Forces and momentum

A sketch of **one object as a dot**, with an **arrow for every force acting ON it** (and nothing it pushes on other things).

The **net (resultant) force is zero**, so the object stays at rest or moves at **constant velocity**.

A **vector** — it has a size (in newtons) **and** a direction.

Horizontal $A_{H} = A\cos\theta$, vertical $A_{V} = A\sin\theta$. **Given** in the data booklet.

Split it into a **horizontal** and a **vertical** part that together do the same job.

**cos** = the side **next to** the angle (horizontal); **sin** = the side **opposite** it (vertical).

A **pull along a rope or string**, acting on the object **away** from it along the rope.

Resolve every force, then set the total to **zero in each direction separately** (left = right, up = down).

Only its **small vertical part** ($A\sin\theta$) holds the weight, so the **full tension** must be huge.

$F_g = mg$ — mass × gravitational field strength (g = 9.8 N kg⁻¹). **Given** in the data booklet.

**No.** At rest is one case; moving at **constant velocity** is also equilibrium (net force still zero).

The **horizontal** ($A\cos 50°$) is slightly larger, since cos 50° > sin 50° — but check the angle's reference each time.

With **zero net force**, an object stays at rest or keeps moving at **constant velocity**. (Motion needs no force — only a change in motion does.)

The **net force** equals mass × acceleration: **F = ma**, with the acceleration in the same direction as the net force.

If A exerts a force on B, then **B exerts an equal and opposite force on A**. The pair acts on **different objects**.

The **newton (N)**. 1 N = 1 kg m s⁻² (the force that gives a 1 kg mass an acceleration of 1 m s⁻²).

The **net (resultant)** force — every force on the object added together, with direction.

Because they act on **different objects**. Two forces only cancel when they act on the **same** object.

The **same acceleration** — connected bodies move together.

Apply **F = ma** to **one** of the masses on its own: tension = that mass × the shared acceleration.

**Bigger** — the cable must support the weight **and** provide the extra net force to accelerate it up (T − mg = ma).

$F = ma = \dfrac{\Delta p}{\Delta t}$ — net force = mass × acceleration = rate of change of momentum.

a = F ÷ m = 20 ÷ 5.0 = **4.0 m s⁻²**.

A **single** force is just one push/pull; the **net** force is all of them combined. Only the net force goes into F = ma.

The force that **resists sliding** between two surfaces in contact; it always **opposes the motion** (or attempted motion).

**Static** acts while the object is **still** (grows to match the push, up to μ_s R). **Dynamic** acts while it is **sliding** (a fixed μ_d R).

$F_f \le \mu_s R$ — friction can be anything up to a maximum of μ_s R.

$F_f = \mu_d R$ — a fixed value while the object moves.

The **support force** from the surface, perpendicular to it. On flat ground **R = mg**. Also written F_N.

**μ_s R** — you must beat the **maximum static** friction (use μ_s, not μ_d).

The **maximum static** friction — that's why it's harder to start something moving than to keep it moving.

μ = F_f ÷ R is a **force ÷ a force**, so the newtons cancel — it has **no unit** (a pure number).

R = mg = 98 N, so F_f = μ_d R = 0.20 × 98 = 19.6 ≈ 20 N.

**No** (in this model) — it depends on μ and the normal force R, not on how big the contact patch is.

Usually between **0 and 1** (e.g. ~0.3 for many everyday surfaces); it can exceed 1 for very grippy surfaces.

The **buoyancy (upthrust) force** on an object equals the **weight of the fluid it pushes aside** (displaces).

The **upward force** a fluid exerts on an object, because the fluid presses harder underneath than on top.

$F_b = \rho V g$ — fluid density × displaced (submerged) volume × g. **Given** in the data booklet.

The **fluid's** density — not the object's.

The **submerged** volume — the volume of fluid pushed aside.

When it is **less dense** than the fluid, so the buoyancy can balance its weight.

Buoyancy = weight: $\rho_{fluid} V_{sub}\, g = \rho_{obj} V_{total}\, g$.

The **density ratio**: ρ_object ÷ ρ_fluid.

$\rho = \dfrac{m}{V}$ — mass ÷ volume. **Given** in the data booklet.

Buoyancy ∝ submerged volume (F_b = ρVg), so the ratio of upthrusts = ratio of submerged volumes.

Ice (≈9.2 × 10²) is only slightly less dense than seawater (≈1.03 × 10³), so the submerged fraction ≈ 0.89.

