Chapter 5 · Work, Energy and Power

5.3 Power and Collisions

Power is the rate of doing work, $P=\mathbf{F}\cdot\mathbf{v}$. In every collision momentum is conserved; kinetic energy is conserved only in elastic collisions — and in one dimension that fixes both final velocities.

Two cranes on a building site can lift the same $1$-tonne load to the same height: they do exactly the same work. But one does it in twenty seconds and the other in two minutes, and the builder will pay much more for the first. What distinguishes them is not work but the rate of doing work — power. Power is what an engine, a motor, a power station or an athlete is rated by, and what an electricity bill is ultimately about.

The second half of this section applies everything in this chapter and in Chapter 4 to collisions: balls on a snooker table, railway wagons coupling, a bullet burying itself in a block, cars in an accident, neutrons slowing down in a nuclear reactor. Collisions are short and violent, and the forces during them are almost impossible to measure. Yet two conservation laws — momentum always, kinetic energy sometimes — predict the outcome without knowing those forces at all.

Power

📘 Definition 5.8 — Power

If a force does work $W$ in a time $t$, its average power is $P_\text{av}=\dfrac{W}{t}$. The instantaneous power is the rate of doing work at an instant:

$$P=\lim_{\Delta t\to0}\frac{\Delta W}{\Delta t}=\frac{dW}{dt}.$$

Power is a scalar. Its SI unit is the watt: $1\ \text{W}=1\ \text{J/s}$, dimensions $[\text{M}\text{L}^2\text{T}^{-3}]$. Engines are still often rated in horsepower: $1\ \text{hp}=746\ \text{W}$.

In a short time $\Delta t$ a force $\mathbf{F}$ moves its point of application through $\Delta\mathbf{r}$ and does work $\mathbf{F}\cdot\Delta\mathbf{r}$. Dividing by $\Delta t$ and letting $\Delta t\to0$ gives a very useful formula.

📐 Result 5.8 — Power in terms of force and velocity

The instantaneous power delivered by a force $\mathbf{F}$ acting on a body moving with velocity $\mathbf{v}$ is

$$P=\mathbf{F}\cdot\mathbf{v}=Fv\cos\theta .$$

A force perpendicular to the velocity delivers no power. For a body moving at constant speed against a resistance $F_\text{r}$, the driving force equals $F_\text{r}$ and the power needed is $F_\text{r}v$.

Electricity companies sell energy, not power. Their unit, the kilowatt-hour (the "unit" on an Indian electricity bill), is the energy delivered in one hour at $1\ \text{kW}$: $1\ \text{kWh}=1000\ \text{W}\times3600\ \text{s}=3.6\times10^6\ \text{J}$. It is a unit of energy, not of power.

✏️ Example 1 — The power of a lift motor

A lift car with its passengers has a total mass of $1800\ \text{kg}$. It rises at a steady $2.0\ \text{m/s}$, and a frictional force of $4000\ \text{N}$ opposes its motion. Find the power the motor must deliver, in watts and in horsepower.

💡 Show solution▶

Force Steady speed means zero net force: the motor's upward pull is $F=mg+f=1800\times9.8+4000=21\,640\ \text{N}$.

Power $P=Fv=21\,640\times2.0=4.3\times10^4\ \text{W}=43\ \text{kW}$.

Horsepower $\dfrac{43\,280}{746}=58\ \text{hp}$.

Check Every second the lift gains $mgh=1800\times9.8\times2.0=35\,280\ \text{J}$ of potential energy and $8000\ \text{J}$ is lost to friction: $43\,280\ \text{J}$ per second ✓.

✏️ Example 2 — A crane motor, a heater and a staircase

(a) A crane motor raises a $200\ \text{kg}$ load at a steady $0.50\ \text{m/s}$. Its efficiency (useful power out ÷ electrical power in) is $80\%$. Find the useful power, the electrical power drawn, and the electrical energy used to raise the load $10\ \text{m}$. (b) A $1.5\ \text{kW}$ heater runs $4.0$ hours a day for $30$ days. How many units (kWh) does it use, and what does that cost at $\text{₹}\,7$ per unit? (c) A $60\ \text{kg}$ student runs up stairs $12\ \text{m}$ high in $15\ \text{s}$. Find her average power output.

