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 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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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.
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, …