Get Ready: Module 3 — Energy & Momentum
Before we balance collisions, let's make sure the algebra underneath is solid — one small skill at a time. Energy and momentum are just \(p = mv\) and \(KE = \tfrac12 mv^2\) with the order of operations and a careful eye on signs. Go at your own pace; nothing to prove here.
Skills to Build First
Four small building blocks. Master these and Module 3 will feel like the next natural step, not a wall.
Squaring a Number & the Order of Operations in \(\tfrac12 mv^2\)
What it is & why you need it: Kinetic energy is \(KE = \tfrac12 mv^2\). The exponent applies only to \(v\), so you must square the speed first, then multiply by the mass, then halve. Get the order wrong and the energy is off by a lot — because squaring is where most of the size comes from.
Worked example: evaluate \(\tfrac12 mv^2\) for \(m = 3\text{ kg}\), \(v = 4\text{ m/s}\)
- Square the velocity first. The exponent binds tightest: \(v^2 = 4^2 = 16\). (Not \((\tfrac12 \cdot 3 \cdot 4)^2\) — only \(v\) is squared.)
- Multiply by the mass. \(m \cdot v^2 = 3 \times 16 = 48\).
- Take half. \(\tfrac12 \times 48 = 24\).
- Attach the unit. Energy is in joules: \(KE = 24\text{ J}\).
Watch the trap: \(\tfrac12 m v^2\) means \(\tfrac12 \cdot m \cdot (v^2)\), not \(\tfrac12 \cdot (mv)^2\). Square \(v\) alone, every time.
1. Evaluate \(\tfrac12 mv^2\) for \(m = 2\text{ kg}\), \(v = 5\text{ m/s}\).
2. A \(4\text{ kg}\) object moves at \(3\text{ m/s}\). Find its kinetic energy.
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2. \(3^2 = 9\); \(4 \times 9 = 36\); \(\tfrac12(36) = 18\). So \(KE = 18\text{ J}\).
Solving a Formula for a Chosen Variable
What it is & why you need it: Physics formulas come written one way, but the unknown you want is often somewhere else in the equation. Solving \(p = mv\) for \(v\), or \(P = \dfrac{W}{t}\) for \(t\), is the same balance-scale algebra you already know: whatever you do to one side, do to the other.
Worked example: solve \(p = mv\) for \(v\)
- Start with the formula: \(p = mv\). You want \(v\) alone.
- Undo the multiplication by \(m\). Divide both sides by \(m\): \(\dfrac{p}{m} = \dfrac{mv}{m}\).
- Simplify the right side: \(\dfrac{mv}{m} = v\), so \(v = \dfrac{p}{m}\).
- Use it. If \(p = 30\text{ kg·m/s}\) and \(m = 6\text{ kg}\), then \(v = \dfrac{30}{6} = 5\text{ m/s}\).
Same idea for \(KE = \tfrac12 mv^2\) solved for \(v\): multiply by 2, divide by \(m\), then take a square root: \(v = \sqrt{\dfrac{2\,KE}{m}}\).
1. Solve \(W = mg\) for \(g\), then find \(g\) if \(W = 98\text{ N}\) and \(m = 10\text{ kg}\).
2. Solve \(P = \dfrac{W}{t}\) for \(t\), then find \(t\) if \(P = 50\text{ W}\) and \(W = 200\text{ J}\).
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2. Multiply both sides by \(t\): \(Pt = W\); divide by \(P\): \(t = \dfrac{W}{P} = \dfrac{200}{50} = 4\text{ s}\).
Tracking Signs in 1-D Motion
What it is & why you need it: In one dimension, direction is a sign. Pick a positive direction (say, right \(=+\)); then anything moving left gets a minus. Momentum is a vector, so when you add two momenta you must respect those signs — that's how head-on collisions partly cancel.
Worked example: total momentum of two carts moving toward each other
- Choose a positive direction. Let right be \(+\). Cart A (\(2\text{ kg}\)) moves right at \(4\text{ m/s}\): \(v_A = +4\).
- Assign the other sign. Cart B (\(3\text{ kg}\)) moves left at \(2\text{ m/s}\), so \(v_B = -2\) (left is negative).
- Add the momenta with their signs. \(p = m_A v_A + m_B v_B = 2(+4) + 3(-2) = 8 - 6\).
- Read the result. \(p = +2\text{ kg·m/s}\). Positive, so the system's net motion is to the right.
A negative answer isn't "wrong" — it just means the quantity points in your negative direction. Keep the sign; it carries the direction.
