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Work, Power and Energy · Class 11 Notes

Complete Class 11 notes on Work, Power and Energy: 11 diagrams, 15 worked numericals to JEE Advanced depth, and a one-page formula sheet. Read free online or download the PDF.

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In short

Work is done only when a force moves something along its own direction: W = Fd cosθ. The work done by all forces equals the change in kinetic energy, which is the work-energy theorem. When no friction acts, kinetic and potential energy simply trade places, so total mechanical energy stays constant. Power is how fast that work happens.

Contents

·How to Read This Set
1Work what counts as work, and its sign
2Work by a Changing Force
3Kinetic Energy and the Work-Energy Theorem
4Potential Energy gravity, springs, and conservative forces
5Conservation of Mechanical Energy
6Power how fast the work is done
7Collisions elastic, inelastic, and restitution
8Important Questions 15 worked numericals
9JEE Advanced Challenge 6 problems at real Advanced level
★Work, Power & Energy · Fact Sheet

How to Read This Set

This chapter is worth marks in all three exams. So it is built to JEE Advanced depth.
Each section carries a badge. It tells a NEET student what they still need.
Most of the text is in points, not paragraphs. Read the points, then study the drawing.
Every numerical answer here was checked by computer before printing.

Think it through

One idea runs through the whole chapter

Energy is never created or destroyed.
It only changes form, or moves from one body to another.
So most questions become: where did the energy start, and where did it end up?
Once you can answer that, the algebra is easy.

Work

In NEET and JEE

In physics, work needs two things: a force, and movement along that force.
Push a wall all day. Nothing moves. You have done zero work.
Work is a scalar. It has size and sign, but no direction.
The SI unit is the joule (J). 1 J = 1 N × 1 m.

Only the part of the force along the movement does work. The sideways part does none.

Work done by a constant force

W = \vec{F}\cdot\vec{d} = F\,d\cos\theta

θ is the angle between the force and the displacement.

The sign tells you the story

Three angles, three signs. The angle between force and movement decides everything.

Positive work adds energy to the body. It speeds up.
Negative work takes energy away. It slows down.
Zero work means the force is at right angles to the motion.
Friction usually does negative work. It points against the sliding.

Trap alert

Three forces that often do zero work

The normal force on a block sliding along the floor. It is at 90°.
The tension in a string as a stone whirls in a circle. Also 90°.
Gravity on a bag you carry level across a room.
Carrying a heavy bag feels tiring, but the physics work is zero.

Not sure when work is zero?

Turn the force and watch the work change sign.

Drag the arrow around and see W go from positive, through zero, to negative. The cosθ in the formula stops being something you memorise.

Play this concept in the app

Work by a Changing Force

JEE Main and Advanced

The formula W = Fd cosθ works only when F is constant.
Springs, gravity far from Earth, and many real forces change as you move.
Then you must add up the work in tiny steps.
On a force against distance graph, that sum is simply the area under the curve.

Split the area into shapes you know. A rectangle plus a triangle here gives 40 J.

Work done by a variable force

W = \int_{x_1}^{x_2} F\,dx = \text{area under the } F\text{-}x \text{ graph}

Think it through

How to read the graph safely

Area above the x-axis is positive work.
Area below the x-axis is negative work.
Add them with their signs. Do not just add the sizes.
For straight lines, break the shape into rectangles and triangles.

Kinetic Energy and the Work-Energy Theorem

In NEET and JEE

Kinetic energy is the energy a body has because it is moving.
It is always positive. It never has a direction.
Double the speed and the kinetic energy becomes four times larger.
This is why stopping distance grows so fast with speed.

Kinetic energy

K = \frac{1}{2}mv^2 = \frac{p^2}{2m}

The net work done on a body equals its change in kinetic energy. Nothing else is needed.

Work-energy theorem

W_{net} = K_f - K_i = \Delta K

This is the big shortcut of the chapter.
You do not need to know the force at every instant.
You do not need the time taken either.
You only need the speed at the start and at the end.

