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 45 km/h brakes with a constant deceleration of 3 m/s² until it stops. Find the braking distance s.

v₀ = 45 km/ha = −3 m/s²v = 0

s = m

2.KinematicsHard

A stone is dropped from rest into a well. The splash is heard 2.7 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 = 2.7 sv_sound = 343 m/sg = 9.81 m/s²

d = m

3.Newton’s lawsMedium

A 50 kg box is pushed up a ramp at 30° to the horizontal at constant speed, by a force parallel to the ramp. The coefficient of kinetic friction is 0.2. Use g = 9.81 m/s². Find the pushing force F.

m = 50 kgθ = 30°μ = 0.2g = 9.81 m/s²

F = N

4.Newton’s lawsHard

A block is given a push so that it starts sliding up a ramp at 35° with a speed of 10 m/s. The coefficient of kinetic friction is 0.1. Use g = 9.81 m/s². How far d along the ramp does it slide before it stops?

v₀ = 10 m/sθ = 35°μ = 0.1g = 9.81 m/s²

d = m

5.Projectile motionMedium

A ball is launched from level ground at 75° above the horizontal and lands 15 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v₀.

θ = 75°R = 15 mg = 9.81 m/s²

v₀ = m/s

6.Projectile motionHard

A ball is thrown from ground level at 35 m/s and 70° above the horizontal. It lands on the flat roof of a building 11 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 = 35 m/sθ = 70°h_roof = 11 mg = 9.81 m/s²

x = m

7.Circular motionMedium

A car takes a flat curve of radius 110 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 = 110 mμ_s = 0.3g = 9.81 m/s²

v_max = m/s

8.Circular motionHard

A 1.8 kg ball on a light string 1.2 m long is swung in a vertical circle. At the lowest point its speed is 10 m/s. Ignore air resistance and use g = 9.81 m/s². Find the tension T_top in the string at the highest point.

m = 1.8 kgr = 1.2 mv_bottom = 10 m/sg = 9.81 m/s²

T_top = N

9.Work & energyMedium

A 5 kg block slides 7 m down a slope at 35° to the horizontal, with a coefficient of kinetic friction of 0.1. Find the work W_f done against friction.

m = 5 kgθ = 35°μ = 0.1d = 7 mg = 9.81 m/s²

W_f = J

10.Work & energyHard

A spring of stiffness 5000 N/m at the bottom of a 25° slope is compressed by 20 cm against a 1 kg block and released. The coefficient of kinetic friction between block and slope is 0.15. How far d up the slope does the block travel from the release point before it stops? (Ignore the short distance over which the spring pushes, i.e. take friction and gravity over the whole distance d.)

m = 1 kgk = 5000 N/mx = 20 cmθ = 25°μ = 0.15g = 9.81 m/s²

d = m

11.Momentum & collisionsMedium

An 8 g bullet is fired horizontally into a 4 kg wooden block hanging on light strings, and stays inside it. The block swings up until it is 5 cm higher than at the start. Use g = 9.81 m/s². Find the speed v of the bullet before impact.

m = 8 gM = 4 kgh = 5 cmg = 9.81 m/s²

v = m/s

12.Momentum & collisionsHard

A 0.6 kg ball hits a wall at 28 m/s and bounces straight back at 19 m/s. The contact lasts 50 ms. Find the magnitude of the impulse J the wall gives the ball.

m = 0.6 kgv_in = 28 m/sv_out = 19 m/sΔt = 50 ms

J = N·s

13.Springs & pendulumsMedium

How long must a simple pendulum be to have a period of 1.7 s? Use g = 9.81 m/s². Find L.

T = 1.7 sg = 9.81 m/s²

L = m

14.Springs & pendulumsHard

A 2.1 kg mass is attached to the lower end of an unstretched vertical spring with k = 270 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 = 270 N/mg = 9.81 m/s²

v_max = m/s

15.Waves, sound & opticsMedium

An object 1.5 cm tall stands 10 cm in front of a thin diverging lens with a focal length of 20 cm. Find the image height h_i (negative for an inverted image).

h_o = 1.5 cmd_o = 10 cmf = −20 cm

h_i = cm

16.Waves, sound & opticsHard

An object is 5 cm in front of a concave mirror with a radius of curvature of 30 cm. Find the magnification m (negative for an inverted image).

d_o = 5 cmR = 30 cm

m =

3Solutions

1.Kinematics

Answer: s = 26.04 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 = 45/3.6 = 12.5 m/s

