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 plane needs a speed of 280 km/h to take off and the usable runway is 1800 m long. Starting from rest, what constant acceleration a does it need?

v₀ = 0v = 280 km/hs = 1800 m

a = m/s²

2.KinematicsHard

A ball is thrown straight up at 9 m/s from the edge of a roof 10 m above the ground. Ignore air resistance and use g = 9.81 m/s². Find the greatest height h_max above the ground.

v₀ = 9 m/sh₀ = 10 mg = 9.81 m/s²

h_max = m

3.Newton’s lawsMedium

A 40 kg box is pushed up a ramp at 15° to the horizontal at constant speed, by a force parallel to the ramp. The coefficient of kinetic friction is 0.25. Use g = 9.81 m/s². Find the pushing force F.

m = 40 kgθ = 15°μ = 0.25g = 9.81 m/s²

F = N

4.Newton’s lawsHard

A 1.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 5.5 kg mass. The coefficient of kinetic friction between block and table is 0.3. Use g = 9.81 m/s². Find the acceleration a of the system.

m₁ = 1.5 kgm₂ = 5.5 kgμ = 0.3g = 9.81 m/s²

a = m/s²

5.Projectile motionMedium

A ball is launched from level ground at 34 m/s and must land 95 m away. Ignore air resistance and use g = 9.81 m/s². Find the smaller of the two possible launch angles θ.

v₀ = 34 m/sR = 95 mg = 9.81 m/s²

θ = °

6.Projectile motionHard

A ball is launched at 45° above horizontal from the top of a 50 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.

v₀ = 26 m/sθ = 45°h = 50 mg = 9.81 m/s²

t = s

7.Circular motionMedium

A car takes a flat curve of radius 105 m. The coefficient of static friction between tyres and road is 0.5. Use g = 9.81 m/s². Find the highest speed v_max at which the car can take the curve without skidding.

r = 105 mμ_s = 0.5g = 9.81 m/s²

v_max = m/s

8.Circular motionHard

A 0.4 kg ball on a string 1.4 m long is swung in a vertical circle. At the lowest point it moves at 8.5 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.

m = 0.4 kgr = 1.4 mv = 8.5 m/sg = 9.81 m/s²

T = N

9.Work & energyMedium

A 2.5 kg block slides 20 m down a slope at 25° to the horizontal, with a coefficient of kinetic friction of 0.25. Find the work W_f done against friction.

m = 2.5 kgθ = 25°μ = 0.25d = 20 mg = 9.81 m/s²

W_f = J

10.Work & energyHard

A 10 kg block is released from rest at height 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 = 1500 N/m) on a smooth surface. Find the maximum compression x of the spring.

m = 10 kgh = 5 mL = 2.5 mμ = 0.25k = 1500 N/mg = 9.81 m/s²

x = cm

11.Momentum & collisionsMedium

An 18 kg projectile moving horizontally at 16 m/s explodes into two fragments that move along the same line. The 4 kg fragment continues forwards at 21 m/s. Find the velocity v′₂ of the 14 kg fragment (positive in the original direction).

M = 18 kgu = 16 m/sm₁ = 4 kgv′₁ = 21 m/sm₂ = 14 kg

v′₂ = m/s

12.Momentum & collisionsHard

A 1800 kg car runs into a parked 1000 kg car. The wrecks lock together and skid 15.5 m before stopping. The coefficient of kinetic friction between tyres and road is 0.75. Use g = 9.81 m/s². Find the speed v₁ of the first car just before the collision.

m₁ = 1800 kgm₂ = 1000 kgd = 15.5 mμ = 0.75g = 9.81 m/s²

v₁ = m/s

13.Springs & pendulumsMedium

How long must a simple pendulum be to have a period of 1 s? Use g = 9.81 m/s². Find L.