💡 Show solution▶

(a) power Useful: $P_\text{out}=mgv=200\times9.8\times0.50=980\ \text{W}$. Drawn: $P_\text{in}=\dfrac{980}{0.80}=1.2\times10^3\ \text{W}$ ($1225\ \text{W}$).

(a) energy Time to rise $10\ \text{m}$: $\tfrac{10}{0.50}=20\ \text{s}$. Energy drawn $=1225\times20=2.5\times10^4\ \text{J}$, of which $mgh=19\,600\ \text{J}$ is stored as potential energy.

(b) $1.5\ \text{kW}\times4.0\ \text{h}\times30=180\ \text{kWh}$ $=6.5\times10^8\ \text{J}$. Cost $=180\times\text{₹}\,7=\text{₹}\,1260$.

Check Part (a) is the default setting of Interactive 1 ✓.

A crane motor lifts a load at constant speed. On the right, the graph shows the energy delivered against time: the green line is the useful work ($mgh$), the red line the electrical energy drawn. Both are straight lines, and their slopes are the useful power $P=mgv$ and the input power $P/\eta$. Change the mass, the speed or the efficiency and watch the slopes change.

load $m$:200 kgspeed $v$:0.50 m/sefficiency $\eta$:80%

useful power $mgv$—

input power—

time to lift 10 m—

energy drawn (10 m)—

Collisions: what is conserved

In a collision two bodies push on each other with large forces for a short time. If no external force acts (or external forces such as gravity and friction are small compared with the collision forces during the brief contact), the two bodies form an isolated system and, by Chapter 4, their total momentum is conserved — in every collision, whatever happens to the bodies. Total energy is conserved too, but some kinetic energy may be turned into heat, sound, or permanent deformation.

📘 Definition 5.9 — Elastic and inelastic collisions

A collision is elastic if the total kinetic energy after the collision equals that before. (During contact, kinetic energy is stored briefly as elastic potential energy and then fully returned.)

It is inelastic if the total kinetic energy decreases. It is perfectly (completely) inelastic if the bodies stick together and move off with a common velocity; this loses the most kinetic energy that momentum conservation allows.

Momentum is conserved in all three cases. In a head-on (one-dimensional) collision all velocities lie along one line; otherwise the collision is oblique (two-dimensional).

Collisions between hard steel balls or billiard balls are nearly elastic; collisions of atoms and subatomic particles can be exactly elastic. A lump of clay hitting the floor, a bullet stopping inside a block and two railway wagons coupling are perfectly inelastic.

Collisions in one dimension

Perfectly inelastic. Masses $m_1$ and $m_2$ with velocities $u_1$, $u_2$ (signs give the direction along the line) stick and move off with common velocity $v$. Momentum alone fixes $v$.

📐 Result 5.9 — Perfectly inelastic collision in one dimension

$$v=\frac{m_1u_1+m_2u_2}{m_1+m_2}.$$

The kinetic energy lost is

$$\Delta K=\frac{m_1m_2}{2(m_1+m_2)}\,(u_1-u_2)^2 ,$$

which is never negative and is zero only if $u_1=u_2$ (no collision at all).

✏️ Example 3 — Coupling railway wagons

A $2.0\times10^4\ \text{kg}$ wagon moving at $3.0\ \text{m/s}$ runs into a stationary $1.0\times10^4\ \text{kg}$ wagon and they couple. Find (a) their common speed, (b) the kinetic energy before and after, and the fraction lost.

💡 Show solution▶

(a) $v=\dfrac{(2.0\times10^4)(3.0)+0}{3.0\times10^4}=2.0\ \text{m/s}$.

(b) $K_i=\tfrac12(2.0\times10^4)(3.0)^2=9.0\times10^4\ \text{J}$; $\ K_f=\tfrac12(3.0\times10^4)(2.0)^2=6.0\times10^4\ \text{J}$. Lost: $3.0\times10^4\ \text{J}$, one third.

Check Result 5.9: $\Delta K=\dfrac{(2.0\times10^4)(1.0\times10^4)}{2(3.0\times10^4)}(3.0)^2=3.0\times10^4\ \text{J}$ ✓. The lost energy goes into the coupling buffers (as heat) and noise.

