Physics 1: Mechanics: cram pack

Formula sheets, the key method steps for every topic and 16 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

Waves, sound & optics

Wave speed: v = fλ, T = 1/f
String: v = √(T/μ), μ = m/L
Sound level: β = 10·log₁₀(I/I₀), I₀ = 10⁻¹² W/m²
Point source: I = P/(4πr²)
Standing waves: string or open pipe f_n = n·v/(2L); closed pipe f_n = n·v/(4L), n odd
Doppler effect: f′ = f·(v ± v_o)/(v ∓ v_s)
Refractive index: n = c/v, λ = λ₀/n
Snell’s law: n₁sin θ₁ = n₂sin θ₂
Critical angle: sin θ_c = n₂/n₁
Optical fibre: NA = sin θ_a = √(n₁² − n₂²)
Lenses and mirrors: 1/f = 1/d_o + 1/d_i, m = −d_i/d_o, mirror f = R/2
Double slit: d·sin θ = mλ, y_m = mλL/d
Grating: d = 1/N, d·sin θ = mλ, m_max ≤ d/λ
Single slit: a·sin θ = mλ (dark), w = 2λL/a

Method

Identify the kind of problem: wave speed, sound level, standing wave, Doppler, refraction, lens or mirror, or interference.
Write the relation first: v = fλ, β = 10·log(I/I₀), fₙ = n·v/(2L) or n·v/(4L), f′ = f(v ± v_o)/(v ∓ v_s), n₁sin θ₁ = n₂sin θ₂, 1/f = 1/d₀ + 1/dᵢ, d·sin θ = mλ.
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 car driving at 50 km/h brakes with a constant deceleration of 7.5 m/s² until it stops. How long does braking take? Find t.

v₀ = 50 km/ha = −7.5 m/s²v = 0

t = s

2.KinematicsHard

A model rocket is launched straight up from rest. Its engine gives it a constant upward acceleration of 26 m/s² for 2 s, after which the engine stops and the rocket moves freely. Ignore air resistance and use g = 9.81 m/s². Find the greatest height h_max the rocket reaches.

a = 26 m/s²t₁ = 2 sg = 9.81 m/s²

h_max = m

3.Newton’s lawsMedium

A box slides across a level floor with an initial speed of 4 m/s and comes to rest after 2 m. Use g = 9.81 m/s². Find the coefficient of kinetic friction μ.

v₀ = 4 m/sd = 2 mg = 9.81 m/s²

μ =

4.Newton’s lawsHard

A 9 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 5 kg mass. The table is frictionless. Use g = 9.81 m/s². Find the acceleration a of the system.

m₁ = 9 kgm₂ = 5 kgμ = 0g = 9.81 m/s²

a = m/s²

5.Projectile motionMedium

A ball is launched at 45° above horizontal from level ground. Ignore air resistance. Find the horizontal distance x travelled before the ball hits the ground.

v₀ = 14 m/sθ = 45°g = 9.81 m/s²

x = m

6.Projectile motionHard

A ball is launched at 40° above horizontal from the top of a 20 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.

v₀ = 32 m/sθ = 40°h = 20 mg = 9.81 m/s²

t = s

7.Circular motionMedium

A car takes a flat curve of radius 70 m. The coefficient of static friction between tyres and road is 0.4. Use g = 9.81 m/s². Find the highest speed v_max at which the car can take the curve without skidding.

r = 70 mμ_s = 0.4g = 9.81 m/s²

v_max = m/s

8.Circular motionHard

A ball on a string 1.5 m long is swung in a vertical circle. Use g = 9.81 m/s². What is the smallest speed v_min at the top for which the string just stays tight?

r = 1.5 mg = 9.81 m/s²

v_min = m/s

9.Work & energyMedium

A 1 kg ball hangs on a light string of length 3 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.

m = 1 kgL = 3 mθ = 50°g = 9.81 m/s²

T = N

10.Work & energyHard

A 0.5 kg cart is released from rest at a height of 9 m on a smooth track that leads into a vertical circular loop of radius 2 m. Find the normal force N the track exerts on the cart at the top of the loop.

m = 0.5 kgh = 9 mR = 2 mg = 9.81 m/s²

N = N

11.Momentum & collisionsMedium

A 2.5 kg glider moving at 4 m/s makes a perfectly elastic head-on collision with a 5 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 = 2.5 kgm₂ = 5 kgv₁ = 4 m/sv₂ = 0

v′₁ = m/s

12.Momentum & collisionsHard

A 0.45 kg ball hits a wall at 25 m/s and bounces straight back at 5 m/s. The contact lasts 50 ms. Find the average force F_avg of the wall on the ball.

m = 0.45 kgv_in = 25 m/sv_out = 5 m/sΔt = 50 ms

F_avg = N

13.Springs & pendulumsMedium

How long must a simple pendulum be to have a period of 2.1 s? Use g = 9.81 m/s². Find L.

T = 2.1 sg = 9.81 m/s²

L = m

14.Springs & pendulumsHard

A 2.8 kg mass on a spring with k = 80 N/m is pulled 15 cm from equilibrium and released. Find its maximum speed v_max.

m = 2.8 kgk = 80 N/mA = 15 cm

v_max = m/s

15.Waves, sound & opticsMedium

An object 3 cm tall stands 40 cm in front of a thin diverging lens with a focal length of 5 cm. Find the image height h_i (negative for an inverted image).

h_o = 3 cmd_o = 40 cmf = −5 cm

h_i = cm

16.Waves, sound & opticsHard

Light of wavelength 500 nm falls at right angles on a grating with 500 lines per millimetre. Find the angle θ₂ of the order-2 maximum.