Trap alert

Use the NET work, not one force

W(net) means the work of every force added together.
Include friction. Include gravity. Include the applied force.
A common slip is to use only the applied force and forget friction.
If the body speeds up, W(net) must come out positive.

Potential Energy

In NEET and JEE

Potential energy is stored energy. It depends on position, not on speed.
Lift a book and you store energy in it. Let go and that energy returns.
Stretch a spring and you store energy in the spring.
Potential energy is always measured from a level you choose.

The two you need

U_{gravity} = mgh \qquad U_{spring} = \frac{1}{2}kx^2

The spring force grows as you stretch it. The stored energy is the triangular area under the line.

A spring pulls back with force F = kx, where k is the spring constant.
The stored energy is ½kx², not kx².
Stretching from 10 cm to 20 cm costs three times more than the first 10 cm.
That is because energy depends on x squared, not on x.

Conservative and non-conservative forces

Gravity does the same work along both paths. Only the change in height matters.

A conservative force does the same work whichever path you take.
Gravity and the spring force are conservative. So we can define a potential energy for them.
A non-conservative force depends on the path. Friction is the main one.
A longer path always means more heat lost to friction.

Think it through

Why friction has no potential energy

Potential energy means the energy can be fully returned.
A stretched spring gives back everything you put in.
Friction turns the energy into heat and sound.
That heat spreads out and cannot be pulled back. So there is nothing to store.

Conservation of Mechanical Energy

In NEET and JEE

Mechanical energy is kinetic plus potential: E = K + U.
If only conservative forces act, E stays the same throughout the motion.
So kinetic energy and potential energy simply trade places.
This turns many hard problems into one line of algebra.

A pendulum trades potential for kinetic and back again. The two bars always add to the same total.

Conservation of mechanical energy

K_i + U_i = K_f + U_f

valid only when no friction or other non-conservative force acts

A moving block stops when all its kinetic energy has gone into the spring.

Falling from height h: v = √(2gh). The mass cancels out.
A block hitting a spring: ½mv² = ½kx² at maximum compression.
At maximum compression the block is momentarily at rest.
A pendulum released from height h reaches the same speed as a free fall from h.

Trap alert

When friction is present, energy is NOT conserved

Mechanical energy drops. The lost part becomes heat.
Use this instead: K_i + U_i + W_friction = K_f + U_f.
W(friction) is negative, so it eats into the total.
Total energy of the universe is still conserved. Only the mechanical part falls.

Energy bars not clicking?

Slide the pendulum and watch the two bars swap.

Move the bob to any angle and see kinetic and potential energy trade in real time. The total bar never changes height, which is the whole idea.

Play this concept in the app

Power

In NEET and JEE

Power is the rate of doing work. It answers: how fast?
Two people can do the same work. The faster one has more power.
The SI unit is the watt (W). 1 W = 1 J/s.
1 horsepower = 746 W. 1 kWh = 3.6 × 10⁶ J, and it is a unit of energy, not power.

The same job done in half the time needs twice the power.

Power, three ways

P_{avg} = \frac{W}{t} \qquad P_{inst} = \frac{dW}{dt} = \vec{F}\cdot\vec{v}

Use P = Fv when a body moves at steady speed against a resisting force.
A car at constant speed: the engine power equals resistance × speed.
For a pump lifting water: useful power = mgh / t.
If efficiency is given, input power = useful power / efficiency.

Trap alert

kWh is energy, not power

Your electricity bill is in kilowatt-hours.
A kWh is power × time, so it measures energy used.
A 1000 W heater run for 1 hour uses 1 kWh.
Calling it a unit of power is a common exam trap.

Collisions

In NEET and JEE

In every collision, momentum is conserved. This is always true.
Kinetic energy is a different matter. It may or may not be conserved.
That single difference splits collisions into two types.
Always write the momentum equation first. It never fails you.

Momentum is conserved in both. Kinetic energy survives only the elastic one.