T = 1 sg = 9.81 m/s²

L = m

14.Springs & pendulumsHard

A 2 kg mass is attached to the lower end of an unstretched vertical spring with k = 80 N/m and released from rest. Use g = 9.81 m/s². Find the greatest speed v_max of the mass.

m = 2 kgk = 80 N/mg = 9.81 m/s²

v_max = m/s

3Solutions

1.Kinematics

Answer: a = 1.68 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 = 280/3.6 = 77.78 m/s

Acceleration

a = v²/(2s) = 77.78²/(2·1800) = 1.68 m/s²

Answer

a = 1.68 m/s²

2.Kinematics

Answer: h_max = 14.13 m

Rise above the roof

0 = v₀² − 2·g·Δh → Δh = 9²/(2·9.81) = 4.128 m

Height above the ground

h_max = 10 + 4.128 = 14.13 m

Answer

h_max = 14.13 m

3.Newton’s laws

Answer: F = 196.3 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.

Forces along the slope

Constant speed: the forces balance.

F = m·g·sin θ + μ·m·g·cos θ

Pushing force

F = 101.6 + 94.76 = 196.3 N

Answer

F = 196.3 N

4.Newton’s laws

Answer: a = 7.077 m/s²

Acceleration of the system

a = (m₂·g − μ·m₁·g)/(m₁ + m₂) = (53.96 − 4.414)/7 = 7.077 m/s²

Answer

a = 7.077 m/s²

5.Projectile motion

Answer: θ = 26.86 °

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.

Range formula

R = v₀²·sin 2θ/g → sin 2θ = 9.81·95/34² = 0.8062

Angle

2θ = 53.72° → θ = 26.86°

Answer

θ = 26.86°

The other solution is 90° − θ = 63.14°.

6.Projectile motion

Answer: t = 5.576 s

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 = 26·cos 45° = 18.38 m/s

v_y = v₀·sin θ = 26·sin 45° = 18.38 m/s

Time in the air

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

0 = 50 + 18.38·t − 4.905·t²

t = (v_y + √(v_y² + 2gh))/g = (18.38 + √(338 + 981))/9.81 = 5.576 s

Answer

t = 5.576 s

7.Circular motion

Answer: v_max = 22.69 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) = √(0.5·9.81·105) = 22.69 m/s

Answer

v_max = 22.69 m/s (81.7 km/h)

8.Circular motion

Answer: T = 24.57 N

Forces at the lowest point

Towards the centre (up) is positive.

T − m·g = m·v²/r

T = 0.4·9.81 + 0.4·8.5²/1.4 = 3.924 + 20.64 = 24.57 N

Answer

T = 24.57 N

9.Work & energy

Answer: W_f = 111.1 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 = 24.53·0.9063 = 22.23 N

F = μN = 0.25·22.23 = 5.557 N

Work against friction

W_f = F·d = 5.557·20 = 111.1 J

Answer

W_f = 111.1 J

10.Work & energy

Answer: x = 75.65 cm

Energy left after the rough strip

The ramp is smooth, so only the strip removes energy.

mgh − μmgL = 98.1·(5 − 0.25·2.5) = 429.2 J

Maximum compression

½kx² = 429.2 → x = √(2·429.2/1500) = 0.7565 m = 75.65 cm

Answer

x = 75.65 cm

11.Momentum & collisions

Answer: v′₂ = 14.57 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 = 18·16 = 288 kg·m/s

Momentum after

288 = 4·21 + 14·v′₂

v′₂ = (288 − 84)/14 = 14.57 m/s

Answer

v′₂ = 14.57 m/s

12.Momentum & collisions

Answer: v₁ = 23.49 m/s

Speed after the collision

V = √(2·μ·g·d) = √(2·0.75·9.81·15.5) = 15.1 m/s

Momentum in the collision

m₁·v₁ = (m₁ + m₂)·V → v₁ = 2800·15.1/1800 = 23.49 m/s

Answer

v₁ = 23.49 m/s

(84.57 km/h)

13.Springs & pendulums

Answer: L = 0.2485 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·(1/2π)² = 0.2485 m

Answer

L = 0.2485 m

14.Springs & pendulums

Answer: v_max = 1.551 m/s

Equilibrium and amplitude

x₀ = m·g/k = 2·9.81/80 = 0.2453 m = A

Greatest speed

The speed is largest at equilibrium.

ω = √(80/2) = 6.325 rad/s

v_max = A·ω = 0.2453·6.325 = 1.551 m/s

Answer

v_max = 1.551 m/s

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