Classical-Mechanics · Unit 17 · Video 1 · Interactive Practice

The Perfect Trade: Why Mechanical Energy Stays Constant

IKey Formulas

Formula	Name	When it applies
$\Delta E_m = \Delta K_{\text{sys}} + \Delta U_{\text{sys}}$	Change in mechanical energy	Always — it is only a definition
$\Delta E_m = 0 \;\Longleftrightarrow\; \Delta K_{\text{sys}} = -\Delta U_{\text{sys}}$	Conservation of mechanical energy	Closed system, conservative internal forces only
$K_f + U_f = K_i + U_i$	The working equation	Any two states of a conserving system
$E_m = \tfrac{1}{2}m_o v^2 + m_o g y$	Near-earth mechanical energy	Earth $+$ object system, $U^g(0) = 0$ at the surface

Key Insight: Neither condition is optional, and the equation carries no memory of the route between the two states. Evaluate $K + U$ once at the start, once at the end, set them equal — the journey in between never enters.

IIVisualization 1 — The Trade, Instant by Instant

A $2.0\ \text{kg}$ ball leaves your hand at $14\ \text{m/s}$ : one quantity holds still for the entire flight.

Time t (s) 0.00

IIIVisualization 2 — When the Sum Stops Being Constant

Three ways to draw the system around the same falling ball; only one of them keeps $K + U$ fixed.

System Fallen d (m) 0.00

💡 Nothing is destroyed in the middle case: the missing joules are thermal energy in the ball and the air, and widening the account to include them is exactly what the coming friction videos do.

IVVisualization 3 — The Working Equation, Step by Step

Released from rest at $10.0\ \text{m}$ , how fast is the ball moving as it passes $2.0\ \text{m}$ ?

Step 1

Step 1 — Choose the system and check the two conditions

System

=

ball

+

earth, so gravity is an internal conservative force; nothing crosses the boundary and no drag acts inside.

\Delta E_m = 0 \quad\Longrightarrow\quad K_f + U_f = K_i + U_i

Step 2 — Evaluate the initial state

K_i = \tfrac{1}{2}m v_i^2 = 0 \qquad (\text{released from rest})

U_i = mgy_i = (2.0)(9.8)(10.0) = 196.0\ \text{J}

Step 3 — Evaluate the final state

U_f = mgy_f = (2.0)(9.8)(2.0) = 39.2\ \text{J}

The total is already fixed at

196.0\ \text{J}

, so the kinetic part is whatever is left:

K_f = (K_i + U_i) - U_f = 196.0 - 39.2 = 156.8\ \text{J}

Step 4 — Solve for the speed

\tfrac{1}{2}mv_f^2 = 156.8\ \text{J} \quad\Longrightarrow\quad v_f = \sqrt{\frac{2(156.8)}{2.0}} = 12.5\ \text{m/s}

In symbols the mass cancels before any number is used:

\tfrac{1}{2}mv_f^2 = mg(y_i - y_f) \quad\Longrightarrow\quad v_f = \sqrt{2g(y_i - y_f)} = \sqrt{2(9.8)(8.0)}

VQuiz Questions

Problem 1 · Straight Drop

Given: A $0.50\ \text{kg}$ ball is released from rest at $y_i = 5.0\ \text{m}$ above the ground, with $U^g = 0$ at the ground and $g = 9.8\ \text{m/s}^2$ . Air resistance is negligible. Find its speed as it reaches the ground.

✅ Correct! All

24.5\ \text{J}

of potential energy has become kinetic:

v_f = \sqrt{2gy_i} = \sqrt{98} = 9.9\ \text{m/s}

❌ Check the factor of 2.

\sqrt{gy_i} = \sqrt{49} = 7.0

drops the

2

that comes from inverting

K = \tfrac{1}{2}mv^2

v_f^2 = 2K_f/m = 2gy_i

❌ The mass was used twice.

mgy_i

already contains

m

, and dividing by

m

again gives

\sqrt{2gy_i/m} = \sqrt{196}

. Cancel

m

instead — it drops out completely.

❌ That is $v_f^2$ , not $v_f$ .

2gy_i = 98\ \text{m}^2/\text{s}^2

; take the square root to get a speed.

❌ Not quite. Set

K_f + U_f = K_i + U_i

with

K_i = 0

and

U_f = 0

, then solve

\tfrac{1}{2}mv_f^2 = mgy_i

Show solution

Take the system to be the ball $+$ earth, so gravity is internal and conservative and mechanical energy is conserved:

K_i + U_i = K_f + U_f

At the start the ball is at rest at $5.0\ \text{m}$ ; at the end it is at the zero of potential energy:

0 + mgy_i = \tfrac{1}{2}mv_f^2 + 0

Divide through by $m$ — this is where the mass disappears:

v_f = \sqrt{2gy_i} = \sqrt{2(9.8)(5.0)} = \sqrt{98} = 9.9\ \text{m/s}

Check the energies: $U_i = (0.50)(9.8)(5.0) = 24.5\ \text{J}$ and $K_f = \tfrac{1}{2}(0.50)(9.9)^2 = 24.5\ \text{J}$ — the same joules, relabelled.

Problem 2 · Which Way Does the Trade Run?

Given: A $1.5\ \text{kg}$ ball leaves your hand and rises $4.0\ \text{m}$ to its highest point, with $g = 9.8\ \text{m/s}^2$ and no air resistance. For the ball $+$ earth system over that rise, find $\Delta U^g$ and $\Delta K$ .

✅ Correct! The separation grows, so

U^g

gains

58.8\ \text{J}

, and

\Delta K = -\Delta U^g

pays for every one of them.

❌ The signs are swapped. The ball ends higher, so

y_f - y_i = +4.0\ \text{m}

and

\Delta U^g = mg(y_f - y_i)

is positive. A rising ball also slows down, so

\Delta K

cannot be positive.

