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 plane starts from rest and accelerates uniformly at 3.75 m/s² along a runway. Find its speed v after 1100 m.

v₀ = 0a = 3.75 m/s²s = 1100 m

v = m/s

2.KinematicsHard

A ball is thrown straight up at 11 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 = 11 m/sh₀ = 30 mg = 9.81 m/s²

t = s

3.Newton’s lawsMedium

A 30 kg sled is pulled across level snow by a rope at 25° above the horizontal with a force of 190 N. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². Find the acceleration a of the sled.

m = 30 kgF = 190 Nθ = 25°μ = 0.1g = 9.81 m/s²

a = m/s²

4.Newton’s lawsHard

A block is given a push so that it starts sliding up a ramp at 30° with a speed of 5 m/s. The coefficient of kinetic friction is 0.15. Use g = 9.81 m/s². How far d along the ramp does it slide before it stops?

v₀ = 5 m/sθ = 30°μ = 0.15g = 9.81 m/s²

d = m

5.Projectile motionMedium

A ball is launched at 30° above horizontal from level ground. Ignore air resistance. Find the time of flight t until the ball hits the ground.

v₀ = 16 m/sθ = 30°g = 9.81 m/s²

t = s

6.Projectile motionHard

A ball is thrown from ground level at 38 m/s and 45° above the horizontal. It lands on the flat roof of a building 4 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.

v₀ = 38 m/sθ = 45°h_roof = 4 mg = 9.81 m/s²

x = m

7.Circular motionMedium

A car takes a flat curve of radius 35 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 = 35 mμ_s = 0.4g = 9.81 m/s²

v_max = m/s

8.Circular motionHard

A 3.5 kg cart is released from rest at a height of 48 m on a frictionless track and runs through a vertical loop of radius 10 m. Use g = 9.81 m/s². Find the normal force N_top of the track on the cart at the top of the loop.

m = 3.5 kgh = 48 mR = 10 mg = 9.81 m/s²

N_top = N

9.Work & energyMedium

A 2 kg block slides 12 m down a slope at 30° to the horizontal, with a coefficient of kinetic friction of 0.2. Find the work W_f done against friction.

m = 2 kgθ = 30°μ = 0.2d = 12 mg = 9.81 m/s²

W_f = J

10.Work & energyHard

An 8 kg block starts from rest at the bottom of a 25° slope and is pulled up the slope by a constant force of 65 N parallel to the slope. The coefficient of kinetic friction is 0.1. Find the speed v of the block after it has moved 6 m up the slope.

m = 8 kgF = 65 Nθ = 25°μ = 0.1d = 6 mg = 9.81 m/s²

v = m/s

11.Momentum & collisionsMedium

A 4 kg glider moving at 10 m/s makes a perfectly elastic head-on collision with a 3.5 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 = 4 kgm₂ = 3.5 kgv₁ = 10 m/sv₂ = 0

v′₁ = m/s

12.Momentum & collisionsHard

A 10 g bullet is fired horizontally into a 1.5 kg block hanging from light strings 0.6 m long, and stays in it. The strings swing out to a largest angle of 35° from the vertical. Use g = 9.81 m/s². Find the speed v of the bullet before impact.

m = 10 gM = 1.5 kgL = 0.6 mθ = 35°g = 9.81 m/s²

v = m/s

13.Springs & pendulumsMedium

A simple pendulum has a length of 2.3 m. Use g = 9.81 m/s². Find its period T for small swings.

L = 2.3 mg = 9.81 m/s²

T = s

14.Springs & pendulumsHard

A 0.9 kg mass is attached to the lower end of an unstretched vertical spring with k = 320 N/m and released from rest. Use g = 9.81 m/s². Find the greatest extension x_max of the spring in cm.

m = 0.9 kgk = 320 N/mg = 9.81 m/s²

x_max = cm

15.Waves, sound & opticsMedium

A small loudspeaker radiates 5 W of sound power equally in all directions. Find the sound intensity level β at 5 m.

P = 5 Wr = 5 mI₀ = 10⁻¹² W/m²

β = dB

16.Waves, sound & opticsHard

Light of wavelength 550 nm falls on two slits 0.15 mm apart. The screen is 3 m away. Find the distance y₄ from the central bright fringe to bright fringe number 4.

λ = 550 nmd = 0.15 mmL = 3 mm = 4

y₄ = mm

3Solutions

1.Kinematics

Answer: v = 90.83 m/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.

Speed after the distance

v = √(2·a·s) = √(2·3.75·1100) = 90.83 m/s

Answer

v = 90.83 m/s

(327 km/h)

2.Kinematics

Answer: t = 3.837 s

Equation of motion

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

−30 = 11·t − ½·9.81·t²

4.905·t² − 11·t − 30 = 0

Solve the quadratic

Only the positive root is physical.

t = (11 + √(11² + 2·9.81·30))/9.81 = 3.837 s

Answer

t = 3.837 s

3.Newton’s laws

Answer: a = 5.027 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.

