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 5.5 m/s² until it stops. Find the braking distance s.

v₀ = 50 km/ha = −5.5 m/s²v = 0

s = m

2.KinematicsHard

A stone is dropped from rest into a well. The splash is heard 1.7 s after the stone is released. The speed of sound is 343 m/s; ignore air resistance and use g = 9.81 m/s². Find the depth d of the well.

t_total = 1.7 sv_sound = 343 m/sg = 9.81 m/s²

d = m

3.Newton’s lawsMedium

Masses of 6 kg and 4 kg hang on either side of a light, frictionless pulley, connected by a light string. The system is released from rest. Use g = 9.81 m/s². Find the tension T in the string.

m₁ = 6 kgm₂ = 4 kgg = 9.81 m/s²

T = N

4.Newton’s lawsHard

A 4.5 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 4 kg mass. The coefficient of kinetic friction between block and table is 0.25. Use g = 9.81 m/s². Find the tension T in the string.

m₁ = 4.5 kgm₂ = 4 kgμ = 0.25g = 9.81 m/s²

T = N

5.Projectile motionMedium

A ball is launched from level ground at 70° above the horizontal and lands 110 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v₀.

θ = 70°R = 110 mg = 9.81 m/s²

v₀ = m/s

6.Projectile motionHard

A ball is launched at 14 m/s and 40° above the horizontal from the top of a 35 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.

v₀ = 14 m/sθ = 40°h = 35 mg = 9.81 m/s²

H = m

7.Circular motionMedium

A car takes a flat curve of radius 140 m. The coefficient of static friction between tyres and road is 0.3. Use g = 9.81 m/s². Find the highest speed v_max at which the car can take the curve without skidding.

r = 140 mμ_s = 0.3g = 9.81 m/s²

v_max = m/s

8.Circular motionHard

A 1.1 kg ball on a string 1.1 m long is swung in a vertical circle. At the lowest point it moves at 5 m/s. Use g = 9.81 m/s². Find the tension T in the string at the lowest point.

m = 1.1 kgr = 1.1 mv = 5 m/sg = 9.81 m/s²

T = N

9.Work & energyMedium

A 0.5 kg block slides 19 m down a slope at 30° to the horizontal, with a coefficient of kinetic friction of 0.1. Find the work W_f done against friction.

m = 0.5 kgθ = 30°μ = 0.1d = 19 mg = 9.81 m/s²

W_f = J

10.Work & energyHard

A 10 kg block starts from rest at the bottom of a 25° slope and is pulled up the slope by a constant force of 95 N parallel to the slope. The coefficient of kinetic friction is 0.2. Find the speed v of the block after it has moved 8 m up the slope.

m = 10 kgF = 95 Nθ = 25°μ = 0.2d = 8 mg = 9.81 m/s²

v = m/s

11.Momentum & collisionsMedium

A 2.5 kg glider moving at 8 m/s makes a perfectly elastic head-on collision with a 1.5 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 = 1.5 kgv₁ = 8 m/sv₂ = 0

v′₁ = m/s

12.Momentum & collisionsHard

A 0.5 kg ball hits a wall at 24 m/s and bounces straight back at 18 m/s. The contact lasts 5 ms. Find the average force F_avg of the wall on the ball.

m = 0.5 kgv_in = 24 m/sv_out = 18 m/sΔt = 5 ms

F_avg = N

13.Springs & pendulumsMedium

How long must a simple pendulum be to have a period of 0.9 s? Use g = 9.81 m/s². Find L.

T = 0.9 sg = 9.81 m/s²

L = m

14.Springs & pendulumsHard

A 2.6 kg mass on a spring with k = 470 N/m is pulled 16 cm from equilibrium and released. Find its maximum acceleration a_max.

m = 2.6 kgk = 470 N/mA = 16 cm

a_max = m/s²

15.Waves, sound & opticsMedium

An object stands 4 cm in front of a thin diverging lens with a focal length of 5 cm. Find the image distance d_i. Use the sign convention 1/f = 1/d_o + 1/d_i (f < 0 for a diverging lens, d_i < 0 for a virtual image).

d_o = 4 cmf = −5 cm

d_i = cm

16.Waves, sound & opticsHard

An object is 20 cm in front of a concave mirror with a radius of curvature of 20 cm. Find the magnification m (negative for an inverted image).

d_o = 20 cmR = 20 cm

m =

3Solutions

1.Kinematics

Answer: s = 17.54 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.