Elastic: kinetic energy is conserved. The bodies bounce apart.
Inelastic: some kinetic energy becomes heat and sound.
Perfectly inelastic: the bodies stick and move as one. Energy loss is largest here.
Real collisions are almost always somewhere in between.

Head-on elastic collision, one dimension

v_1 = \frac{m_1-m_2}{m_1+m_2}u_1 + \frac{2m_2}{m_1+m_2}u_2

v_2 = \frac{m_2-m_1}{m_1+m_2}u_2 + \frac{2m_1}{m_1+m_2}u_1

Think it through

Three cases worth memorising

Equal masses: they simply swap velocities.
Heavy hits light at rest: the heavy one carries on, the light one shoots off at nearly 2u.
Light hits heavy at rest: the light one bounces straight back at nearly the same speed.
These three cover a large share of the objective questions asked.

Coefficient of restitution

Each bounce reaches e squared times the height before it. The pattern never changes.

Coefficient of restitution

e = \frac{\text{speed of separation}}{\text{speed of approach}} = \frac{v_2-v_1}{u_1-u_2}

e = 1 is perfectly elastic. e = 0 is perfectly inelastic.
A ball dropped from h rebounds to e²h.
After n bounces the height is e²ⁿh.
Total distance travelled before stopping is h(1+e²)/(1−e²).

Energy lost in a perfectly inelastic collision

\Delta K = \frac{m_1 m_2}{2(m_1+m_2)}(u_1-u_2)^2

Important Questions

Fifteen numericals, tagged by exam. They rise in difficulty. Try each one before reading the working.

Four forces, one block

NEET

A 10 kg block is pulled 10 m along a rough floor by a 50 N force at 37° above the horizontal. Take μ = 0.2 and g = 10. Find the work done by each force, and the final speed from rest. (sin 37° = 0.6, cos 37° = 0.8)

CoreFind the normal force first. The upward part of F lightens the block, so N is not mg.

Working

N = mg - F\sin\theta = 100 - 30 = 70\ \text{N} \Rightarrow f = 0.2(70) = 14\ \text{N}

W_F = (50\cos 37^\circ)(10) = \mathbf{400\ \text{J}}, \qquad W_{friction} = \mathbf{-140\ \text{J}}

W_{gravity} = W_{normal} = 0 \Rightarrow W_{net} = 260\ \text{J}

\frac{1}{2}mv^2 = 260 \Rightarrow v = \mathbf{7.2\ \text{m/s}}

TrapN is 70 N, not 100 N. The lifting part of the force reduces it. Using mg gives f = 20 N and a wrong answer throughout.

Work from a graph

JEE MAIN

A force acts along the x-axis. It stays at 10 N from x = 0 to x = 2 m. It then falls in a straight line to zero at x = 6 m. Find the total work done.

CoreWork is the area under the force against distance graph. Split it into a rectangle and a triangle.

Working

\text{rectangle} = 10 \times 2 = 20\ \text{J}

\text{triangle} = \frac{1}{2} \times 4 \times 10 = 20\ \text{J}

W = \mathbf{40\ \text{J}}

TrapYou cannot use W = Fd here, because F changes. Taking an average force works only when the change is linear, which it is in the second part but not overall.

Stopping a bullet

NEET

A 10 g bullet moving at 200 m/s is stopped by a wooden block after going 20 cm into it. Find the average resistive force.

CoreUse the work-energy theorem. The resistive force removes all the kinetic energy.

Working

K_i = \frac{1}{2}(0.01)(200)^2 = 200\ \text{J}

W_{resist} = -200 = -F(0.20) \Rightarrow F = \mathbf{1000\ \text{N}}

TrapConvert grams to kilograms and centimetres to metres first. Leaving the mass as 10 makes the answer 1000 times too large.

The second stretch costs more

JEE MAIN

A spring has k = 200 N/m. Find the work needed to stretch it from 0 to 10 cm, then from 10 cm to 20 cm. Compare them.

CoreUse U = ½kx² and take the difference. Energy depends on x squared.