❌ That is $\Delta K = +\Delta U$ . Conservation says

\Delta K + \Delta U = 0

, so

\Delta K = -\Delta U

. If both were

+58.8\ \text{J}

the system would have gained

117.6\ \text{J}

from nowhere.

❌ Conservation pins the sum, not each term.

K

and

U

are free to change as much as they like, provided the changes cancel — and the ball measurably slows as it climbs.

❌ Not quite. Compute

\Delta U^g = mg(y_f - y_i)

first, then use

\Delta K = -\Delta U^g

Show solution

Near the surface $U^g = mgy$ , so the rise costs

\Delta U^g = mg(y_f - y_i) = (1.5)(9.8)(4.0) = +58.8\ \text{J}

The system is closed and gravity is its only internal force doing work, so $\Delta E_m = \Delta K + \Delta U^g = 0$ :

\Delta K = -\Delta U^g = -58.8\ \text{J}

Sanity check: the ball must have carried at least $58.8\ \text{J}$ of kinetic energy when it left your hand, i.e.

v_i = \sqrt{\frac{2(58.8)}{1.5}} = \sqrt{78.4} = 8.9\ \text{m/s}

which is exactly the launch speed that just reaches $4.0\ \text{m}$ .

Problem 3 · The Video's Ball, From Hand to Apex

Given: A $2.0\ \text{kg}$ ball is thrown straight up at $14\ \text{m/s}$ from your hand, where $y = 0$ and $U^g = 0$ . Take $g = 9.8\ \text{m/s}^2$ and neglect air resistance.

What is the system's mechanical energy?

How high does it rise?

✅ Correct!

196\ \text{J}

of pure kinetic energy at the hand becomes

196\ \text{J}

of pure potential energy at

10.0\ \text{m}

— the same number, twice.

❌ $U^g = 0$ there, but $K$ is not. The zero of potential energy is a choice about where, not a claim that the system holds no energy: the ball is moving at

14\ \text{m/s}

❌ Check the mechanical energy. At the hand

U^g = mgy = 0

, so

E_m = K = \tfrac{1}{2}mv^2

with

m = 2.0\ \text{kg}

and

v = 14\ \text{m/s}

— keep both the

\tfrac{1}{2}

and the mass.

❌ Check the height. At the top the ball is momentarily at rest, so

K_f = 0

and the whole

196\ \text{J}

sits in

U^g = mgy_{\max} = 19.6\,y_{\max}

Show solution

Step 1 — mechanical energy at the hand. With $y = 0$ the potential term vanishes and only the kinetic term survives:

E_m = \tfrac{1}{2}mv^2 + mgy = \tfrac{1}{2}(2.0)(14)^2 + 0 = 196\ \text{J}

Step 2 — apply the working equation to the apex. At the highest point the ball is instantaneously at rest, so $K_f = 0$ :

K_f + U_f = K_i + U_i \;\Longrightarrow\; 0 + mgy_{\max} = 196\ \text{J}

y_{\max} = \frac{196}{(2.0)(9.8)} = \frac{196}{19.6} = 10.0\ \text{m}

In symbols: $\tfrac{1}{2}mv^2 = mgy_{\max}$ gives $y_{\max} = v^2/2g = 196/19.6$ — the mass cancels again, and the common slip is to write $v^2/g = 20.0\ \text{m}$ by forgetting the $\tfrac{1}{2}$ .

Problem 4 · The Earth's Share of the Kinetic Energy

Given: A $2.0\ \text{kg}$ ball falls toward the earth, $m_e = 6.0 \times 10^{24}\ \text{kg}$ . The two gravitational forces are equal and opposite and act for the same time, so both bodies acquire the same momentum magnitude $p$ . Using $K = p^2/2m$ , find the ratio $K_e/K_o$ of the kinetic energies gained by the earth and by the object.

✅ Correct! At equal momentum the heavier body carries the smaller share, and

3 \times 10^{-25}

is why

\Delta K_{\text{sys}} \cong \Delta K_o

for anything you can drop.

❌ The ratio is upside down.

K = p^2/2m

divides by the mass, so at equal

p

the heavier body carries less energy. The earth's share must be the tiny one.

❌ Equal momentum is not equal energy. The third-law pair does guarantee

|\Delta p_e| = |\Delta p_o|

, but

K = p^2/2m

then splits that momentum into energies in inverse proportion to the masses.

❌ No square root appears.

K

is proportional to

1/m

, not

1/\sqrt{m}

: forming

K_e/K_o

cancels

p^2/2

exactly and leaves

m_o/m_e

❌ Not quite. Write

K_e = p^2/2m_e

and

K_o = p^2/2m_o

and divide — everything except the masses cancels.

Show solution

Newton's third law makes the forces equal and opposite, and they act over the same interval, so the momentum magnitudes match: $|\Delta p_e| = |\Delta p_o| = p$ . Writing each kinetic energy in terms of that shared momentum,

K_e = \frac{p^2}{2m_e}, \qquad K_o = \frac{p^2}{2m_o}

\frac{K_e}{K_o} = \frac{p^2/2m_e}{p^2/2m_o} = \frac{m_o}{m_e} = \frac{2.0}{6.0 \times 10^{24}} = 3.3 \times 10^{-25}

More than twenty orders of magnitude, so the earth's term is utterly negligible:

\Delta K_{\text{sys}} = \Delta K_e + \Delta K_o \cong \Delta K_o = \tfrac{1}{2}m_o v_{o,f}^2 - \tfrac{1}{2}m_o v_{o,i}^2

Note what this does not say: the earth is still part of the system, and $U^g$ still belongs to the pair. Only the earth's share of the kinetic bookkeeping is dropped.

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