Units

v₀ = 50/3.6 = 13.89 m/s

Braking distance

0 = v₀² − 2·a·s → s = v₀²/(2a) = 13.89²/(2·5.5) = 17.54 m

Answer

s = 17.54 m

2.Kinematics

Answer: d = 13.53 m

Two parts of the time

½·9.81·t₁² = 343·(1.7 − t₁)

4.905·t₁² + 343·t₁ − 583.1 = 0

Fall time

Only the positive root is physical.

t₁ = (−343 + √(343² + 4·4.905·583.1))/(2·4.905) = 1.661 s

Depth

d = ½·9.81·1.661² = 13.53 m

Answer

d = 13.53 m

3.Newton’s laws

Answer: T = 47.09 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.

Acceleration

The net driving force is the difference of the weights; it moves the total mass.

a = (6 − 4)·9.81/(6 + 4) = 1.962 m/s²

Tension from the lighter mass

T − m₂·g = m₂·a → T = 4·(9.81 + 1.962) = 47.09 N

Answer

T = 47.09 N

4.Newton’s laws

Answer: T = 25.97 N

Acceleration of the system

a = (m₂·g − μ·m₁·g)/(m₁ + m₂) = (39.24 − 11.04)/8.5 = 3.318 m/s²

Tension from the block on the table

T − μ·m₁·g = m₁·a

T = 4.5·3.318 + 11.04 = 25.97 N

Answer

T = 25.97 N

5.Projectile motion

Answer: v₀ = 40.97 m/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.

Solve the range formula for v₀

v₀ = √(g·R/sin 2θ) = √(9.81·110/sin 140°) = 40.97 m/s

Answer

v₀ = 40.97 m/s

6.Projectile motion

Answer: H = 39.13 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.

Vertical launch speed

v_y = 14·sin 40° = 8.999 m/s

Rise above the cliff

Δh = v_y²/(2g) = 8.999²/(2·9.81) = 4.128 m

Height above the ground

H = 35 + 4.128 = 39.13 m

Answer

H = 39.13 m

7.Circular motion

Answer: v_max = 20.3 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) = √(0.3·9.81·140) = 20.3 m/s

Answer

v_max = 20.3 m/s (73.07 km/h)

8.Circular motion

Answer: T = 35.79 N

Forces at the lowest point

Towards the centre (up) is positive.

T − m·g = m·v²/r

T = 1.1·9.81 + 1.1·5²/1.1 = 10.79 + 25 = 35.79 N

Answer

T = 35.79 N

9.Work & energy

Answer: W_f = 8.071 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 = 4.905·0.866 = 4.248 N

F = μN = 0.1·4.248 = 0.4248 N

Work against friction

W_f = F·d = 0.4248·19 = 8.071 J

Answer

W_f = 8.071 J

10.Work & energy

Answer: v = 7.564 m/s

Forces along the slope

m·g = 10·9.81 = 98.1 N

mg·sin θ = 41.46 N

μ·mg·cos θ = 17.78 N

Work–energy

½mv² = (95 − 59.24)·8 = 286.1 J

v = √(2·286.1/10) = 7.564 m/s

Answer

v = 7.564 m/s

11.Momentum & collisions

Answer: v′₁ = 2 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 − 1.5)/(2.5 + 1.5)·8 = 2 m/s

v′₂ = 2·2.5/(2.5 + 1.5)·8 = 10 m/s

Answer

v′₁ = 2 m/s

12.Momentum & collisions

Answer: F_avg = 4200 N

Change in momentum

J = m·Δv = 0.5·(18 + 24) = 21 N·s

Average force

F_avg = J/Δt = 21/0.005 = 4200 N

Answer

F_avg = 4200 N

13.Springs & pendulums

Answer: L = 0.2013 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·(0.9/2π)² = 0.2013 m

Answer

L = 0.2013 m

14.Springs & pendulums

Answer: a_max = 28.92 m/s²

Angular frequency

ω = √(470/2.6) = 13.45 rad/s

Maximum acceleration

a_max = A·ω² = 0.16·13.45² = 28.92 m/s²

Answer

a_max = 28.92 m/s²

15.Waves, sound & optics

Answer: d_i = −2.222 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) − 1/4

d_i = −2.222 cm

Ray diagram

virtual, on the same side as the object

Answer

d_i = −2.222 cm

16.Waves, sound & optics

Answer: m = −1

Focal length

f = R/2 = 10 cm

Mirror equation

1/d_i = 1/f − 1/d_o = 1/10 − 1/20

d_i = 20 cm

Magnification

m = −20/20 = −1

Answer

m = −1

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