N = 500 mm⁻¹λ = 500 nmm = 2

θ₂ = °

3Solutions

1.Kinematics

Answer: t = 1.852 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 = 50/3.6 = 13.89 m/s

Braking time

0 = v₀ − a·t → t = v₀/a = 13.89/7.5 = 1.852 s

Answer

t = 1.852 s

2.Kinematics

Answer: h_max = 189.8 m

Powered phase

v₁ = 26·2 = 52 m/s

h₁ = ½·26·2² = 52 m

Free flight up

At the top the velocity is zero.

Δh = v₁²/(2g) = 52²/(2·9.81) = 137.8 m

h_max = 52 + 137.8 = 189.8 m

Answer

h_max = 189.8 m

3.Newton’s laws

Answer: μ = 0.4077

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.

Deceleration

a = v₀²/(2d) = 4²/(2·2) = 4 m/s²

Friction coefficient

μ·m·g = m·a → μ = a/g = 4/9.81 = 0.4077

Answer

μ = 0.4077

4.Newton’s laws

Answer: a = 3.504 m/s²

Acceleration of the system

a = (m₂·g − μ·m₁·g)/(m₁ + m₂) = (49.05 − 0)/14 = 3.504 m/s²

Answer

a = 3.504 m/s²

5.Projectile motion

Answer: x = 19.98 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 = 14·cos 45° = 9.899 m/s

v_y = v₀·sin θ = 14·sin 45° = 9.899 m/s

Time in the air

Starting and landing at the same height: t = 2v_y/g.

t = 2·9.899/9.81 = 2.018 s

Horizontal distance

x = v_x·t = 9.899·2.018 = 19.98 m

Answer

x = 19.98 m

6.Projectile motion

Answer: t = 5.008 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 = 32·cos 40° = 24.51 m/s

v_y = v₀·sin θ = 32·sin 40° = 20.57 m/s

Time in the air

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

0 = 20 + 20.57·t − 4.905·t²

t = (v_y + √(v_y² + 2gh))/g = (20.57 + √(423.1 + 392.4))/9.81 = 5.008 s

Answer

t = 5.008 s

7.Circular motion

Answer: v_max = 16.57 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) = √(0.4·9.81·70) = 16.57 m/s

Answer

v_max = 16.57 m/s (59.66 km/h)

8.Circular motion

Answer: v_min = 3.836 m/s

Tension zero at the top

m·g = m·v²/r → v = √(g·r) = √(9.81·1.5) = 3.836 m/s

Answer

v_min = 3.836 m/s

9.Work & energy

Answer: T = 16.82 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.

Speed at the bottom

v² = 2gL(1 − cos θ) = 2·9.81·3·(1 − cos 50°) = 21.03 m²/s²

Forces at the bottom

m·g = 1·9.81 = 9.81 N

T = mg + mv²/L = 9.81 + 1·21.03/3 = 16.82 N

Answer

T = 16.82 N

Check: T = mg(3 − 2cos θ) = 16.82 N.

10.Work & energy

Answer: N = 19.62 N

Speed at the top

v² = 2g(h − 2R) = 2·9.81·(9 − 4) = 98.1 m²/s²

Forces at the top

m·g = 0.5·9.81 = 4.905 N

N = mv²/R − mg = 0.5·98.1/2 − 4.905 = 19.62 N

Answer

N = 19.62 N

The cart just makes it round when h = 2.5R; then N = 0 at the top.

11.Momentum & collisions

Answer: v′₁ = −1.333 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 = (2.5 − 5)/(2.5 + 5)·4 = −1.333 m/s

v′₂ = 2·2.5/(2.5 + 5)·4 = 2.667 m/s

Answer

v′₁ = −1.333 m/s

The minus sign means the first glider bounces back.

12.Momentum & collisions

Answer: F_avg = 270 N

Change in momentum

J = m·Δv = 0.45·(5 + 25) = 13.5 N·s

Average force

F_avg = J/Δt = 13.5/0.05 = 270 N

Answer

F_avg = 270 N

13.Springs & pendulums

Answer: L = 1.096 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·(2.1/2π)² = 1.096 m

Answer

L = 1.096 m

14.Springs & pendulums

Answer: v_max = 0.8018 m/s

Angular frequency

ω = √(80/2.8) = 5.345 rad/s

Maximum speed

v_max = A·ω = 0.15·5.345 = 0.8018 m/s

Answer

v_max = 0.8018 m/s

15.Waves, sound & optics

Answer: h_i = 0.3333 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 θ_c = arcsin(n₂/n₁) all light is reflected: total internal reflection. Thin lenses and spherical mirrors obey 1/f = 1/d_o + 1/d_i with the magnification m = −d_i/d_o. 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; d_i > 0 for a real image, d_i < 0 for a virtual one, and m < 0 for an inverted image.

Thin-lens equation

1/d_i = 1/f − 1/d_o = 1/(−5) − 1/40

d_i = −4.444 cm

Magnification and height

m = −(−4.444)/40 = 0.1111

h_i = 0.1111·3 = 0.3333 cm

Answer

h_i = 0.3333 cm

16.Waves, sound & optics

Answer: θ₂ = 30 °

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, …

Line spacing

d = 1/500 mm = 2 × 10⁻⁶ m

Grating equation

sin θ = mλ/d = 2·500 × 10⁻⁹/(2 × 10⁻⁶) = 0.5

θ = 30°

Answer

θ₂ = 30°

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