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 90 km/h brakes with a constant deceleration of 4.5 m/s² until it stops. Find the braking distance s.
v₀ = 90 km/ha = −4.5 m/s²v = 0
s = m
2.KinematicsHard
A ball is thrown straight up at 9 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 = 9 m/sh₀ = 25 mg = 9.81 m/s²
hmax = m
3.Newton’s lawsMedium
A box slides across a level floor with an initial speed of 2 m/s and comes to rest after 1.4 m. Use g = 9.81 m/s². Find the coefficient of kinetic friction μ.
v₀ = 2 m/sd = 1.4 mg = 9.81 m/s²
μ =
4.Newton’s lawsHard
A 9 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 3.5 kg mass. The coefficient of kinetic friction between block and table is 0.3. Use g = 9.81 m/s². Find the tension T in the string.
m1 = 9 kgm2 = 3.5 kgμ = 0.3g = 9.81 m/s²
T = N
5.Projectile motionMedium
A ball is launched from level ground at 38 m/s and must land 135 m away. Ignore air resistance and use g = 9.81 m/s². Find the smaller of the two possible launch angles θ.
v₀ = 38 m/sR = 135 mg = 9.81 m/s²
θ = °
6.Projectile motionHard
A ball is launched at 33 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 = 33 m/sθ = 30°h = 50 mg = 9.81 m/s²
H = m
7.Circular motionMedium
A small ball on a string of length 1.9 m moves in a horizontal circle so that the string makes a constant angle of 20° with the vertical (a conical pendulum). Use g = 9.81 m/s². Find the time T for one revolution.
L = 1.9 mθ = 20°g = 9.81 m/s²
T = s
8.Circular motionHard
A curve of radius 150 m is banked at 20°, and the coefficient of static friction between tyres and road is 0.25. Use g = 9.81 m/s². Find the highest speed vmax at which a car can take the curve without sliding up and out.
r = 150 mθ = 20°μs = 0.25g = 9.81 m/s²
vmax = m/s
9.Work & energyMedium
A toy launcher has a vertical spring of stiffness 500 N/m. A 0.15 kg ball is placed on it and the spring is compressed by 4 cm, then released. Ignoring air resistance and the spring’s mass, find the height h the ball rises, measured from its position when the spring was compressed.
m = 0.15 kgk = 500 N/mx = 4 cmg = 9.81 m/s²
h = m
10.Work & energyHard
A 1 kg block is released from rest at height 3 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 4 m long (μ = 0.25) and then runs onto a spring (k = 5000 N/m) on a smooth surface. Find the maximum compression x of the spring.
m = 1 kgh = 3 mL = 4 mμ = 0.25k = 5000 N/mg = 9.81 m/s²
x = cm
11.Momentum & collisionsMedium
A 4 kg projectile moving horizontally at 13 m/s explodes into two fragments that move along the same line. The 2 kg fragment continues forwards at 27 m/s. Find the velocity v′2 of the 2 kg fragment (positive in the original direction).
M = 4 kgu = 13 m/sm1 = 2 kgv′1 = 27 m/sm2 = 2 kg
v′2 = m/s
12.Momentum & collisionsHard
A 0.45 kg ball hits a wall at 26 m/s and bounces straight back at 18 m/s. The contact lasts 20 ms. Find the magnitude of the impulse J the wall gives the ball.
m = 0.45 kgvin = 26 m/svout = 18 m/sΔt = 20 ms
J = N·s
13.Springs & pendulumsMedium
An astronaut on another planet finds that a simple pendulum 1.8 m long has a period of 4.38 s for small swings. Find the gravitational acceleration g on that planet.
L = 1.8 mT = 4.38 s
g = m/s²
14.Springs & pendulumsHard
A 1.7 kg mass on a spring with k = 110 N/m is pulled 10 cm from equilibrium and released. Find its maximum speed vmax.
m = 1.7 kgk = 110 N/mA = 10 cm
vmax = m/s
15.Waves, sound & opticsMedium
One machine gives a sound level of 80 dB at a certain point. Find the level β at that point when 8 identical machines run at the same time (incoherent sources).