Elastic. Now both momentum and kinetic energy are conserved:

$$m_1u_1+m_2u_2=m_1v_1+m_2v_2,$$ $$\tfrac12m_1u_1^2+\tfrac12m_2u_2^2=\tfrac12m_1v_1^2+\tfrac12m_2v_2^2 .$$

Rearrange both with the $m_1$ terms on the left: $m_1(u_1-v_1)=m_2(v_2-u_2)$ and $m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2)$. Factorise the second as $m_1(u_1-v_1)(u_1+v_1)$ $=m_2(v_2-u_2)(v_2+u_2)$ and divide by the first (for a real collision $u_1\ne v_1$): $u_1+v_1=u_2+v_2$, i.e. $v_2-v_1=u_1-u_2$. Solving this together with the momentum equation gives the final velocities.

📐 Result 5.10 — Elastic collision in one dimension

In a head-on elastic collision, the relative speed of separation equals the relative speed of approach: $v_2-v_1=u_1-u_2$, and

$$v_1=\frac{(m_1-m_2)\,u_1+2m_2u_2}{m_1+m_2},$$ $$v_2=\frac{(m_2-m_1)\,u_2+2m_1u_1}{m_1+m_2}.$$

If $m_2$ is initially at rest ($u_2=0$): $v_1=\dfrac{m_1-m_2}{m_1+m_2}u_1$ and $v_2=\dfrac{2m_1}{m_1+m_2}u_1$. Special cases: equal masses exchange velocities (the moving ball stops dead); a very heavy body hitting a light one at rest keeps going at almost $u_1$ and the light one flies off at almost $2u_1$; a light body hitting a very heavy one at rest bounces back with almost $-u_1$.

✏️ Example 4 — Two carts on an air track

A $2.0\ \text{kg}$ cart moving at $3.0\ \text{m/s}$ collides elastically and head-on with a $1.0\ \text{kg}$ cart at rest on a frictionless track. Find the velocities after the collision and check both conservation laws.

💡 Show solution▶

$v_1$ $v_1=\dfrac{2.0-1.0}{3.0}\times3.0=1.0\ \text{m/s}$ (same direction as before).

$v_2$ $v_2=\dfrac{2\times2.0}{3.0}\times3.0=4.0\ \text{m/s}$.

Momentum Before $6.0\ \text{kg m/s}$; after $2.0(1.0)+1.0(4.0)=6.0\ \text{kg m/s}$ ✓.

Energy Before $\tfrac12(2.0)(9.0)=9.0\ \text{J}$; after $1.0+8.0=9.0\ \text{J}$ ✓. Separation speed $4.0-1.0=3.0\ \text{m/s}$ equals approach speed ✓.

Check A numerical simulation with a very stiff spring between the carts gives the same $1.0$ and $4.0\ \text{m/s}$; it is also the default setting of Interactive 2.

✏️ Example 5 — The ballistic pendulum

To measure a bullet's speed, a $10\ \text{g}$ bullet is fired horizontally into a $2.0\ \text{kg}$ wooden block hanging from long strings. The bullet stays in the block, which swings up so that its centre rises $20\ \text{cm}$. Find (a) the speed of block-plus-bullet just after impact, (b) the bullet's speed, (c) the percentage of the bullet's kinetic energy that was lost in the impact.

💡 Show solution▶

Two stages The impact is perfectly inelastic: use momentum (not energy). The swing afterwards has only gravity doing work (the string tension is perpendicular): use energy.

(a) $\tfrac12(m+M)V^2=(m+M)gh$: $V=\sqrt{2gh}=\sqrt{2\times9.8\times0.20}=1.98\ \text{m/s}$.

(b) $mu=(m+M)V$: $u=\dfrac{2.01}{0.010}\times1.98=4.0\times10^2\ \text{m/s}$ ($398$).

(c) $\dfrac{K_f}{K_i}=\dfrac{\tfrac12(m+M)V^2}{\tfrac12mu^2}=\dfrac{m}{m+M}=\dfrac{0.010}{2.01}=0.5\%$, so $99.5\%$ is lost — mostly as heat in the wood.

Check Using energy conservation for the impact would give $u=\sqrt{\tfrac{2.01}{0.010}\times2\times9.8\times0.20}=28\ \text{m/s}$ — absurdly low for a bullet. Energy is not conserved in the impact ✓.

