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 plane needs a speed of 200 km/h to take off and the usable runway is 2400 m long. Starting from rest, what constant acceleration a does it need?
v₀ = 0v = 200 km/hs = 2400 m
a = m/s²
2.KinematicsHard
A tram starts from rest and accelerates at 3 m/s² for 5 s, then runs at constant speed for 27 s, and finally brakes at 2.5 m/s² until it stops. Find the total travel time t.
a1 = 3 m/s²t1 = 5 st2 = 27 sa3 = −2.5 m/s²
t = s
3.Newton’s lawsMedium
A 3 kg block slides down a slope at 45° to the horizontal. The coefficient of kinetic friction is 0.25. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
m = 3 kgθ = 45°μ = 0.25g = 9.81 m/s²
a = m/s²
4.Newton’s lawsHard
A 1 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 1 kg mass. The coefficient of kinetic friction between block and table is 0.25. Use g = 9.81 m/s². Find the tension T in the string.
m1 = 1 kgm2 = 1 kgμ = 0.25g = 9.81 m/s²
T = N
5.Projectile motionMedium
A ball is launched from level ground at 65° above the horizontal and lands 90 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
θ = 65°R = 90 mg = 9.81 m/s²
v0 = m/s
6.Projectile motionHard
A ball is launched at 25 m/s and 40° above the horizontal from the top of a 45 m cliff. Ignore air resistance and use g = 9.81 m/s². Find the speed v with which it hits the ground.
v₀ = 25 m/sθ = 40°h = 45 mg = 9.81 m/s²
v = m/s
7.Circular motionMedium
A small ball on a string of length 1 m moves in a horizontal circle so that the string makes a constant angle of 60° with the vertical (a conical pendulum). Use g = 9.81 m/s². Find the time T for one revolution.
L = 1 mθ = 60°g = 9.81 m/s²
T = s
8.Circular motionHard
A 0.6 kg ball on a string 0.8 m long is swung in a vertical circle. At the lowest point it moves at 4 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.
m = 0.6 kgr = 0.8 mv = 4 m/sg = 9.81 m/s²
T = N
9.Work & energyMedium
A 0.5 kg block slides 6 m down a slope at 45° to the horizontal, with a coefficient of kinetic friction of 0.2. Find the work Wf done against friction.
m = 0.5 kgθ = 45°μ = 0.2d = 6 mg = 9.81 m/s²
Wf = J
10.Work & energyHard
A 2.5 kg block is released from rest at height 4 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 1.5 m long (μ = 0.25) and then runs onto a spring (k = 800 N/m) on a smooth surface. Find the maximum compression x of the spring.
m = 2.5 kgh = 4 mL = 1.5 mμ = 0.25k = 800 N/mg = 9.81 m/s²
x = cm
11.Momentum & collisionsMedium
A 3.5 kg glider moving at 12 m/s makes a perfectly elastic head-on collision with a 2 kg glider at rest on a frictionless track. Find the velocity v′1 of the first glider after the collision (positive in its original direction).
m1 = 3.5 kgm2 = 2 kgv1 = 12 m/sv2 = 0
v′1 = m/s
12.Momentum & collisionsHard
A 0.3 kg ball hits a wall at 30 m/s and bounces straight back at 13 m/s. The contact lasts 5 ms. Find the magnitude of the impulse J the wall gives the ball.
m = 0.3 kgvin = 30 m/svout = 13 m/sΔt = 5 ms
J = N·s
13.Springs & pendulumsMedium
How long must a simple pendulum be to have a period of 1.8 s? Use g = 9.81 m/s². Find L.
