Physics 1A: Energy, Work, Power & Momentum

Work — a force acting through a distance

In physics, work has a very specific meaning that only partially matches the everyday word. Work is done on an object when a force pushes or pulls it through a distance in the direction of the force. In symbols:

W = F · d · cos θ

Where F is the magnitude of the force, d is the magnitude of the displacement, and θ is the angle between them. The unit is the joule (J), defined so that 1 J = 1 N·m. Some CBE-relevant edge cases:

• If the force is parallel to the displacement (θ = 0°), then cos θ = 1, and W = F·d. Push straight, work is maximal.
• If the force is perpendicular to the displacement (θ = 90°), then cos θ = 0, and W = 0. A satellite in a circular orbit does zero work because gravity always points perpendicular to its motion. Similarly, the normal force on a block sliding across a horizontal floor does zero work — it points up, motion is horizontal.
• If the force is opposite to the displacement (θ = 180°), then cos θ = −1, and W = −F·d. Friction on a moving box does negative work — it takes energy away.
• If the object does not move, work is zero, no matter how hard you push. Holding a heavy box still is exhausting, but from the physicist’s perspective, no work is being done on the box.

Kinetic energy — the energy of motion

Kinetic energy (KE) is the energy an object has because it is moving:

KE = ½ · m · v²

Mass in kg, velocity in m/s, energy in joules. Two features of this formula the CBE loves to test:

• The factor of ½. Missing it turns a correct calculation into a "twice-the-answer" distractor. Read your working carefully.
• The v is squared. Doubling the speed of a car quadruples its kinetic energy. Tripling it multiplies KE by 9. This is why highway crashes are so much more damaging than parking-lot bumps: energy scales with v², not v.

Potential energy — energy stored by position

Gravitational potential energy (PE) is the energy an object has because of its height in a gravitational field:

PE = m · g · h

Where h is the height above a reference level of your choice. The reference matters: PE at the top of a building is different if you measure h from the ground floor or from the roof of the neighboring building. What is meaningful is the change in PE, ΔPE = mg·Δh — differences in height, not absolute heights.

Physics 1A treats gravity as constant near Earth’s surface, so PE = mgh works. When you get into orbital or astronomical problems, a more general formula involving G is required — but not for this course.

The work-energy theorem

Here is a beautiful result that unifies work and kinetic energy:

W_net = ΔKE = KE_f − KE_i

The net work done on an object equals its change in kinetic energy. If you know the net force and the distance over which it acts, you know how much the object’s KE changed — and therefore what its final speed is. This shortcut often lets you solve problems that would take multiple kinematic equations to unwind. If a 2 kg object at rest is pushed with 10 N over 3 m on a frictionless surface, the net work is 30 J, so its final KE is 30 J, so ½ · 2 · v² = 30, giving v² = 30, v ≈ 5.48 m/s. No time variable ever entered.

Conservation of energy — the master rule

In a system with no external work done against friction or air resistance, mechanical energy — the sum of KE and PE — is conserved. What is lost as PE is gained as KE, and vice versa. In symbols:

KE_i + PE_i = KE_f + PE_f

This principle solves roller-coaster problems, pendulums, dropped objects, and springs. Pick a starting point where you know KE and PE, pick an ending point where you know one of them, and solve for the other. Distance and time drop out.

Power — how fast work is done

Power is the rate at which work is done or energy transferred:

P = W / t     or equivalently     P = F · v

Unit is the watt (W): 1 W = 1 J/s. A student who lifts a 20 kg backpack 1 m in 2 seconds does 20 · 9.8 · 1 ≈ 196 J of work at a rate of 98 W. A car engine delivering 100 hp (about 74,600 W) is doing over 700 times as much work per second as that student.

Watch the CBE distractor pattern here: a problem might give you mass, height, and time and ask for power. It is easy to compute mgh (the work) and report that as the answer, forgetting the last division by t. Power is a rate — always divide by time at the end.

Momentum — a different currency

Momentum is a vector quantity that captures how difficult it is to stop a moving object:

p = m · v

Unit: kg·m/s. A truck moving at 3 m/s and a bullet moving at 800 m/s might have similar momentum — the truck’s mass is enormous while the bullet’s speed is huge.

Momentum is a vector: it has direction. A ball moving right with momentum +12 kg·m/s and a ball moving left with momentum −12 kg·m/s have equal magnitudes but opposite directions. When you add them, they cancel.