Two carts on a frictionless track collide head-on. Choose the masses, the initial velocities (negative = moving left) and the type of collision, then press Play. The bars compare the total momentum (purple) and total kinetic energy (amber) before and after: momentum never changes, kinetic energy is kept only in an elastic collision. The "partly elastic" option uses the coefficient of restitution $e=0.5$ from the 🚀 box below.

type:$m_1$:2.0 kg$m_2$:1.0 kg$u_1$:3.0 m/s$u_2$:0 m/s

$v_1$ after—

$v_2$ after—

momentum before / after—

KE before / after—

Collisions in two dimensions

When a snooker ball strikes another off-centre, the two move off at angles to the original line. Momentum is a vector, so it is conserved in each direction separately. Taking $x$ along the incoming velocity of $m_1$ (with $m_2$ at rest), and angles $\theta_1$, $\theta_2$ on opposite sides of the $x$-axis:

$$x:\quad m_1u_1=m_1v_1\cos\theta_1+m_2v_2\cos\theta_2,$$ $$y:\quad 0=m_1v_1\sin\theta_1-m_2v_2\sin\theta_2 .$$

If the collision is elastic, add $\tfrac12m_1u_1^2=\tfrac12m_1v_1^2+\tfrac12m_2v_2^2$. That is three equations for four unknowns ($v_1$, $v_2$, $\theta_1$, $\theta_2$): one of them must be measured (it depends on exactly where the balls touch).

Equal masses, one at rest, elastic. Then momentum gives $\mathbf{u}_1=\mathbf{v}_1+\mathbf{v}_2$ and energy gives $u_1^2=v_1^2+v_2^2$. Squaring the first with the scalar product, $u_1^2=v_1^2+v_2^2+2\,\mathbf{v}_1\cdot\mathbf{v}_2$, so $\mathbf{v}_1\cdot\mathbf{v}_2=0$: the two balls move off at right angles (unless the collision is head-on). Snooker players use this every game.

✏️ Example 6 — A glancing collision of two pucks

An ice-hockey puck moving at $3.0\ \text{m/s}$ strikes an identical puck at rest. The collision is elastic, and the first puck moves off at $30^\circ$ to its original direction. Find the direction of the second puck and both speeds.

💡 Show solution▶

Direction Equal masses, elastic, one at rest: the pucks separate at $90^\circ$, so the second moves at $60^\circ$ on the other side of the original line.

Speeds $\mathbf{u}_1=\mathbf{v}_1+\mathbf{v}_2$ with $\mathbf{v}_1\perp\mathbf{v}_2$: a right-angled triangle with hypotenuse $u_1$. So $v_1=u_1\cos30^\circ=2.6\ \text{m/s}$ and $v_2=u_1\sin30^\circ=1.5\ \text{m/s}$.

Check $x$ $2.60\cos30^\circ+1.50\cos60^\circ=2.25+0.75=3.0\ \text{m/s}$ ✓ (masses cancel).

Check $y$, KE $2.60\sin30^\circ-1.50\sin60^\circ=1.30-1.30=0$ ✓; $\ 2.60^2+1.50^2=6.75+2.25=9.0=3.0^2$ ✓.

⚠️ Common mistakes

1. Conserving kinetic energy in an inelastic collision. Momentum is conserved in every collision; kinetic energy only in elastic ones. When bodies stick, never write $\tfrac12m_1u_1^2=\tfrac12(m_1+m_2)v^2$ (see the Check in Example 5).

2. Dropping the signs. In one dimension, velocities to the left are negative. A ball that bounces back has a negative final velocity, and its momentum change is $m(v-u)$, not $m(v+u)$ written carelessly.

3. Mixing up power and energy. The kilowatt is a unit of power; the kilowatt-hour is a unit of energy. A "$2\ \text{kW}$ heater" uses $2\ \text{kWh}$ in one hour.

4. Adding momenta in two dimensions as numbers. Resolve into components and conserve $p_x$ and $p_y$ separately.

5. "Kinetic energy is destroyed." In an inelastic collision the lost kinetic energy becomes heat, sound and deformation; the total energy is conserved.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — the coefficient of restitution

In JEE Main, NEET and JEE Advanced (as part of elastic and inelastic collisions); not required by the CBSE syllabus.

Real collisions are neither perfectly elastic nor perfectly inelastic. JEE and NEET describe them with the coefficient of restitution, defined for a head-on collision as

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

with $e=1$ for an elastic collision, $e=0$ for a perfectly inelastic one, and $0\lt e\lt1$ in between. Combining it with momentum conservation gives $v_1=\dfrac{m_1u_1+m_2u_2-m_2e(u_1-u_2)}{m_1+m_2}$ and $v_2=v_1+e(u_1-u_2)$. For the carts of Example 4 with $e=0.5$: $v_1=\dfrac{6.0-1.5}{3.0}=1.5\ \text{m/s}$ and $v_2=3.0\ \text{m/s}$ (try it in Interactive 2).