Working

W_1 = \frac{1}{2}(200)(0.1)^2 = \mathbf{1\ \text{J}}

W_2 = \frac{1}{2}(200)\left[(0.2)^2 - (0.1)^2\right] = \mathbf{3\ \text{J}}

TrapThe second 10 cm costs three times as much, not the same. Equal stretches never need equal work, because the force keeps growing.

Block meets spring

JEE MAIN

A 2 kg block slides at 4 m/s on a smooth floor and hits a spring of k = 800 N/m. Find the maximum compression.

CoreAt maximum compression the block has stopped. All its kinetic energy is now in the spring.

Working

\frac{1}{2}mv^2 = \frac{1}{2}kx^2 \Rightarrow x = v\sqrt{\frac{m}{k}}

x = 4\sqrt{\frac{2}{800}} = \mathbf{0.2\ \text{m}}

TrapThe block is at rest at maximum compression, but its acceleration is largest there. Zero speed does not mean zero force.

Pendulum at the bottom

NEET

A pendulum of length 2 m is released from rest with the string horizontal. Find the speed of the bob at the lowest point. Take g = 10.

CoreOnly gravity and tension act. Tension does no work, so mechanical energy is conserved.

Working

mgL = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gL}

v = \sqrt{2(10)(2)} = \mathbf{6.3\ \text{m/s}}

TrapThe mass cancels, so a heavy bob and a light bob arrive at the same speed. The tension never appears, because it is always at 90° to the motion.

Down a rough slope

JEE MAIN

A block starts from rest and slides 5 m down a 30° incline with μ = 0.25. Find its speed at the bottom. Take g = 10.

CoreMechanical energy is not conserved here. Subtract the work done against friction.

Working

mgL\sin\theta - \mu mgL\cos\theta = \frac{1}{2}mv^2

v = \sqrt{2gL(\sin\theta - \mu\cos\theta)} = \sqrt{2(10)(5)(0.5 - 0.2165)}

v = \mathbf{5.3\ \text{m/s}}

TrapWithout friction the answer would be 7.07 m/s. Forgetting the friction term is the standard error. Note the mass cancels again.

Power of a pump

NEET

A pump lifts 200 kg of water through 10 m in 20 s. Its efficiency is 80%. Find the useful power and the input power. Take g = 10.

CoreUseful power is the work against gravity divided by time. Input power is larger because of losses.

Working

P_{useful} = \frac{mgh}{t} = \frac{200(10)(10)}{20} = \mathbf{1000\ \text{W}}

P_{input} = \frac{1000}{0.8} = \mathbf{1250\ \text{W}}

TrapDivide by the efficiency, never multiply. The input must always be bigger than the useful output, so if your answer is smaller, you have divided the wrong way.

Car at steady speed

JEE MAIN

A car moves at a constant 20 m/s against a total resistance of 500 N. Find the power delivered by the engine.

CoreConstant speed means zero acceleration. So the driving force exactly equals the resistance.

Working

F = 500\ \text{N}, \qquad P = Fv = 500 \times 20 = \mathbf{10\ \text{kW}}

TrapThe mass of the car is not needed at all. It is given only to tempt you into using it. At constant speed the net force is zero.

Q10

Power grows with time

JEE MAIN

A constant 20 N force pulls a 5 kg block from rest on a smooth floor. Find the instantaneous power at t = 3 s, and the average power over those 3 s.

CoreForce is constant but speed is not. So instantaneous power keeps rising.

Working

a = \frac{F}{m} = 4\ \text{m/s}^2 \Rightarrow v = at = 12\ \text{m/s}

P_{inst} = Fv = 20 \times 12 = \mathbf{240\ \text{W}}

P_{avg} = \frac{W}{t} = \frac{F \cdot \frac{1}{2}at^2}{t} = \mathbf{120\ \text{W}}

TrapAverage power is not the same as instantaneous power. Here the average is exactly half, because the speed rose linearly from zero.

