AP Physics 1 practice problems: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

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1Formulas and methods

Kinematics

Velocity: v = v₀ + a·t
Displacement: s = v₀·t + ½·a·t²
Without time: v² = v₀² + 2·a·s
Units: 1 m/s = 3.6 km/h

Method

List what you know: v₀, v, a, t and s, with units.
Pick the equation that contains your three knowns and the unknown.
Choose one positive direction and give opposing quantities a minus sign.
Solve and check that the answer is physically sensible.

Newton’s laws

Second law: ΣF = m·a
Weight: W = m·g
Kinetic friction: F_f = μ·N
On a slope: along: m·g·sin θ, normal: m·g·cos θ

Method

Draw a free-body diagram with every force on the body.
Choose a positive direction along the motion.
Split forces into components along and perpendicular to the motion (on a slope: m·g·sin θ along, m·g·cos θ into the slope).
Apply ΣF = m·a along the motion; friction is μ·N against the motion.

Projectile motion

Components: v_x = v₀cos θ, v_y = v₀sin θ
Position: x = v_xt, y = h + v_yt − ½gt²
Maximum height: H = h + v_y²/(2g)
Range (level ground): R = v₀²·sin 2θ/g

Method

Split the launch velocity into components: v_x = v₀·cos θ and v_y = v₀·sin θ.
The horizontal motion has constant speed; the vertical motion has acceleration −g.
Find the time of flight from the vertical motion.
Use that time in the horizontal motion to find the range.

Circular motion

Centripetal acceleration: a = v²/r = ω²·r
Centripetal force: F = m·v²/r
Angular speed: ω = v/r = 2π/T
Flat curve, friction: v_max = √(μ·g·r)

Work & energy

Kinetic energy: E_k = ½mv²
Potential energy: E_p = mgh
Energy balance: E_k1 + E_p1 + W = E_k2 + E_p2 + losses
Work of friction: W = μ·N·d
Power: P = W/t

Momentum & collisions

Momentum: p = m·v
Conservation: Σm·v (before) = Σm·v (after)
Perfectly inelastic: v′ = (m₁v₁ + m₂v₂)/(m₁ + m₂)
Impulse: J = F·Δt = m·Δv

Method

Choose a positive direction and give each velocity its sign.
Write the total momentum before: Σm·v.
For a perfectly inelastic collision the bodies move together after; for an elastic one kinetic energy is also conserved.
Set momentum before equal to momentum after and solve.

Springs & pendulums

Mass on a spring: T = 2π·√(m/k)
Pendulum: T = 2π·√(L/g)
Angular frequency: ω = √(k/m) = 2π/T
Maximum speed and acceleration: v_max = A·ω, a_max = A·ω²
Energy: E = ½·k·A²

2Practice problems

1.KinematicsMedium

A car driving at 72 km/h brakes with a constant deceleration of 7 m/s² until it stops. How long does braking take? Find t.

v₀ = 72 km/ha = −7 m/s²v = 0

t = s

2.KinematicsHard

A ball is thrown straight up at 7 m/s from the edge of a roof 30 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.

v₀ = 7 m/sh₀ = 30 mg = 9.81 m/s²

t = s

3.Newton’s lawsMedium

A 20 kg block slides down a slope at 20° 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 = 20 kgθ = 20°μ = 0.25g = 9.81 m/s²

a = m/s²

4.Newton’s lawsHard

A 1 kg block on a ramp at 30° is connected by a light string over a frictionless pulley at the top of the ramp to a hanging 1 kg mass. The hanging mass moves down. The ramp is frictionless. Use g = 9.81 m/s². Find the tension T in the string.

m₁ = 1 kgm₂ = 1 kgθ = 30°μ = 0g = 9.81 m/s²

T = N

5.Projectile motionMedium

A ball is launched at 60° above horizontal from level ground. Ignore air resistance. Find the maximum height H the ball reaches.

v₀ = 28 m/sθ = 60°g = 9.81 m/s²

H = m

6.Projectile motionHard

A ball is launched at 15° above horizontal from the top of a 40 m cliff. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.

