KinematicsHard
v₀ = 16 m/s25 m

A ball is thrown straight up at 16 m/s from the edge of a roof 25 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height hmax above the ground.

v₀ = 16 m/sh₀ = 25 mg = 9.81 m/s²
hmax = m
3.Newton’s lawsMedium
boxv₀ = 4.5 m/sμ = ?

A box slides across a level floor with an initial speed of 4.5 m/s and comes to rest after 6.9 m. Use g = 9.81 m/s². Find the coefficient of kinetic friction μ.

v₀ = 4.5 m/sd = 6.9 mg = 9.81 m/s²
μ =
4.Newton’s lawsHard
11 kg3 kgTF = 135 Nμ = 0.25

Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 135 N pulls the front 3 kg block; the 11 kg block is towed behind. The coefficient of kinetic friction is 0.25 for both blocks. Use g = 9.81 m/s². Find the acceleration a of the blocks.

m1 = 11 kgm2 = 3 kgF = 135 Nμ = 0.25g = 9.81 m/s²
a = m/s²
5.Projectile motionMedium
v0​ = ?65°R = 30 m

A ball is launched from level ground at 65° above the horizontal and lands 30 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.

θ = 65°R = 30 mg = 9.81 m/s²
v0 = m/s
6.Projectile motionHard
h = 30 mv0​ = 28 m/s25°range

A ball is launched at 25° above horizontal from the top of a 30 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.

v0 = 28 m/sθ = 25°h = 30 mg = 9.81 m/s²
t = s
7.Circular motionMedium
r = 45 mvmax​ = ?

A car takes a flat curve of radius 45 m. The coefficient of static friction between tyres and road is 0.5. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.

r = 45 mμs = 0.5g = 9.81 m/s²
vmax = m/s
8.Circular motionHard
R = 4 mh = ?

A small cart is released from rest on a frictionless track and runs into a vertical loop of radius 4 m. From what minimum height hmin above the bottom of the loop must it start to stay on the track at the top of the loop?

R = 4 mv₀ = 0
hmin = m
9.Work & energyMedium
20°d = 7 mμ = 0.2

A 2.5 kg block slides 7 m down a slope at 20° to the horizontal, with a coefficient of kinetic friction of 0.2. Find the work Wf done against friction.

m = 2.5 kgθ = 20°μ = 0.2d = 7 mg = 9.81 m/s²
Wf = J
10.Work & energyHard
h = 3.5 mL = 2.5 mμ = 0.25k = 2000 N/m

A 1 kg block is released from rest at height 3.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 2.5 m long (μ = 0.25) and then runs onto a spring (k = 2000 N/m) on a smooth surface. Find the maximum compression x of the spring.

m = 1 kgh = 3.5 mL = 2.5 mμ = 0.25k = 2000 N/mg = 9.81 m/s²
x = cm
11.Momentum & collisionsMedium
5 kg82 m/s9 kgv′2​ = ?

A 14 kg projectile moving horizontally at 34 m/s explodes into two fragments that move along the same line. The 5 kg fragment continues forwards at 82 m/s. Find the velocity v′2 of the 9 kg fragment (positive in the original direction).

M = 14 kgu = 34 m/sm1 = 5 kgv′1 = 82 m/sm2 = 9 kg
v′2 = m/s
12.Momentum & collisionsHard
19 m/s15 m/s

A 0.45 kg ball hits a wall at 19 m/s and bounces straight back at 15 m/s. The contact lasts 20 ms. Find the average force Favg of the wall on the ball.

m = 0.45 kgvin = 19 m/svout = 15 m/sΔt = 20 ms
Favg = N
13.Springs & pendulumsMedium
L = 2 mT = 4.61 s

An astronaut on another planet finds that a simple pendulum 2 m long has a period of 4.61 s for small swings. Find the gravitational acceleration g on that planet.

L = 2 mT = 4.61 s
g = m/s²
14.Springs & pendulumsHard
0.6 kgk = 370 N/mA = 4 cm

A 0.6 kg mass on a spring with k = 370 N/m is pulled 4 cm from equilibrium and released. Find the total energy E of the oscillation.

m = 0.6 kgk = 370 N/mA = 4 cm
E = J
15.Waves, sound & opticsMedium
L = 1.5 mn = 2

A pipe open at both ends is 1.5 m long. The speed of sound is 343 m/s. Find the frequency f2 of its harmonic number 2.

v = 343 m/sn = 2L = 1.5 m
f2 = Hz
16.Waves, sound & opticsHard
λ = 633 nmd = 0.4 mmL = 1.5 m

Light of wavelength 633 nm falls on two slits 0.4 mm apart. The screen is 1.5 m away. Find the spacing Δy between neighbouring bright fringes.

λ = 633 nmd = 0.4 mmL = 1.5 m
Δy = mm

3Solutions

1.Kinematics

Answer: d = 484 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.