1. A \(5\text{ kg}\) cart moves left at \(3\text{ m/s}\). With right \(=+\), what is its momentum (sign included)?
2. Cart A (\(1\text{ kg}\), \(+6\text{ m/s}\)) and cart B (\(2\text{ kg}\), \(-6\text{ m/s}\)). Find the total momentum.
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2. \(p = 1(+6) + 2(-6) = 6 - 12 = -6\text{ kg·m/s}\). The net momentum points left (cart B's larger mass wins).
The Units: Joules (J), Watts (W), and kg·m/s
What it is & why you need it: Every quantity in this module wears a unit, and the unit tells you which quantity it is. Energy is in joules (J), power (energy per second) in watts (W), and momentum in kilogram-metres per second (kg·m/s). Matching the unit to the formula is a built-in error check.
Worked example: name each quantity from its unit
- Joule (J) = energy. It comes from \(KE = \tfrac12 mv^2\) (kg·(m/s)\(^2\)) or work \(W = Fd\) (N·m). So \(1\text{ J} = 1\text{ kg·m}^2/\text{s}^2\).
- Watt (W) = power = energy per time, \(P = \dfrac{W}{t}\). So \(1\text{ W} = 1\text{ J/s}\). A 60 W bulb uses 60 joules every second.
- kg·m/s = momentum, \(p = mv\) (kilograms times metres-per-second). It is not an energy — different unit, different idea.
- Use the unit as a check. If you computed a "momentum" and the unit came out in joules, you used the wrong formula. The unit is a free safety net.
Quick reads: J = energy, W = energy-per-second (power), kg·m/s = momentum. Same letter "W" is also used for the variable work; context tells them apart.
1. A quantity is measured in kg·m/s. Which physics quantity is it, and which formula produces it?
2. A machine delivers \(600\text{ J}\) of energy in \(4\text{ s}\). What is its power, and in what unit?
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2. Power \(P = \dfrac{W}{t} = \dfrac{600}{4} = 150\). The unit is the watt (W), since \(1\text{ W} = 1\text{ J/s}\). So \(P = 150\text{ W}\).
Quick Readiness Check
Six short questions across all four skills. Try each one first, then peek at the answer.
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Evaluate \(\tfrac12 mv^2\) for \(m = 6\text{ kg}\), \(v = 2\text{ m/s}\).
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Square first: \(2^2 = 4\). Then \(6 \times 4 = 24\). Half: \(12\). So \(KE = 12\text{ J}\). -
Solve \(p = mv\) for \(m\). Then find \(m\) if \(p = 24\text{ kg·m/s}\) and \(v = 8\text{ m/s}\).
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Divide both sides by \(v\): \(m = \dfrac{p}{v} = \dfrac{24}{8} = 3\text{ kg}\). -
A \(4\text{ kg}\) cart moves left at \(5\text{ m/s}\). With right \(=+\), state its momentum with sign.
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Left is negative: \(v = -5\). \(p = 4(-5) = -20\text{ kg·m/s}\) (pointing left). -
Cart A (\(2\text{ kg}\), \(+3\text{ m/s}\)) and cart B (\(2\text{ kg}\), \(-3\text{ m/s}\)). Total momentum?
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\(p = 2(+3) + 2(-3) = 6 - 6 = 0\text{ kg·m/s}\). The equal-and-opposite momenta cancel. -
\(900\text{ J}\) of work is done in \(3\text{ s}\). Find the power and give its unit.
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\(P = \dfrac{W}{t} = \dfrac{900}{3} = 300\text{ W}\) (watts, since 1 W = 1 J/s). -
Find the momentum of a \(7\text{ kg}\) object moving at \(4\text{ m/s}\), with the correct unit.
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\(p = mv = 7 \times 4 = 28\text{ kg·m/s}\).
If these feel comfortable, you're ready for the module. If a couple felt tricky, that's fine — revisit that skill card above and drill a little with the free links below. No shame, just steps.
Practice More — Free
Want extra reps? These free sites let you drill each skill until it clicks. (Links open in a new tab.)
Squaring & \(\tfrac12 mv^2\)
Solving a formula for a variable
Signed numbers in 1-D
Units: J, W, kg·m/s
When You're Ready →
No rush. When the skills above feel solid, step into Module 3.
Physics · enrichment — algebra-based mechanics offered for curiosity and cross-training. Not a TEKS-graded mathematics course and not scored toward any STAAR EOC. Foundations support to make the conservation laws approachable for any scholar who wants the stretch.