Q11

Equal masses, elastic

NEET

A 1 kg ball moving at 6 m/s hits an identical ball at rest head-on. The collision is perfectly elastic. Find both velocities afterwards.

CoreFor equal masses in a perfectly elastic head-on collision, the two simply exchange velocities.

Working

v_1 = \frac{m_1-m_2}{m_1+m_2}u_1 = 0, \qquad v_2 = \frac{2m_1}{m_1+m_2}u_1 = u_1

\mathbf{v_1 = 0\ \text{m/s}, \quad v_2 = 6\ \text{m/s}}

TrapThe first ball stops completely. This is why a Newton's cradle works. Both momentum and kinetic energy check out, which no other pair of answers does.

Q12

Unequal masses, elastic

JEE MAIN

A 2 kg body moving at 10 m/s hits a 3 kg body at rest head-on and elastically. Find both velocities afterwards.

CoreApply the standard elastic formulas, then verify with momentum and kinetic energy.

Working

v_1 = \frac{2-3}{5}(10) = \mathbf{-2\ \text{m/s}}, \qquad v_2 = \frac{2(2)}{5}(10) = \mathbf{8\ \text{m/s}}

\text{check } p: \ 20 = 2(-2)+3(8) = 20 \ \checkmark

\text{check } K: \ 100 = \frac{1}{2}(2)(4) + \frac{1}{2}(3)(64) = 100 \ \checkmark

TrapThe minus sign means the lighter body bounces back. A lighter body always rebounds when it hits a heavier one at rest. Always run the two checks.

Q13

Counting the bounces

JEE MAIN

A ball is dropped from 5 m onto a floor with e = 0.8. Find the height after the first and second bounces, and the total distance it travels before coming to rest.

CoreEach bounce multiplies the height by e squared. The total path is a geometric series.

Working

h_1 = e^2 h = 0.64(5) = \mathbf{3.2\ \text{m}}, \qquad h_2 = e^4 h = \mathbf{2.05\ \text{m}}

\text{total} = h\,\frac{1+e^2}{1-e^2} = 5\left(\frac{1.64}{0.36}\right) = \mathbf{22.8\ \text{m}}

TrapThe height falls by e² each time, not by e. The total distance is finite even though the number of bounces is infinite, which surprises most students.

Q14

How much energy is lost

JEE MAIN

A 4 kg body moving at 5 m/s strikes a 6 kg body at rest. They stick together. Find their common velocity and the kinetic energy lost.

CoreMomentum is conserved. Kinetic energy is not, so find it before and after and take the difference.

Working

v = \frac{m_1 u_1}{m_1+m_2} = \frac{4(5)}{10} = \mathbf{2\ \text{m/s}}

K_i = 50\ \text{J}, \qquad K_f = \frac{1}{2}(10)(4) = 20\ \text{J}

\Delta K = \mathbf{30\ \text{J lost, which is } 60\%}

TrapNever apply conservation of kinetic energy to a sticking collision. The shortcut formula m₁m₂(u₁−u₂)²/2(m₁+m₂) gives the same 30 J.

Q15

The ballistic pendulum

JEE MAIN

A 20 g bullet moving at 300 m/s embeds itself in a 2 kg block hanging from a string. Find how high the block rises. Take g = 10.

CoreThis is two separate stages. Momentum for the collision, then energy for the swing. Never mix them.

Working

\text{stage 1, momentum: } (0.02)(300) = (2.02)V \Rightarrow V = 2.97\ \text{m/s}

\text{stage 2, energy: } \frac{1}{2}(2.02)V^2 = (2.02)gh

h = \frac{V^2}{2g} = \mathbf{0.44\ \text{m}}

TrapUsing energy conservation for stage 1 is wrong, because the bullet embeds and loses energy. Using momentum for stage 2 is also wrong, because gravity acts. Each stage needs its own rule.

JEE Advanced Challenge

JEE Advanced mainly

Six problems at genuine JEE Advanced level. Expect 5 to 8 minutes each.
Each one needs two or more ideas joined together, not one formula.
Two use real Advanced formats: one multiple-correct, one numerical answer.
Every assumption is stated in the question, as a real Advanced paper does.