Bouncing ball. A ball dropped from height $h$ onto a hard floor rebounds with $e$ times its impact speed, so it rises to $e^2h$; after $n$ bounces it reaches $e^{2n}h$. With $e=0.80$ and $h=5.0\ \text{m}$: first rebound $3.2\ \text{m}$, third rebound $0.8^6\times5.0=1.3\ \text{m}$.

Exercises

Take $g=9.8\ \text{m/s}^2$ unless told otherwise. Collisions are on frictionless surfaces unless stated.

Power.

1 A $1.0\ \text{kW}$ motor runs for $2.0$ hours. How much energy does it use, in joules and in kWh?
2 A pump lifts $600\ \text{kg}$ of water through $15\ \text{m}$ in $2.0$ minutes. Find its useful power output in watts and in horsepower.
3 If the pump of part 2 is $70\%$ efficient, what electrical power does it draw?
4 A car moves at a steady $20\ \text{m/s}$ against a total resistance of $800\ \text{N}$. What power does its engine deliver to the wheels?
5 A force $\mathbf{F}=(2\hat{\mathbf{i}}+3\hat{\mathbf{j}})\ \text{N}$ acts on a body moving with velocity $\mathbf{v}=(4\hat{\mathbf{i}}-\hat{\mathbf{j}})\ \text{m/s}$. Find the power delivered by the force.
6 A $50\ \text{kg}$ student runs up $30$ steps, each $18\ \text{cm}$ high, in $9.0\ \text{s}$. Find her average power.
7 A $2.0\ \text{kg}$ body starts from rest under a constant net force of $4.0\ \text{N}$. Find the instantaneous power at $t=3.0\ \text{s}$ and the average power over the first $3.0\ \text{s}$.
8 A $1000\ \text{kg}$ car accelerates uniformly from rest to $20\ \text{m/s}$ in $10\ \text{s}$ (neglect resistance). Find the average power and the power at the end of the $10\ \text{s}$.

💡 Show solutions▶

1 $1000\times7200=7.2\times10^6\ \text{J}=2.0\ \text{kWh}$.

2 $P=\dfrac{mgh}{t}=\dfrac{600\times9.8\times15}{120}=735\ \text{W}\approx7.4\times10^2\ \text{W}$, i.e. $0.99\ \text{hp}$.

3 $\dfrac{735}{0.70}=1.05\times10^3\ \text{W}$.

4 $P=Fv=800\times20=1.6\times10^4\ \text{W}=16\ \text{kW}$.

5 $P=\mathbf{F}\cdot\mathbf{v}=8-3=5\ \text{W}$.

6 $h=30\times0.18=5.4\ \text{m}$; $P=\dfrac{50\times9.8\times5.4}{9.0}=294\ \text{W}\approx2.9\times10^2\ \text{W}$.

7 $a=2.0\ \text{m/s}^2$, $v(3.0)=6.0\ \text{m/s}$: $P=Fv=24\ \text{W}$. Work in $3.0\ \text{s}$ $=\Delta K=\tfrac12(2.0)(6.0)^2=36\ \text{J}$, so $P_\text{av}=12\ \text{W}$ — half the final power, because $P\propto v\propto t$.

8 $P_\text{av}=\dfrac{\tfrac12(1000)(20)^2}{10}=2.0\times10^4\ \text{W}$. $F=ma=1000\times2.0=2000\ \text{N}$; at the end $P=Fv=4.0\times10^4\ \text{W}$.

Collisions in one dimension.