Using the **object's** density for ρ, or the **whole** volume when only part is submerged.

A **resistive force** a fluid (air or liquid) exerts on an object moving through it. It points **against the motion** and **grows with speed**.

The **steady (constant) speed** a falling object reaches when the **drag balances the weight**, so the net force — and the acceleration — is zero.

How **thick or sticky** a fluid is (symbol η, unit Pa s). Honey has high viscosity; water has low viscosity.

$F_d = 6\pi\eta r v$ — drag grows with viscosity η, radius r and speed v. **Given** in the data booklet.

**Weight = drag**: $mg = 6\pi\eta r v$ (net force zero, so steady speed).

About **g** — there's no drag yet because the speed is zero.

It **starts near g and decreases to zero** as drag builds up — it is **not** constant.

The **terminal velocity** — speed constant, acceleration zero, drag = weight.

**v ∝ r²** — weight ∝ r³ and Stokes drag ∝ r, so doubling the radius gives **4×** the terminal velocity.

Assuming the acceleration is **constant** while falling. It isn't — it falls from ≈ g to zero as drag grows.

Constant speed ⇒ no acceleration ⇒ net force = 0, so the upward drag exactly balances the downward weight.

The **net (resultant) force** that points **toward the centre** of a circle and keeps an object moving in that circle.

**Toward the centre**, along the radius — never along the direction of motion.

**Yes** — its direction keeps changing, so its velocity changes (it accelerates toward the centre).

$F_c = \dfrac{mv^2}{r}$ — from $F = ma$ with $a = \dfrac{v^2}{r}$.

$a = \dfrac{v^2}{r} = \omega^2 r = \dfrac{4\pi^2 r}{T^2}$ (in the data booklet).

$v = \dfrac{2\pi r}{T} = \omega r$ (in the data booklet).

It becomes **4× bigger** — because $F_c \propto v^2$.

$T - mg = \dfrac{mv^2}{r}$, so $T = mg + \dfrac{mv^2}{r}$ — the tension is **greater** than the weight.

**Friction** between the tyres and the road (pointing toward the centre).

**No** — F_c is the **net** of the real forces (friction, tension, gravity, normal). Never draw it as a separate arrow.

$F_c = \dfrac{1.5 \times 4.0^2}{2.0} = 12$ N.

The **mass × velocity** of an object — how much motion it has. p = mv, unit kg m s⁻¹. It is a **vector** (has direction).

The **average force × the time** it acts for, J = FΔt. It equals the **change in momentum** (Δp). Unit: N s.

**kg m s⁻¹**. Impulse uses **N s**, which is the same unit.

$p = mv$ — mass × velocity (given in the data booklet).

$J = F\Delta t = \Delta p$ — force × time = change in momentum (given).

$F = \dfrac{\Delta p}{\Delta t}$ — the change in momentum ÷ the contact time (given form of Newton's 2nd law).

The **impulse** — which equals the **change in momentum**.

**No** — the direction flips, so Δp = m(v + u) = 2mu. Bouncing changes momentum more than stopping.

They **increase the contact time** Δt. Since F = Δp/Δt, a longer time means a **smaller force** for the same change in momentum.

Δp = J = 6.0 kg m s⁻¹, so v = p/m = 6.0 ÷ 0.50 = 12 m s⁻¹.

Impulse gives momentum p = J; then $E_k = \dfrac{p^2}{2m} = \dfrac{1}{2}mv^2$ once you have the speed.

**Momentum p = mv** — mass × velocity. It is a **vector** (has direction). Unit: kg m s⁻¹.

If no external force acts, the **total momentum before = total momentum after** a collision or explosion.

**Yes** — momentum is conserved in **every** collision (with no outside force), elastic or inelastic.

One where the **total kinetic energy is also conserved** (KE before = KE after). Objects bounce cleanly.

One where the objects **stick together** and move as one. Momentum is conserved, but the **most kinetic energy is lost** (to heat/sound).

Compare **total KE before** and **total KE after** (E_k = ½mv²). If they're equal, it's elastic.

Velocity has direction — objects moving opposite ways get opposite signs, or the momentum total is wrong.

As **one combined mass** at one common velocity: (m₁ + m₂)v.

$p = mv$ (given in the data booklet).

$E_k = \tfrac{1}{2}mv^2$ (given) — used to test elasticity.

**No** — only in an **elastic** collision. In an inelastic one some KE becomes heat/sound.

(KE before − KE after) ÷ KE before. It's never zero for a sticking (perfectly inelastic) collision.