β1 = 80 dBn = 8
β = dB
16.Waves, sound & opticsHard
An optical fibre has a core with n₁ = 1.46 and a cladding with n₂ = 1.43. Find its numerical aperture NA.
n1 = 1.46n2 = 1.43
NA =
3Solutions
1.Kinematics
Answer: s = 69.44 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 = 90/3.6 = 25 m/s
Braking distance
0 = v₀² − 2·a·s → s = v₀²/(2a) = 25²/(2·4.5) = 69.44 m
Answer
s = 69.44 m
2.Kinematics
Answer: hmax = 29.13 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 = 9²/(2·9.81) = 4.128 m
Height above the ground
hmax = 25 + 4.128 = 29.13 m
Answer
hmax = 29.13 m
3.Newton’s laws
Answer: μ = 0.1456
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) = 2²/(2·1.4) = 1.429 m/s²
Friction coefficient
μ·m·g = m·a → μ = a/g = 1.429/9.81 = 0.1456
Answer
μ = 0.1456
4.Newton’s laws
Answer: T = 32.14 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.
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.
Range formula
R = v₀²·sin 2θ/g → sin 2θ = 9.81·135/38² = 0.9171
Angle
2θ = 66.51° → θ = 33.26°
Answer
θ = 33.26°
The other solution is 90° − θ = 56.74°.
6.Projectile motion
Answer: H = 63.88 m
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.
Vertical launch speed
vy = 33·sin 30° = 16.5 m/s
Rise above the cliff
Δh = vy²/(2g) = 16.5²/(2·9.81) = 13.88 m
Height above the ground
H = 50 + 13.88 = 63.88 m
Answer
H = 63.88 m
7.Circular motion
Answer: T = 2.68 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 = 1.9·sin 20° = 0.6498 m
Speed
Tension and gravity together give the inward force: tan θ = v²/(r·g).
v = √(r·g·tan θ) = √(0.6498·9.81·tan 20°) = 1.523 m/s
Period
T = 2π·r/v = 2π·0.6498/1.523 = 2.68 s
Answer
T = 2.68 s
8.Circular motion
Answer: vmax = 31.53 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.
Force balances
Friction points down the slope.
N·(sin θ + μ·cos θ) = m·v²/r
N·(cos θ − μ·sin θ) = m·g
Speed
v = √(150·9.81·0.5769/0.8542) = 31.53 m/s
Answer
vmax = 31.53 m/s
(113.5 km/h)
9.Work & energy
Answer: h = 0.2718 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).
Spring energy
x = 4 cm = 0.04 m
½kx² = ½·500·0.04² = 0.4 J
Height
m·g = 0.15·9.81 = 1.472 N
h = ½kx²/(mg) = 0.4/1.472 = 0.2718 m
Answer
h = 0.2718 m
10.Work & energy
Answer: x = 8.859 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 − 0.25·4) = 19.62 J
Maximum compression
½kx² = 19.62 → x = √(2·19.62/5000) = 0.08859 m = 8.859 cm
Answer
x = 8.859 cm
11.Momentum & collisions
Answer: v′2 = −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.
Momentum before
p = M·u = 4·13 = 52 kg·m/s
Momentum after
52 = 2·27 + 2·v′2
v′2 = (52 − 54)/2 = −1 m/s
Answer
v′2 = −1 m/s
The minus sign means the second fragment flies backwards.
12.Momentum & collisions
Answer: J = 19.8 N·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.
Change in momentum
J = m·Δv = 0.45·(18 + 26) = 19.8 N·s
Answer
J = 19.8 N·s
13.Springs & pendulums
Answer: g = 3.704 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 = 4π²·1.8/4.38² = 3.704 m/s²
Answer
g = 3.704 m/s²
14.Springs & pendulums
Answer: vmax = 0.8044 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 = √(110/1.7) = 8.044 rad/s
Maximum speed
vmax = A·ω = 0.1·8.044 = 0.8044 m/s
Answer
vmax = 0.8044 m/s
15.Waves, sound & optics
Answer: β = 89.03 dB
Waves: a periodic wave moves one wavelength in one period, so v = fλ with T = 1/f. On a stretched string v = √(T/μ), with T the tension and μ = m/L the mass per metre. Sound in air travels at about 343 m/s at 20 °C, light in vacuum at c = 3.00 × 10⁸ m/s. Loudness is measured on a logarithmic scale: β = 10·log₁₀(I/I₀) dB with I₀ = 10⁻¹² W/m², so 10 times the intensity adds 10 dB. A point source spreads its power P over a sphere: I = P/(4πr²).
Total intensity
I = 8·I₁
Level
β = 80 + 10·log₁₀(8) = 89.03 dB
Answer
β = 89.03 dB
16.Waves, sound & optics
Answer: NA = 0.2944
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 θc = arcsin(n₂/n₁) all light is reflected: total internal reflection. Thin lenses and spherical mirrors obey 1/f = 1/do + 1/di with the magnification m = −di/do. 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; di > 0 for a real image, di < 0 for a virtual one, and m < 0 for an inverted image.