Think it through

What makes a problem Advanced, not Main

A Main problem tells you which idea to use. You then do the algebra.
An Advanced problem hides the idea. Finding it is the problem.
In A2 below, the whole question turns on one hidden fact: at maximum compression both blocks move at the same speed.
Spot that and it takes two lines. Miss it and there is no way in.

The engine that cannot pull hard at speed

JEE ADVNUMERICAL

A car of mass 1000 kg has an engine giving a constant power of 40 kW. It starts from rest on a level road. Ignore friction and air resistance. Find its speed 25 s later. Give your answer in m/s, correct to two decimal places.

CoreConstant power means the force is not constant. As v grows, F = P/v falls. So you cannot use the equations of motion. Start from Newton's second law and separate the variables.

Working

P = Fv = m\frac{dv}{dt}v \Rightarrow \int_0^v mv\,dv = \int_0^t P\,dt

\frac{1}{2}mv^2 = Pt \Rightarrow v = \sqrt{\frac{2Pt}{m}}

v = \sqrt{\frac{2(40000)(25)}{1000}} = \sqrt{2000} = \mathbf{44.72\ \text{m/s}}

TrapUsing v = u + at is the trap, and it needs a constant force. Note the neat side result:

\frac{1}{2}mv^2 = Pt

says the whole energy delivered has become kinetic energy, which you could also have written down directly from the work-energy theorem.

A spring caught between two blocks

JEE ADVANCED

A 2 kg block slides at 6 m/s along a frictionless floor. A light spring of k = 1000 N/m is fixed to its front face. It strikes a stationary 4 kg block. Find the maximum compression of the spring.

CoreCompression is largest at the instant the two blocks stop approaching each other. At that moment they move with the same velocity. Use momentum to find it, then energy for the rest.

Working

\text{momentum: } (2)(6) = (2+4)v \Rightarrow v = 2\ \text{m/s}

\text{energy: } \frac{1}{2}(2)(6)^2 = \frac{1}{2}(6)(2)^2 + \frac{1}{2}kx^2

36 = 12 + 500x^2 \Rightarrow x = \mathbf{0.22\ \text{m}}

TrapSetting the first block's whole 36 J into the spring gives 0.27 m and is wrong, because the second block is moving too and keeps 12 J. This is the same hidden step as a perfectly inelastic collision, except here the energy is stored instead of lost.

The chain that pulls itself off the table

JEE ADVANCED

A uniform chain of length 3 m lies on a frictionless table. One third of it hangs over the edge. It is released from rest. Find the speed of the chain as the last link leaves the table. Take g = 10.

CoreThe mass is spread out, so track the centre of mass of the hanging part, not one point. Compute the potential energy at the start and at the end, and let energy conservation do the rest.

Working

\text{start: hanging mass } \frac{M}{3}, \text{ its centre } \frac{L}{6} \text{ below the edge}

U_i = -\frac{M}{3}g\frac{L}{6} = -\frac{MgL}{18}, \qquad U_f = -Mg\frac{L}{2}

\frac{1}{2}Mv^2 = \frac{MgL}{2} - \frac{MgL}{18} = \frac{4MgL}{9}

v = \sqrt{\frac{8gL}{9}} = \sqrt{\frac{8(10)(3)}{9}} = \mathbf{5.16\ \text{m/s}}

TrapTreating the hanging third as a point mass at the edge gives zero drop and no answer at all. The mass M cancels, so the chain's weight never mattered. Only its length did.

Slide down, then squash a spring

JEE ADVANCED

A 2 kg block is released from rest on a rough incline of 30°, with μ = 0.25. It slides 4 m along the incline before touching an unstretched spring of k = 500 N/m lying along the slope. Find the maximum compression. Take g = 10.

CoreFriction acts over the whole path, including the part while the spring is being squashed. So the distance in the friction term is (4 + x), not 4. That makes the equation a quadratic.

