Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
A driver moving at 80 km/h sees an obstacle. Her reaction time is 1 s, after which the car brakes with a constant deceleration of 4 m/s². Find the total stopping distance d from the moment she sees the obstacle.
A ball is thrown straight up at 7 m/s from the edge of a roof 35 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 9 kg block slides down a slope at 30° to the horizontal. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
Two blocks on a level floor are joined by a light horizontal string. A horizontal force of 105 N pulls the front 10 kg block; the 12 kg block is towed behind. The coefficient of kinetic friction is 0.1 for both blocks. Use g = 9.81 m/s². Find the tension T in the connecting string.
A ball is launched at 30° above horizontal from level ground. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.
A ball is launched at 19 m/s and 55° 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.
A car takes a flat curve of radius 110 m. The coefficient of static friction between tyres and road is 0.6. Use g = 9.81 m/s². Find the highest speed vmax at which the car can take the curve without skidding.
A 1.3 kg ball on a light string 0.9 m long is swung in a vertical circle. At the lowest point its speed is 9.5 m/s. Ignore air resistance and use g = 9.81 m/s². Find the tension Ttop in the string at the highest point.
A 2 kg ball hangs on a light string of length 0.8 m. It is pulled aside until the string makes 50° with the vertical and released from rest. Find the tension T in the string when the ball passes the lowest point.
A 5 kg cart is released from rest at a height of 3.2 m on a smooth track that leads into a vertical circular loop of radius 1 m. Find the normal force N the track exerts on the cart at the top of the loop.
A 12 g bullet is fired horizontally into a 1 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 16 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.
A 0.25 kg ball hits a wall at 25 m/s and bounces straight back at 16 m/s. The contact lasts 40 ms. Find the magnitude of the impulse J the wall gives the ball.
A 2.3 kg mass on a spring with k = 210 N/m is pulled 13 cm in the positive direction and released from rest at t = 0. Find its displacement x from equilibrium at t = 0.25 s (with sign).
A body oscillates in simple harmonic motion with period 0.7 s and amplitude 14 cm. What is the shortest time t it needs to move from equilibrium to a displacement of 8.4 cm?
Answer: d = 83.95 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.
Answer: t = 3.478 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 = 4.055 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.
Down the slope is positive.
Answer: T = 57.27 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.
With the same μ on both blocks, friction drops out: T = F·m₁/(m₁ + m₂).
Answer: x = 17.3 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.
Horizontal velocity stays constant; vertical motion has a constant acceleration g = 9.81 m/s² downward.
Starting and landing at the same height: t = 2vy/g.
Answer: H = 62.35 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.
Answer: vmax = 25.45 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.
Answer: Ttop = 66.6 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.
Kinetic energy is traded for potential energy over a height 2r.
Tension and gravity both point down, towards the centre.
Answer: T = 33.64 N
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). On a circular path the net force towards the centre must equal mv²/r.
Check: T = mg(3 − 2cos θ) = 33.64 N.
Answer: N = 68.67 N
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). On a circular path the net force towards the centre must equal mv²/r.
The cart just makes it round when h = 2.5R; then N = 0 at the top.
Answer: v = 149.4 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: J = 10.25 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.
Answer: x = −9.487 cm
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: t = 0.07169 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.
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