Braking distance

0 = v₀² − 2·a·s → s = v₀²/(2a) = 12.5²/(2·3) = 26.04 m

Answer

s = 26.04 m

2.Kinematics

Answer: d = 33.24 m

Two parts of the time

½·9.81·t₁² = 343·(2.7 − t₁)

4.905·t₁² + 343·t₁ − 926.1 = 0

Fall time

Only the positive root is physical.

t₁ = (−343 + √(343² + 4·4.905·926.1))/(2·4.905) = 2.603 s

Depth

d = ½·9.81·2.603² = 33.24 m

Answer

d = 33.24 m

3.Newton’s laws

Answer: F = 330.2 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 = 245.2 + 84.96 = 330.2 N

Answer

F = 330.2 N

4.Newton’s laws

Answer: d = 7.776 m

Deceleration

Both forces point down the slope; the mass cancels.

a = 9.81·(sin 35° + 0.1·cos 35°) = 6.43 m/s²

Distance

d = v₀²/(2a) = 10²/(2·6.43) = 7.776 m

Answer

d = 7.776 m

5.Projectile motion

Answer: v₀ = 17.16 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) = √(9.81·15/sin 150°) = 17.16 m/s

Answer

v₀ = 17.16 m/s

6.Projectile motion

Answer: x = 76.04 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 = 11.97 m/s, v_y = 32.89 m/s

Time to land on the roof

Larger root: the ball is descending.

11 = 32.89·t − 4.905·t²

t = (32.89 + √(32.89² − 2·9.81·11))/9.81 = 6.352 s

Horizontal distance

x = v_x·t = 11.97·6.352 = 76.04 m

Answer

x = 76.04 m

7.Circular motion

Answer: v_max = 17.99 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) = √(0.3·9.81·110) = 17.99 m/s

Answer

v_max = 17.99 m/s (64.77 km/h)

8.Circular motion

Answer: T_top = 61.71 N

Speed at the top

Kinetic energy is traded for potential energy over a height 2r.

v_top² = 10² − 4·9.81·1.2 = 52.91 m²/s²

Forces at the top

Tension and gravity both point down, towards the centre.

T + m·g = m·v_top²/r

T = 1.8·52.91/1.2 − 1.8·9.81 = 61.71 N

Answer

T_top = 61.71 N

9.Work & energy

Answer: W_f = 28.13 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 = 49.05·0.8192 = 40.18 N

F = μN = 0.1·40.18 = 4.018 N

Work against friction

W_f = F·d = 4.018·7 = 28.13 J

Answer

W_f = 28.13 J

10.Work & energy

Answer: d = 18.25 m

Spring energy

x = 20 cm = 0.2 m

½kx² = ½·5000·0.2² = 100 J

Energy balance along the slope

m·g = 1·9.81 = 9.81 N

½kx² = mg·d·(sin θ + μ·cos θ)

d = 100/(9.81·(0.4226 + 0.15·0.9063)) = 18.25 m

Answer

d = 18.25 m

11.Momentum & collisions

Answer: v = 496.2 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.

Swing after impact

V = √(2·g·h) = √(2·9.81·0.05) = 0.9905 m/s

Collision

m·v = (m + M)·V → v = (0.008 + 4)·0.9905/0.008 = 496.2 m/s

Answer

v = 496.2 m/s

12.Momentum & collisions

Answer: J = 28.2 N·s

Change in momentum

J = m·Δv = 0.6·(19 + 28) = 28.2 N·s

Answer

J = 28.2 N·s

13.Springs & pendulums

Answer: L = 0.7181 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·(1.7/2π)² = 0.7181 m

Answer

L = 0.7181 m

14.Springs & pendulums

Answer: v_max = 0.8652 m/s

Equilibrium and amplitude

x₀ = m·g/k = 2.1·9.81/270 = 0.0763 m = A

Greatest speed

The speed is largest at equilibrium.

ω = √(270/2.1) = 11.34 rad/s

v_max = A·ω = 0.0763·11.34 = 0.8652 m/s

Answer

v_max = 0.8652 m/s

15.Waves, sound & optics

Answer: h_i = 1 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/(−20) − 1/10

d_i = −6.667 cm

Magnification and height

m = −(−6.667)/10 = 0.6667

h_i = 0.6667·1.5 = 1 cm

Answer

h_i = 1 cm

16.Waves, sound & optics

Answer: m = 1.5

Focal length

f = R/2 = 15 cm

Mirror equation

1/d_i = 1/f − 1/d_o = 1/15 − 1/5

d_i = −7.5 cm

Magnification

m = −(−7.5)/5 = 1.5

Answer

m = 1.5

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