1 A $4.0\ \text{kg}$ ball moving at $5.0\ \text{m/s}$ hits a $6.0\ \text{kg}$ ball at rest and they stick together. Find their common velocity and the kinetic energy lost.
2 A $1.0\ \text{kg}$ ball moving at $6.0\ \text{m/s}$ collides elastically and head-on with a $2.0\ \text{kg}$ ball at rest. Find both final velocities.
3 Two identical balls move towards each other at $3.0\ \text{m/s}$ and $2.0\ \text{m/s}$ and collide elastically head-on. Find their velocities afterwards.
4 In a nuclear reactor, fast neutrons are slowed by elastic head-on collisions with nuclei at rest. What fraction of a neutron's kinetic energy is transferred to a carbon nucleus (mass $12$ times the neutron's)? To a hydrogen nucleus (about the same mass as a neutron)?
5 Verify, for your answer to part 2, that the speed of separation equals the speed of approach.
6 A $1.5\ \text{kg}$ trolley moving right at $2.0\ \text{m/s}$ meets a $0.50\ \text{kg}$ trolley moving left at $4.0\ \text{m/s}$; they stick. Find the final velocity and the kinetic energy lost.
7 A $60\ \text{g}$ ball hits a wall at $8.0\ \text{m/s}$ perpendicularly and rebounds elastically. Find its velocity after and the magnitude of its change in momentum.
8 A wagon moving at $6.0\ \text{m/s}$ runs into two identical coupled wagons at rest, and all three couple. Find their speed and the fraction of kinetic energy lost.

💡 Show solutions▶

1 $v=\dfrac{4.0\times5.0}{10}=2.0\ \text{m/s}$; $K_i=50\ \text{J}$, $K_f=\tfrac12(10)(2.0)^2=20\ \text{J}$; lost $30\ \text{J}$.

2 $v_1=\dfrac{1.0-2.0}{3.0}\times6.0=-2.0\ \text{m/s}$ (bounces back); $v_2=\dfrac{2(1.0)}{3.0}\times6.0=4.0\ \text{m/s}$.

3 Equal masses exchange velocities: taking the first ball's direction as positive, $u_1=3.0$, $u_2=-2.0$; after, $v_1=-2.0\ \text{m/s}$ and $v_2=+3.0\ \text{m/s}$.

4 Fraction transferred $=\dfrac{\tfrac12m_2v_2^2}{\tfrac12m_1u_1^2}=\dfrac{4m_1m_2}{(m_1+m_2)^2}$. Carbon: $\dfrac{4\times12}{13^2}=0.28$ ($28\%$). Hydrogen: $\dfrac{4}{4}=1$ — all of it in a head-on collision. That is why hydrogen-rich water is a good moderator.

5 Approach: $u_1-u_2=6.0\ \text{m/s}$; separation: $v_2-v_1=4.0-(-2.0)=6.0\ \text{m/s}$ ✓.

6 $v=\dfrac{1.5(2.0)+0.50(-4.0)}{2.0}=0.50\ \text{m/s}$ (right). $K_i=3.0+4.0=7.0\ \text{J}$, $K_f=\tfrac12(2.0)(0.50)^2=0.25\ \text{J}$; lost $6.75\ \text{J}\approx6.8\ \text{J}$.

7 A very heavy target: $v=-8.0\ \text{m/s}$. $|\Delta p|=0.060\times(8.0+8.0)=0.96\ \text{kg m/s}$.

8 $v=\dfrac{m(6.0)}{3m}=2.0\ \text{m/s}$; $\dfrac{K_f}{K_i}=\dfrac{\tfrac12(3m)(4.0)}{\tfrac12m(36)}=\dfrac13$, so $\tfrac23$ is lost.

III

Applications and two-dimensional collisions.

1 A $20\ \text{g}$ bullet is fired into a $4.98\ \text{kg}$ ballistic pendulum, which rises $5.0\ \text{cm}$. Find the bullet's speed.
2 What percentage of the bullet's kinetic energy is lost in the impact of part 1?
3 A $1.0\ \text{kg}$ block at $6.0\ \text{m/s}$ hits and sticks to a $5.0\ \text{kg}$ block at rest. The pair then compresses a spring with $k=1500\ \text{N/m}$. Find the greatest compression.
4 A $1200\ \text{kg}$ car at $15\ \text{m/s}$ runs into the back of a stationary $1800\ \text{kg}$ car and the two lock together. Find their speed just after and the kinetic energy lost.
5 A puck moving at $5.0\ \text{m/s}$ collides elastically with an identical puck at rest and is deflected through $40^\circ$. Find the direction of the second puck and both speeds.
6 A $1500\ \text{kg}$ car travelling east at $20\ \text{m/s}$ collides at a junction with a $2500\ \text{kg}$ van travelling north at $12\ \text{m/s}$. They lock together. Find the velocity of the wreckage just after the collision.
7 (Modelling) In a "Newton's cradle" of identical steel balls, one ball released from the end knocks out exactly one ball at the other end. Show that two balls leaving at half the speed would conserve momentum but not kinetic energy. What does that tell you about the collisions?