Normal force

N = m·g − F·sin θ = 294.3 − 190·sin 25° = 214 N

Horizontal forces

m·a = F·cos θ − μ·N = 172.2 − 21.4

a = 150.8/30 = 5.027 m/s²

Answer

a = 5.027 m/s²

4.Newton’s laws

Answer: d = 2.023 m

Deceleration

Both forces point down the slope; the mass cancels.

a = 9.81·(sin 30° + 0.15·cos 30°) = 6.179 m/s²

Distance

d = v₀²/(2a) = 5²/(2·6.179) = 2.023 m

Answer

d = 2.023 m

5.Projectile motion

Answer: t = 1.631 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 = 16·cos 30° = 13.86 m/s

v_y = v₀·sin θ = 16·sin 30° = 8 m/s

Time in the air

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

t = 2·8/9.81 = 1.631 s

Answer

t = 1.631 s

6.Projectile motion

Answer: x = 143.1 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

v_x = 26.87 m/s, v_y = 26.87 m/s

Time to land on the roof

Larger root: the ball is descending.

4 = 26.87·t − 4.905·t²

t = (26.87 + √(26.87² − 2·9.81·4))/9.81 = 5.325 s

Horizontal distance

x = v_x·t = 26.87·5.325 = 143.1 m

Answer

x = 143.1 m

7.Circular motion

Answer: v_max = 11.72 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) = √(0.4·9.81·35) = 11.72 m/s

Answer

v_max = 11.72 m/s (42.19 km/h)

8.Circular motion

Answer: N_top = 157.9 N

Speed from energy conservation

v² = 2·9.81·(48 − 2·10) = 549.4 m²/s²

Forces

Towards the centre is positive.

N = m·v²/R − m·g = 3.5·549.4/10 − 34.34 = 157.9 N

Answer

N_top = 157.9 N

9.Work & energy

Answer: W_f = 40.78 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 = 19.62·0.866 = 16.99 N

F = μN = 0.2·16.99 = 3.398 N

Work against friction

W_f = F·d = 3.398·12 = 40.78 J

Answer

W_f = 40.78 J

10.Work & energy

Answer: v = 6.089 m/s

Forces along the slope

m·g = 8·9.81 = 78.48 N

mg·sin θ = 33.17 N

μ·mg·cos θ = 7.113 N

Work–energy

½mv² = (65 − 40.28)·6 = 148.3 J

v = √(2·148.3/8) = 6.089 m/s

Answer

v = 6.089 m/s

11.Momentum & collisions

Answer: v′₁ = 0.6667 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 = (4 − 3.5)/(4 + 3.5)·10 = 0.6667 m/s

v′₂ = 2·4/(4 + 3.5)·10 = 10.67 m/s

Answer

v′₁ = 0.6667 m/s

12.Momentum & collisions

Answer: v = 220.3 m/s

Rise of the block

h = L·(1 − cos θ) = 0.6·(1 − cos 35°) = 0.1085 m

Speed after impact

V = √(2·9.81·0.1085) = 1.459 m/s

Collision

v = (m + M)·V/m = 1.51·1.459/0.01 = 220.3 m/s

Answer

v = 220.3 m/s

13.Springs & pendulums

Answer: T = 3.042 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.

Pendulum period

T = 2π·√(2.3/9.81) = 3.042 s

Answer

T = 3.042 s

14.Springs & pendulums

Answer: x_max = 5.518 cm

Equilibrium and amplitude

x₀ = m·g/k = 0.9·9.81/320 = 0.02759 m = A

Lowest point

x_max = 2A = 0.05518 m = 5.518 cm

Answer

x_max = 5.518 cm

15.Waves, sound & optics

Answer: β = 102 dB

Waves: a periodic wave moves one wavelength in one period, so v = fλ with T = 1/f. On a stretched string v = √(T/μ), with T the tension and μ = m/L the mass per metre. Sound in air travels at about 343 m/s at 20 °C, light in vacuum at c = 3.00 × 10⁸ m/s. Loudness is measured on a logarithmic scale: β = 10·log₁₀(I/I₀) dB with I₀ = 10⁻¹² W/m², so 10 times the intensity adds 10 dB. A point source spreads its power P over a sphere: I = P/(4πr²).

Intensity

I = P/(4πr²) = 0.01592 W/m²

Level

β = 10·log₁₀(0.01592/10⁻¹²) = 102 dB

Answer

β = 102 dB

16.Waves, sound & optics

Answer: y₄ = 44 mm

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, …

Fringe spacing

Δy = λL/d = 550 × 10⁻⁹·3/(0.15 × 10⁻³) = 11 mm

Fringe number 4

y₄ = 4·11 = 44 mm

Answer

y₄ = 44 mm

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