Equal distances

v·t = ½·a·t² → t = 2v/a = 2·22/2 = 22 s

Distance

d = v·t = 22·22 = 484 m

Answer

d = 484 m

At that moment the police car moves at 2v = 44 m/s.

2.Kinematics

Answer: hmax = 38.05 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.

Rise above the roof

0 = v₀² − 2·g·Δh → Δh = 16²/(2·9.81) = 13.05 m

Height above the ground

hmax = 25 + 13.05 = 38.05 m

Answer

hmax = 38.05 m
3.Newton’s laws

Answer: μ = 0.1496

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/(2·6.9) = 1.467 m/s²

Friction coefficient

μ·m·g = m·a → μ = a/g = 1.467/9.81 = 0.1496

Answer

μ = 0.1496
4.Newton’s laws

Answer: a = 7.19 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.

Whole system

a = (F − μ·(m1 + m2)·g)/(m1 + m2) = (135 − 34.34)/14 = 7.19 m/s²

Answer

a = 7.19 m/s²
5.Projectile motion

Answer: v0 = 19.6 m/s

Projectile motion: with no air resistance, x = vxt and y = h + vyt − ½gt². The two directions are independent except that they share the same time.

Solve the range formula for v₀

v₀ = √(g·R/sin 2θ) = √(9.81·30/sin 130°) = 19.6 m/s

Answer

v0 = 19.6 m/s
6.Projectile motion

Answer: t = 3.958 s

Projectile motion: with no air resistance, x = vxt and y = h + vyt − ½gt². The two directions are independent except that they share the same time.

Split the launch velocity

Horizontal velocity stays constant; vertical motion has a constant acceleration g = 9.81 m/s² downward.

vx = v0·cos θ = 28·cos 25° = 25.38 m/s
vy = v0·sin θ = 28·sin 25° = 11.83 m/s

Time in the air

Solve y(t) = h + vy·t − ½gt² = 0 for the positive root.

0 = 30 + 11.83·t − 4.905·t²
t = (vy + √(vy² + 2gh))/g = (11.83 + √(140 + 588.6))/9.81 = 3.958 s

Answer

t = 3.958 s
7.Circular motion

Answer: vmax = 14.86 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) = √(0.5·9.81·45) = 14.86 m/s

Answer

vmax = 14.86 m/s (53.48 km/h)
8.Circular motion

Answer: hmin = 10 m

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.

Condition at the top

m·g = m·v²/R → v² = g·R

Energy conservation

g·h = g·2R + ½·g·R → h = 2.5·R = 2.5·4 = 10 m

Answer

hmin = 10 m
9.Work & energy

Answer: Wf = 32.26 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 = 24.53·0.9397 = 23.05 N
F = μN = 0.2·23.05 = 4.609 N

Work against friction

Wf = F·d = 4.609·7 = 32.26 J

Answer

Wf = 32.26 J
10.Work & energy

Answer: x = 16.79 cm

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 left after the rough strip

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

mgh − μmgL = 9.81·(3.5 − 0.25·2.5) = 28.2 J

Maximum compression

½kx² = 28.2 → x = √(2·28.2/2000) = 0.1679 m = 16.79 cm

Answer

x = 16.79 cm
11.Momentum & collisions

Answer: v′2 = 7.333 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.

Momentum before

p = M·u = 14·34 = 476 kg·m/s

Momentum after

476 = 5·82 + 9·v′2
v′2 = (476 − 410)/9 = 7.333 m/s

Answer

v′2 = 7.333 m/s
12.Momentum & collisions

Answer: Favg = 765 N

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.

Change in momentum

J = m·Δv = 0.45·(15 + 19) = 15.3 N·s

Average force

Favg = J/Δt = 15.3/0.02 = 765 N

Answer

Favg = 765 N
13.Springs & pendulums

Answer: g = 3.715 m/s²

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 g

g = 4π²·L/T² = 4π²·2/4.61² = 3.715 m/s²

Answer

g = 3.715 m/s²
14.Springs & pendulums

Answer: E = 0.296 J

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 = √(370/0.6) = 24.83 rad/s

Energy

E = ½·k·A² = ½·370·0.04² = 0.296 J

Answer

E = 0.296 J
15.Waves, sound & optics

Answer: f2 = 228.7 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, fn = 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, fn = n·v/(4L). When source or observer move along the line between them, f′ = f·(v ± vo)/(v ∓ vs): use the upper signs for motion towards each other. Two waves of nearly equal frequency beat at |f₁ − f₂|.

Wavelength of this mode

λ = 2L/n = 2·1.5/2 = 1.5 m

Frequency

f = v/λ = 343/1.5 = 228.7 Hz

Answer

f2 = 228.7 Hz
16.Waves, sound & optics

Answer: Δy = 2.374 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 = 633 × 10⁻⁹·1.5/(0.4 × 10⁻³) = 2.374 mm

Answer

Δy = 2.374 mm

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