Impulse — the change of momentum

Impulse is what a force delivers over a time interval:

I = F · t = Δp

Impulse equals the change in momentum. This is why airbags and crumple zones save lives: they extend the time over which a car’s momentum drops to zero, which reduces the force felt by the passenger. Same impulse (same Δp), longer t, smaller F. Same principle for catching an egg — pull your hand back as the egg lands to lengthen t.

Conservation of momentum

In a closed system (no external forces along the direction of interest), total momentum is conserved. If two objects collide, the total momentum before the collision equals the total momentum after:

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

This is the master equation for collision problems. Whichever the collision type — elastic or inelastic — momentum is always conserved as long as no external force acts.

Elastic vs inelastic collisions

Two categories the CBE tests separately:

• Elastic collision: both momentum and kinetic energy are conserved. Two billiard balls colliding, or two hard rubber balls bouncing off each other, come close. In the ideal case, the objects bounce apart, with no energy lost to sound, heat, or deformation.
• Inelastic collision: momentum is conserved, but kinetic energy is NOT. Some KE is converted to heat, sound, or deformation of the colliding objects. The extreme case is a perfectly inelastic collision: the two objects stick together after the collision and move as one. Use the special form: m₁v_1i + m₂v_2i = (m₁ + m₂)·v_f.

The CBE will ask you to identify whether a collision is elastic, inelastic, or perfectly inelastic based on the description, and to apply momentum conservation to find missing velocities. Read carefully — "stick together" means perfectly inelastic; "bounce apart with no energy loss" means elastic.

Where students lose points

• Forgetting the ½ in KE. Delivers the "double the answer" distractor.
• Forgetting to square v. KE = ½mv (missing square) is the most common single mistake in this topic.
• Assuming elastic when the problem says otherwise. If objects stick together, KE is not conserved — only momentum is.
• Skipping the division by time in power problems. Reporting work as power.
• Treating momentum as a scalar. Momentum has direction. Momentum going right and equal momentum going left cancel.
• Setting up conservation of energy with friction present. If there is friction, mechanical energy is not conserved — you have to account for the work done by friction separately.

Worked example — perfectly inelastic collision

A 3.0-kg cart moving at 4.0 m/s to the right collides with a stationary 5.0-kg cart. The two carts stick together after the collision. Find the velocity of the combined mass immediately after the collision, and calculate how much kinetic energy was lost.

Step 1 — Apply momentum conservation. m₁v_1i + m₂v_2i = (m₁ + m₂)·v_f.

Step 2 — Plug in. (3.0)(4.0) + (5.0)(0) = (3.0 + 5.0)·v_f → 12 = 8·v_f → v_f = 1.5 m/s to the right.

Step 3 — Compute KE before and after. KE_i = ½(3.0)(4.0)² = 24 J. KE_f = ½(8.0)(1.5)² = 9.0 J.

Step 4 — Energy lost. ΔKE = 24 − 9.0 = 15 J was converted to heat, sound, and deformation of the carts.

This is a classic problem type. Momentum is conserved (12 kg·m/s before, 12 kg·m/s after). Kinetic energy is not (24 J → 9 J). This dual behavior is exactly what "inelastic" means.

Check yourself

1. Write the four equations from this lesson: W, KE, PE, and P.
2. Explain why a normal force on a horizontally sliding block does zero work.
3. State the work-energy theorem in one sentence.
4. What quantity is conserved in an elastic collision that is NOT conserved in an inelastic collision?
5. Give an everyday example of impulse in action (airbag, catching a ball, etc.) and explain in one sentence how it reduces the force felt.
6. A 2 kg ball is dropped from 5 m. Ignoring air resistance, what is its speed just before it hits the ground?

(Answer to #6: PE at start = 2 · 9.8 · 5 = 98 J. At the bottom, all PE has converted to KE, so ½(2)v² = 98, v² = 98, v ≈ 9.9 m/s.)

Practice with CBE-style questions

Energy and momentum questions dominate the second half of the Physics Semester A CBE. Work through the practice bank filtered by Energy, Work, Power & Momentum — every question includes a step-by-step solution and identifies which conceptual mistake each distractor represents.

Independent practice content aligned to Texas Essential Knowledge and Skills (TEKS) §112.39(c)(6)(A)-(D). Not affiliated with TTU K-12, UT High School, UT-Austin, the Texas Education Agency, or any Credit by Examination administrator. Texas CBE™ does not administer any exam or grant academic credit.