T = 1.8 sg = 9.81 m/s²
L = m
14.Springs & pendulumsHard
A 1.8 kg mass on a spring with k = 500 N/m is pulled 18 cm from equilibrium and released. Find its maximum speed vmax.
m = 1.8 kgk = 500 N/mA = 18 cm
vmax = m/s
15.Waves, sound & opticsMedium
A string fixed at both ends is 0.8 m long; waves travel on it at 200 m/s. Find the frequency f1 of its fundamental (first harmonic).
v = 200 m/sn = 1L = 0.8 m
f1 = Hz
16.Waves, sound & opticsHard
Two thin lenses share an axis, 40 cm apart: lens 1 with f₁ = 12 cm and lens 2 with f₂ = 10 cm. An object stands 18 cm in front of lens 1. Find the position di2 of the final image measured from lens 2 (positive beyond lens 2, real-is-positive convention).
f1 = 12 cmf2 = 10 cms = 40 cmdo = 18 cm
di2 = cm
3Solutions
1.Kinematics
Answer: a = 0.643 m/s²
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
v = 200/3.6 = 55.56 m/s
Acceleration
a = v²/(2s) = 55.56²/(2·2400) = 0.643 m/s²
Answer
a = 0.643 m/s²
2.Kinematics
Answer: t = 38 s
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.
Top speed
v = a1·t1 = 3·5 = 15 m/s
Braking time
t3 = v/a3 = 15/2.5 = 6 s
Total time
t = 5 + 27 + 6 = 38 s
Answer
t = 38 s
3.Newton’s laws
Answer: a = 5.203 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.
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) = √(9.81·90/sin 130°) = 33.95 m/s
Answer
v0 = 33.95 m/s
6.Projectile motion
Answer: v = 38.83 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.
Split the launch velocity
vx = 25·cos 40° = 19.15 m/s
vy0 = 25·sin 40° = 16.07 m/s
Vertical speed at impact
Energy or v² = v₀² + 2as in the vertical direction.
vy = √(16.07² + 2·9.81·45) = 33.78 m/s
Impact speed
v = √(19.15² + 33.78²) = 38.83 m/s
Answer
v = 38.83 m/s
Check with energy: v² = v₀² + 2gh.
7.Circular motion
Answer: T = 1.419 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·sin 60° = 0.866 m
Speed
Tension and gravity together give the inward force: tan θ = v²/(r·g).
v = √(r·g·tan θ) = √(0.866·9.81·tan 60°) = 3.836 m/s
Period
T = 2π·r/v = 2π·0.866/3.836 = 1.419 s
Answer
T = 1.419 s
8.Circular motion
Answer: T = 17.89 N
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.
Forces at the lowest point
Towards the centre (up) is positive.
T − m·g = m·v²/r
T = 0.6·9.81 + 0.6·4²/0.8 = 5.886 + 12 = 17.89 N
Answer
T = 17.89 N
9.Work & energy
Answer: Wf = 4.162 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 = 4.905·0.7071 = 3.468 N
F = μN = 0.2·3.468 = 0.6937 N
Work against friction
Wf = F·d = 0.6937·6 = 4.162 J
Answer
Wf = 4.162 J
10.Work & energy
Answer: x = 47.14 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 = 24.53·(4 − 0.25·1.5) = 88.9 J
Maximum compression
½kx² = 88.9 → x = √(2·88.9/800) = 0.4714 m = 47.14 cm
Answer
x = 47.14 cm
11.Momentum & collisions
Answer: v′1 = 3.273 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 = (3.5 − 2)/(3.5 + 2)·12 = 3.273 m/s
v′2 = 2·3.5/(3.5 + 2)·12 = 15.27 m/s
Answer
v′1 = 3.273 m/s
12.Momentum & collisions
Answer: J = 12.9 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 = 0.3·(13 + 30) = 12.9 N·s
Answer
J = 12.9 N·s
13.Springs & pendulums
Answer: L = 0.8051 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·(1.8/2π)² = 0.8051 m
Answer
L = 0.8051 m
14.Springs & pendulums
Answer: vmax = 3 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 = √(500/1.8) = 16.67 rad/s
Maximum speed
vmax = A·ω = 0.18·16.67 = 3 m/s
Answer
vmax = 3 m/s
15.Waves, sound & optics
Answer: f1 = 125 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 = 2·0.8/1 = 1.6 m
Frequency
f = v/λ = 200/1.6 = 125 Hz
Answer
f1 = 125 Hz
16.Waves, sound & optics
Answer: di2 = −6.667 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 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.