Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.
A car speeding at a constant 24 m/s passes a police car standing still. At that moment the police car starts and accelerates uniformly at 3.5 m/s². How far from its starting point does the police car catch up? Find d.
A ball is thrown straight up at 17 m/s from the edge of a roof 5 m above the ground. Ignore air resistance and use g = 9.81 m/s². How long after the throw does it hit the ground? Find t.
A 30 kg sled is pulled across level snow by a rope at 35° above the horizontal with a force of 170 N. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². Find the acceleration a of the sled.
A 7.5 kg block on a ramp at 20° is connected by a light string over a frictionless pulley at the top of the ramp to a hanging 8 kg mass. The hanging mass moves down. The coefficient of kinetic friction between block and ramp is 0.25. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched from level ground at 34 m/s and must land 75 m away. Ignore air resistance and use g = 9.81 m/s². Find the smaller of the two possible launch angles θ.
A ball is thrown from ground level at 38 m/s and 60° above the horizontal. It lands on the flat roof of a building 18 m high, on its way down. Ignore air resistance and use g = 9.81 m/s². Find the horizontal distance x from the launch point to where it lands.
A 900 kg car drives over the top of a hump in the road at 10 m/s. The top of the hump is a circular arc of radius 50 m. Use g = 9.81 m/s². Find the normal force N of the road on the car at the top.
A small cart is released from rest on a frictionless track and runs into a vertical loop of radius 9.5 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?
A 2000 kg car drives up a 6° slope at a constant 72 km/h. Air resistance and rolling resistance together amount to 550 N. Find the power P the wheels must deliver, in kW.
A 4 kg block is released from rest at height 4 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 4.5 m long (μ = 0.25) and then runs onto a spring (k = 1500 N/m) on a smooth surface. Find the speed v of the block when it reaches the spring.
A 7 kg projectile moving horizontally at 23 m/s explodes into two fragments that move along the same line. The 1 kg fragment continues forwards at 31 m/s. Find the velocity v′2 of the 6 kg fragment (positive in the original direction).
A 900 kg car runs into a parked 1300 kg car. The wrecks lock together and skid 24.5 m before stopping. The coefficient of kinetic friction between tyres and road is 0.7. Use g = 9.81 m/s². Find the speed v1 of the first car just before the collision.
A 0.7 kg mass on a spring with k = 70 N/m oscillates on a frictionless surface with an amplitude of 11 cm. Find its speed v when it is 10 cm from equilibrium.
A 3 kg mass on a spring with k = 420 N/m is pulled 17 cm from equilibrium and released. Find its maximum acceleration amax.
An object stands 50 cm in front of a thin converging lens with a focal length of 30 cm. Find the magnification m (negative for an inverted image).
Light of wavelength 450 nm falls at right angles on a grating with 600 lines per millimetre. Find the angle θ1 of the order-1 maximum.
Answer: d = 329.1 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.
At that moment the police car moves at 2v = 48 m/s.
Answer: t = 3.739 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.
Up is positive and the ground is h₀ below the starting point.
Only the positive root is physical.
Answer: a = 2.674 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.
Answer: a = 2.325 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.
Answer: θ = 19.76 °
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.
The other solution is 90° − θ = 70.24°.
Answer: x = 116.1 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.
Larger root: the ball is descending.
Answer: N = 7029 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.
Towards the centre (down) is positive.
The car feels lighter: the road pushes up less than its weight.
Answer: hmin = 23.75 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.
Answer: P = 52.02 kW
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). Power is force times speed: P = F·v.
Answer: v = 7.51 m/s
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).
The ramp is smooth, so only the strip removes energy.
Answer: v′2 = 21.67 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.
Answer: v1 = 44.84 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.
Answer: v = 0.4583 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.
Answer: amax = 23.8 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.
Answer: m = −1.5
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.
Answer: θ1 = 15.66 °
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, …
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