Working

mg\sin\theta(d+x) - \mu mg\cos\theta(d+x) = \frac{1}{2}kx^2

(10 - 4.33)(4+x) = 250x^2 \Rightarrow 250x^2 - 5.67x - 22.68 = 0

x = \mathbf{0.31\ \text{m}}

TrapUsing a friction distance of 4 m instead of (4 + x) is the classic error here. Friction depends on the path length travelled, never on the displacement. Gravity is the opposite, and that contrast is exactly what the question is testing.

Which statements survive

JEE ADVMULTIPLE CORRECT

A ball strikes an identical stationary ball head-on. The coefficient of restitution is e = 0.5. One or more of the following are correct. Identify all of them.

(A) Both balls move forward after the collision
(B) The first ball ends with one third of the speed of the second
(C) The fractional loss of kinetic energy is 37.5%
(D) The first ball rebounds backwards

CoreWrite the two conditions: momentum conservation, and the definition of e. Solve them together, then test each statement against the result.

Working

u = v_1 + v_2 \quad\text{and}\quad v_2 - v_1 = eu = 0.5u

v_1 = 0.25u, \qquad v_2 = 0.75u

\frac{K_f}{K_i} = \frac{(0.25)^2+(0.75)^2}{1} = 0.625 \Rightarrow \text{loss} = 37.5\%

\mathbf{\text{A, B and C are correct}}

Trap(D) fails: both velocities came out positive, so nothing rebounds. A body only rebounds when it strikes something heavier. The general result is worth carrying: for equal masses the fractional energy loss is

(1-e^2)/2

, which gives 37.5% at e = 0.5.

Why they always separate at a right angle

JEE ADVANCED

A moving ball collides obliquely and elastically with an identical stationary ball. Both are smooth. Prove that after the collision the two velocities are at 90° to each other.

CoreWrite momentum as a vector equation and energy as a scalar one. Then square the vector equation and compare the two.

Working

\text{momentum: } \vec{u} = \vec{v_1} + \vec{v_2}

\text{energy (elastic): } u^2 = v_1^2 + v_2^2

\text{square the first: } u^2 = v_1^2 + v_2^2 + 2\,\vec{v_1}\cdot\vec{v_2}

\Rightarrow \vec{v_1}\cdot\vec{v_2} = 0 \Rightarrow \mathbf{\text{the angle is } 90^\circ}

TrapThis holds only for equal masses and only when the collision is perfectly elastic. If e < 1 the angle comes out less than 90°. Snooker players rely on this result without ever deriving it.

★ Work, Power & Energy · Fact Sheet

Every formula for revision day. Print this page alone.

WORK

W = F d cosθ

Scalar, unit joule.
Zero when force is at 90°.

VARIABLE FORCE

W = area under F-x graph

Above axis positive,
below axis negative.

KINETIC ENERGY

K = ½mv² = p²/2m

Always positive.
Double v gives 4 times K.

WORK-ENERGY THEOREM

W(net) = ΔK

Needs every force,
including friction.

POTENTIAL ENERGY

U = mgh (gravity)

U = ½kx² (spring)
Spring force F = kx.

CONSERVATIVE FORCE

Same work on any path.

Gravity and springs yes.
Friction no.

ENERGY CONSERVATION

Kᵢ + Uᵢ = Kᶠ + Uᶠ

Only without friction.
Else add W(friction).

FREE FALL & SPRING

v = √(2gh) from height h

½mv² = ½kx² at full
compression.

POWER

P = W/t = F·v

Unit watt. 1 hp = 746 W.
kWh is ENERGY, not power.

COLLISIONS

Momentum always conserved.

KE conserved only if elastic.
Equal masses swap velocity.

RESTITUTION

e = separation / approach

Rebound height = e²h
After n bounces eⁿ² h.

INELASTIC LOSS

ΔK = m₁m₂(u₁−u₂)² / 2(m₁+m₂)

Largest when they stick.
Momentum still conserved.

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