Get Ready: Module 3 — Energy, Momentum & SHM
Everyone starts somewhere. Before we track energy and momentum through collisions, let's lock in the five algebra moves the whole module leans on — one small skill at a time. Go at your own pace; there's no rush and nothing to prove here.
Skills to Build First
Five small building blocks. Master these and Module 3 will feel like the next natural step, not a wall.
Substituting Into Squared Terms
What it is & why you need it: Kinetic energy is \(KE = \tfrac12 mv^2\) — the velocity is squared. Plugging numbers into a formula with a square means substituting first, then squaring only the velocity, not the whole expression. Get the order right and the energy bookkeeping in this module stays clean.
Worked example: evaluate \(\tfrac12 m v^2\) when \(m = 3\) kg and \(v = 4\) m/s
- Substitute the values, keeping \(v\) in parentheses: \(\tfrac12 (3)(4)^2\).
- Square the velocity first (exponents before multiplying): \((4)^2 = 16\), giving \(\tfrac12 (3)(16)\).
- Multiply left to right: \(\tfrac12 \times 3 = 1.5\), then \(1.5 \times 16 = 24\).
- So \(KE = 24\) J. Only \(v\) was squared — the mass was not.
Tip: always wrap the substituted number in parentheses before squaring — \((4)^2 = 16\), but \(-4^2\) without parentheses is read as \(-(4^2) = -16\), a classic trap.
1. Evaluate \(\tfrac12 m v^2\) when \(m = 2\) kg and \(v = 5\) m/s.
2. Evaluate \(\tfrac12 k x^2\) when \(k = 200\) N/m and \(x = 0.3\) m.
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2. Substitute: \(\tfrac12 (200)(0.3)^2\). Square first: \((0.3)^2 = 0.09\). Then \(\tfrac12 (200)(0.09) = 100 \times 0.09 = 9\) J.
Taking Square Roots
What it is & why you need it: Energy formulas square the speed, so when you solve for the speed you have to undo the square with a square root. \(\tfrac12 m v^2 = E\) rearranges to \(v = \sqrt{\dfrac{2E}{m}}\). Knowing how to isolate a square and then take its root is the heart of "speed at the bottom" energy problems.
Worked example: solve \(\tfrac12 m v^2 = 36\) for \(v\), with \(m = 2\) kg
- Substitute the mass: \(\tfrac12 (2) v^2 = 36\), which simplifies to \(v^2 = 36\).
- Undo the square by taking the square root of both sides: \(v = \sqrt{36}\).
- Evaluate the root: \(\sqrt{36} = 6\) (the number that times itself gives 36).
- So \(v = 6\) m/s. (We take the positive root for a speed.)
Not a perfect square? Estimate or use a calculator: \(\sqrt{20}\approx 4.47\) because \(4.47^2 \approx 20\). The square root always lands between the roots of the nearest perfect squares.
1. Solve \(v^2 = 49\) for the positive value of \(v\).
2. A cart's energy gives \(v = \sqrt{2gh}\) with \(g = 9.8\) and \(h = 5\). Find \(v\) (round to 2 decimals).
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2. Inside first: \(2(9.8)(5) = 98\). Then \(v = \sqrt{98} \approx 9.90\) m/s (because \(9.90^2 \approx 98\)).
Conservation Bookkeeping (Before = After)
What it is & why you need it: A conservation law is just an equation that says one quantity is the same before and after an event. For momentum, \(\big(\text{total } p\big)_{\text{before}} = \big(\text{total } p\big)_{\text{after}}\). The whole skill is: add up the "before" side, add up the "after" side, set them equal, and solve. It's bookkeeping — nothing fancier.
Worked example: a \(2\) kg cart at \(3\) m/s sticks to a \(1\) kg cart at rest — find \(v_f\)
- Add the "before" side. \(p_{\text{before}} = m_1 v_1 + m_2 v_2 = 2(3) + 1(0) = 6\) kg·m/s.
- Write the "after" side. They stick, so it's one combined mass moving at \(v_f\): \(p_{\text{after}} = (2 + 1)\,v_f = 3 v_f\).
- Set before = after: \(6 = 3 v_f\).
- Solve: \(v_f = \dfrac{6}{3} = 2\) m/s.
Watch the signs: velocity to the right is positive, to the left is negative. A cart moving left at 3 m/s contributes \(-3\), not \(+3\), to the "before" total.
1. A \(4\) kg cart at \(2\) m/s sticks to a \(2\) kg cart at rest. Find the common speed \(v_f\).
2. Before a push-apart, total momentum is \(0\). One \(3\) kg piece moves \(+2\) m/s. A \(1\) kg piece must have what momentum so the after-total is still \(0\)?
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2. The two momenta must cancel. The \(3\) kg piece has \(3(2) = +6\) kg·m/s, so the \(1\) kg piece must have \(-6\) kg·m/s (moving the other way at \(6\) m/s).