💡 Show solutions▶

1 $V=\sqrt{2\times9.8\times0.050}=0.99\ \text{m/s}$; $u=\dfrac{5.00}{0.020}\times0.99=2.5\times10^2\ \text{m/s}$.

2 Lost fraction $=\dfrac{M}{m+M}=\dfrac{4.98}{5.00}=99.6\%$.

3 $v=\dfrac{6.0}{6.0}=1.0\ \text{m/s}$, $K=\tfrac12(6.0)(1.0)^2=3.0\ \text{J}$ (energy is conserved after sticking). $\tfrac12(1500)x^2=3.0$: $x=0.063\ \text{m}=6.3\ \text{cm}$.

4 $v=\dfrac{1200\times15}{3000}=6.0\ \text{m/s}$; $K_i=1.35\times10^5\ \text{J}$, $K_f=5.4\times10^4\ \text{J}$; lost $8.1\times10^4\ \text{J}$.

5 The pucks separate at $90^\circ$: the second moves at $50^\circ$ on the other side. $v_1=5.0\cos40^\circ=3.8\ \text{m/s}$, $v_2=5.0\sin40^\circ=3.2\ \text{m/s}$.

6 $p_x=1500\times20=3.0\times10^4$, $p_y=2500\times12=3.0\times10^4\ \text{kg m/s}$. $v=\dfrac{\sqrt{2}\times3.0\times10^4}{4000}=10.6\ \text{m/s}\approx11\ \text{m/s}$ at $45^\circ$ north of east.

7 One ball in at $v$: $p=mv$, $K=\tfrac12mv^2$. Two out at $v/2$: $p=2m(v/2)=mv$ ✓ but $K=2\cdot\tfrac12m(v/2)^2=\tfrac14mv^2$ ✗. Only "one ball out at $v$" conserves both, so the collisions are (nearly) elastic.

Power and energy in context.

1 A $2.0\ \text{MW}$ wind turbine produces, on average over a year, $35\%$ of its rated power. How many kWh does it generate in a year ($8760$ hours)?
2 Water falls through $40\ \text{m}$ at a dam at $50\ \text{m}^3$ per second. Find the power available, and the electrical output if the turbines are $85\%$ efficient.
3 A cyclist working at $250\ \text{W}$ rides at constant speed against a total resistance of $25\ \text{N}$. Find her speed in km/h.
4 Show that a vehicle whose engine delivers a maximum power $P$ against a resistance $F_\text{r}$ has top speed $v_\text{max}=P/F_\text{r}$ (resistance taken as constant). Evaluate for $P=60\ \text{kW}$, $F_\text{r}=1500\ \text{N}$.

💡 Show solutions▶

1 $2000\ \text{kW}\times0.35\times8760\ \text{h}=6.1\times10^6\ \text{kWh}$.

2 Each second $5.0\times10^4\ \text{kg}$ falls $40\ \text{m}$: $P=\dfrac{mgh}{t}=5.0\times10^4\times9.8\times40=2.0\times10^7\ \text{W}$ ($19.6\ \text{MW}$); output $0.85\times19.6=17\ \text{MW}$.

3 $v=\dfrac PF=\dfrac{250}{25}=10\ \text{m/s}=36\ \text{km/h}$.

4 At top speed the acceleration is zero, so the driving force equals $F_\text{r}$ and $P=F_\text{r}v_\text{max}$. $v_\text{max}=\dfrac{60\,000}{1500}=40\ \text{m/s}$ ($144\ \text{km/h}$). (Real air resistance grows with speed, so real top speeds are lower.)

🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks A body of mass $m_1$ moving with velocity $u_1$ collides elastically and head-on with a body of mass $m_2$ at rest. Obtain expressions for their velocities after the collision. Show that if $m_1=m_2$ the bodies exchange velocities.

💡 Show solution▶

Equations Momentum: $m_1u_1=m_1v_1+m_2v_2$. Kinetic energy: $m_1u_1^2=m_1v_1^2+m_2v_2^2$. [1]

Solve Rearranging, $m_1(u_1-v_1)=m_2v_2$ and $m_1(u_1-v_1)(u_1+v_1)=m_2v_2^2$; dividing, $u_1+v_1=v_2$. Substituting: $v_1=\dfrac{m_1-m_2}{m_1+m_2}u_1$, $v_2=\dfrac{2m_1}{m_1+m_2}u_1$. [1]

Equal masses $m_1=m_2$: $v_1=0$, $v_2=u_1$ — the first stops and the second moves off with its velocity. [1]

Exam tip Dividing by $(u_1-v_1)$ is allowed because $v_1\ne u_1$ in a real collision — say so.