v₀ = 21 m/sθ = 15°h = 40 mg = 9.81 m/s²

x = m

7.Circular motionMedium

A road engineer wants cars to take a curve of radius 390 m at 100 km/h without relying on friction. Use g = 9.81 m/s². Find the banking angle θ.

r = 390 mv = 100 km/hg = 9.81 m/s²

θ = °

8.Circular motionHard

A curve of radius 230 m is banked at 20°, and the coefficient of static friction between tyres and road is 0.1. Use g = 9.81 m/s². Find the highest speed v_max at which a car can take the curve without sliding up and out.

r = 230 mθ = 20°μ_s = 0.1g = 9.81 m/s²

v_max = m/s

9.Work & energyMedium

A toy launcher has a vertical spring of stiffness 400 N/m. A 0.5 kg ball is placed on it and the spring is compressed by 8 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 = 0.5 kgk = 400 N/mx = 8 cmg = 9.81 m/s²

h = m

10.Work & energyHard

A 2 kg block starts from rest at the bottom of a 15° slope and is pulled up the slope by a constant force of 15 N parallel to the slope. The coefficient of kinetic friction is 0.15. Find the speed v of the block after it has moved 4 m up the slope.

m = 2 kgF = 15 Nθ = 15°μ = 0.15d = 4 mg = 9.81 m/s²

v = m/s

11.Momentum & collisionsMedium

An 8 kg projectile moving horizontally at 33 m/s explodes into two fragments that move along the same line. The 2 kg fragment continues forwards at 54 m/s. Find the velocity v′₂ of the 6 kg fragment (positive in the original direction).

M = 8 kgu = 33 m/sm₁ = 2 kgv′₁ = 54 m/sm₂ = 6 kg

v′₂ = m/s

12.Momentum & collisionsHard

A 900 kg car runs into a parked 2000 kg car. The wrecks lock together and skid 17 m before stopping. The coefficient of kinetic friction between tyres and road is 0.6. Use g = 9.81 m/s². Find the speed v₁ of the first car just before the collision.

m₁ = 900 kgm₂ = 2000 kgd = 17 mμ = 0.6g = 9.81 m/s²

v₁ = m/s

13.Springs & pendulumsMedium

A 3.8 kg mass on a spring with k = 500 N/m oscillates on a frictionless surface with an amplitude of 13 cm. Find its speed v when it is 7 cm from equilibrium.

m = 3.8 kgk = 500 N/mA = 13 cmx = 7 cm

v = m/s

14.Springs & pendulumsHard

A 2.1 kg mass is attached to the lower end of an unstretched vertical spring with k = 500 N/m and released from rest. Use g = 9.81 m/s². Find the greatest speed v_max of the mass.

m = 2.1 kgk = 500 N/mg = 9.81 m/s²

v_max = m/s

3Solutions

1.Kinematics

Answer: t = 2.857 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.

Units

v₀ = 72/3.6 = 20 m/s

Braking time

0 = v₀ − a·t → t = v₀/a = 20/7 = 2.857 s

Answer

t = 2.857 s

2.Kinematics

Answer: t = 3.288 s

Equation of motion

Up is positive and the ground is h₀ below the starting point.

−30 = 7·t − ½·9.81·t²

4.905·t² − 7·t − 30 = 0

Solve the quadratic

Only the positive root is physical.

t = (7 + √(7² + 2·9.81·30))/9.81 = 3.288 s

Answer

t = 3.288 s

3.Newton’s laws

Answer: a = 1.051 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.