Reading Work: \(W = Fd\cos\theta\)
What it is & why you need it: Work measures the energy a force transfers as it acts over a distance. The \(\cos\theta\) factor only counts the part of the force that points along the motion. Pull straight along the displacement (\(\theta = 0\)) and \(\cos 0 = 1\) — full work. Push sideways (\(\theta = 90^\circ\)) and \(\cos 90^\circ = 0\) — no work at all.
Worked example: a \(50\) N force pulls a sled \(8\) m at \(60^\circ\) above the ground
- Write the formula: \(W = F d \cos\theta\).
- Substitute: \(W = (50)(8)\cos 60^\circ\).
- Use \(\cos 60^\circ = 0.5\): only half the force pulls along the ground.
- Multiply: \(W = 50 \times 8 \times 0.5 = 200\) J.
Key angles: \(\cos 0^\circ = 1\) (force along motion, max work), \(\cos 90^\circ = 0\) (force perpendicular, zero work), \(\cos 180^\circ = -1\) (force opposes motion, negative work — like friction).
1. A \(30\) N force pushes a box \(5\) m straight along the floor (\(\theta = 0^\circ\)). Find the work.
2. You carry a \(40\) N bag horizontally for \(10\) m, holding it straight up (force is vertical, motion is horizontal, \(\theta = 90^\circ\)). How much work do you do on the bag?
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2. \(W = (40)(10)\cos 90^\circ = 400 \times 0 = 0\) J. The lifting force is perpendicular to the horizontal motion, so it does no work — surprising but correct.
Module 1 Kinematics
What it is & why you need it: Energy and momentum problems often hand you a velocity that you first have to find from motion — how fast is the cart going when it reaches the collision? That's kinematics from Module 1: \(v_f = v_i + at\) and \(v_f^2 = v_i^2 + 2ad\). Treat a force-times-distance setup like \(F \times d\) the same way — substitute and multiply.
Worked example: a cart starts at \(2\) m/s and accelerates at \(3\) m/s² for \(4\) s — find \(v_f\)
- Pick the right equation. We have \(v_i\), \(a\), and \(t\), and want \(v_f\): use \(v_f = v_i + a t\).
- Substitute: \(v_f = 2 + (3)(4)\).
- Multiply before adding (order of operations): \(3 \times 4 = 12\), giving \(2 + 12\).
- Add: \(v_f = 14\) m/s. That speed can now feed straight into \(KE = \tfrac12 mv^2\).
Force × distance works the same way: the work or "push" setup \(F \times d\) is just a substitute-and-multiply, e.g. \(20 \times 6 = 120\) — the same arithmetic muscle as a kinematics plug-in.
1. A cart at \(0\) m/s accelerates at \(2\) m/s² for \(5\) s. Find \(v_f\).
2. Evaluate the force-times-distance setup \(15 \times 8\) (e.g. a \(15\) N force over \(8\) m).
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2. \(15 \times 8 = 120\). (That's the kind of quick multiply behind \(W = Fd\) and force-times-distance setups.)
Quick Readiness Check
Six short questions across all five skills. Try each one first, then peek at the answer.
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Evaluate \(\tfrac12 m v^2\) when \(m = 4\) kg and \(v = 3\) m/s.
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Square first: \((3)^2 = 9\). Then \(\tfrac12 (4)(9) = 2 \times 9 = 18\) J. -
Find \(\sqrt{64}\), then \(\sqrt{2(9.8)(2)}\) rounded to 2 decimals.
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\(\sqrt{64} = 8\). Next, inside: \(2(9.8)(2) = 39.2\), so \(\sqrt{39.2} \approx 6.26\). -
A \(3\) kg cart at \(4\) m/s sticks to a \(1\) kg cart at rest. Find \(v_f\).
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Before: \(3(4) + 1(0) = 12\). After: \((3+1)v_f = 4v_f\). So \(12 = 4v_f\), \(v_f = 3\) m/s. -
Find the work done by a \(20\) N force over \(6\) m at \(\theta = 0^\circ\).
Show answer
\(W = (20)(6)\cos 0^\circ = 120 \times 1 = 120\) J. -
A cart starts at \(3\) m/s and accelerates at \(4\) m/s² for \(2\) s. Find \(v_f\).
Show answer
\(v_f = v_i + at = 3 + (4)(2) = 3 + 8 = 11\) m/s. -
How much work does a \(25\) N force do if it acts perpendicular to the motion (\(\theta = 90^\circ\))?
Show answer
\(\cos 90^\circ = 0\), so \(W = (25)(d)(0) = 0\) J — no work for a perpendicular force.
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.)
Kinetic energy & squared terms
Square roots
Conservation of momentum
Work & \(W = Fd\cos\theta\)
Kinematics (Module 1)
When You're Ready →
No rush. When the skills above feel solid, step into Module 3.
Instructor: Dr. Goodluck Ijezie-Desbois, PharmD · Room: TBA
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