Forces along the slope

Down the slope is positive.

m·a = m·g·sin θ − μ·m·g·cos θ

Acceleration

a = 9.81·(sin 20° − 0.25·cos 20°) = 9.81·(0.342 − 0.2349) = 1.051 m/s²

Answer

a = 1.051 m/s²

4.Newton’s laws

Answer: T = 7.357 N

Forces along the string

m₂·g = 9.81 N

m₁·g·sin θ = 4.905 N

μ·m₁·g·cos θ = 0 N

Acceleration of the system

a = (9.81 − 4.905 − 0)/2 = 2.453 m/s²

Tension from the hanging mass

m₂·g − T = m₂·a → T = 1·(9.81 − 2.453) = 7.357 N

Answer

T = 7.357 N

5.Projectile motion

Answer: H = 29.97 m

Projectile motion: with no air resistance, x = v_xt and y = h + v_yt − 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.

v_x = v₀·cos θ = 28·cos 60° = 14 m/s

v_y = v₀·sin θ = 28·sin 60° = 24.25 m/s

Highest point

At the top the vertical velocity is zero: v_y² = 2gH.

H = v_y²/(2g) = 24.25²/(2·9.81) = 29.97 m

Answer

H = 29.97 m

6.Projectile motion

Answer: x = 70.24 m

Projectile motion: with no air resistance, x = v_xt and y = h + v_yt − 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.

v_x = v₀·cos θ = 21·cos 15° = 20.28 m/s

v_y = v₀·sin θ = 21·sin 15° = 5.435 m/s

Time in the air

Solve y(t) = h + v_y·t − ½gt² = 0 for the positive root.

0 = 40 + 5.435·t − 4.905·t²

t = (v_y + √(v_y² + 2gh))/g = (5.435 + √(29.54 + 784.8))/9.81 = 3.463 s

Horizontal distance

x = v_x·t = 20.28·3.463 = 70.24 m

Answer

x = 70.24 m

7.Circular motion

Answer: θ = 11.4 °

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.

Units

v = 100/3.6 = 27.78 m/s

Banking angle

tan θ = v²/(r·g) = 27.78²/(390·9.81) = 0.2017

θ = 11.4°

Answer

θ = 11.4°

8.Circular motion

Answer: v_max = 32.96 m/s

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 = √(230·9.81·0.436/0.9055) = 32.96 m/s

Answer

v_max = 32.96 m/s

(118.7 km/h)

9.Work & energy

Answer: h = 0.261 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 = 8 cm = 0.08 m

½kx² = ½·400·0.08² = 1.28 J

Height

m·g = 0.5·9.81 = 4.905 N

h = ½kx²/(mg) = 1.28/4.905 = 0.261 m

Answer

h = 0.261 m

10.Work & energy

Answer: v = 5.321 m/s

Forces along the slope

m·g = 2·9.81 = 19.62 N

mg·sin θ = 5.078 N

μ·mg·cos θ = 2.843 N

Work–energy

½mv² = (15 − 7.921)·4 = 28.32 J

v = √(2·28.32/2) = 5.321 m/s

Answer

v = 5.321 m/s

11.Momentum & collisions

Answer: v′₂ = 26 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 = 8·33 = 264 kg·m/s

Momentum after

264 = 2·54 + 6·v′₂

v′₂ = (264 − 108)/6 = 26 m/s

Answer

v′₂ = 26 m/s

12.Momentum & collisions

Answer: v₁ = 45.58 m/s

Speed after the collision

V = √(2·μ·g·d) = √(2·0.6·9.81·17) = 14.15 m/s

Momentum in the collision

m₁·v₁ = (m₁ + m₂)·V → v₁ = 2900·14.15/900 = 45.58 m/s

Answer

v₁ = 45.58 m/s

(164.1 km/h)

13.Springs & pendulums

Answer: v = 1.257 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 = √(500/3.8) = 11.47 rad/s

Speed at x

v = ω·√(A² − x²) = 11.47·√(0.13² − 0.07²) = 1.257 m/s

Answer

v = 1.257 m/s

14.Springs & pendulums

Answer: v_max = 0.6358 m/s

Equilibrium and amplitude

x₀ = m·g/k = 2.1·9.81/500 = 0.0412 m = A

Greatest speed

The speed is largest at equilibrium.

ω = √(500/2.1) = 15.43 rad/s

v_max = A·ω = 0.0412·15.43 = 0.6358 m/s

Answer

v_max = 0.6358 m/s

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