E2. NEET A body moving with speed $v$ collides with an identical body at rest and they stick together. The fraction of the initial kinetic energy lost is (a) $\tfrac14$ (b) $\tfrac12$ (c) $\tfrac34$ (d) zero

💡 Show solution▶

Idea Common velocity $v/2$. $K_f=\tfrac12(2m)(v/2)^2=\tfrac14mv^2=\tfrac12K_i$.

Answer (b) $\tfrac12$.

Exam tip In general the fraction lost when $m_1$ hits $m_2$ at rest and they stick is $\dfrac{m_2}{m_1+m_2}$.

E3. JEE Main (Numerical value) A pump draws water from a well $10\ \text{m}$ deep at $20\ \text{kg}$ per second and delivers it at the surface with a speed of $10\ \text{m/s}$. Taking $g=10\ \text{m/s}^2$, find the useful power of the pump in kW.

💡 Show solution▶

Energy per second Potential: $20\times10\times10=2000\ \text{J/s}$. Kinetic: $\tfrac12\times20\times10^2=1000\ \text{J/s}$.

Answer $P=3000\ \text{W}=3\ \text{kW}$.

Exam tip The common slip is to forget the kinetic energy given to the water ($2\ \text{kW}$ is the trap).

E4. JEE Advanced (One or more correct) A body of mass $m_1$ moving with speed $u$ collides elastically and head-on with a body of mass $m_2$ at rest. Which statements are correct? (A) If $m_1=m_2$, the first body comes to rest. (B) If $m_1\gg m_2$, the second body moves off with a speed close to $2u$. (C) If $m_1\ll m_2$, the first body rebounds with a speed close to $u$. (D) The fraction of kinetic energy transferred to $m_2$ is greatest when $m_2=2m_1$.

💡 Show solution▶

(A) $v_1=\dfrac{m_1-m_2}{m_1+m_2}u=0$ ✓.

(B) $v_2=\dfrac{2m_1}{m_1+m_2}u\to2u$ as $m_2/m_1\to0$ ✓.

(D) Fraction transferred $=\dfrac{4m_1m_2}{(m_1+m_2)^2}$, which equals $1$ at $m_2=m_1$ and only $\tfrac89$ at $m_2=2m_1$ ✗.

Answer (A), (B), (C).

E5. NEET A machine gun fires $360$ bullets per minute, each of mass $20\ \text{g}$, with a speed of $500\ \text{m/s}$. The power of the gun is (a) $2.5\ \text{kW}$ (b) $15\ \text{kW}$ (c) $150\ \text{kW}$ (d) $900\ \text{kW}$

💡 Show solution▶

Idea Each bullet gets $\tfrac12(0.020)(500)^2=2500\ \text{J}$; $6$ bullets per second.

Answer $P=6\times2500=15\ \text{kW}$: (b).

Exam tip (a) is the energy of one bullet per second; (d) forgets to convert minutes to seconds.

E6. JEE Main A ball moving at $6.0\ \text{m/s}$ collides elastically with an identical ball at rest. After the collision the first ball moves at $60^\circ$ to its original direction. The speeds of the first and second balls are (a) $3.0$ and $5.2\ \text{m/s}$ (b) $5.2$ and $3.0\ \text{m/s}$ (c) $3.0$ and $3.0\ \text{m/s}$ (d) $6.0$ and $0$

💡 Show solution▶

Idea Equal masses, elastic, one at rest: they separate at $90^\circ$, so the second ball moves at $30^\circ$ on the other side.

Speeds $v_1=6.0\cos60^\circ=3.0\ \text{m/s}$, $v_2=6.0\cos30^\circ=5.2\ \text{m/s}$.

Answer (a).

Exam tip Check: $3.0^2+5.2^2=36$ ✓ (energy). (b) swaps the angles; (c) conserves neither $p_x$ nor KE.

🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. How do accident investigators work out how fast a car was going before a crash, using only the skid marks and the masses of the vehicles?

Why it matters. Traffic-police reconstruction units in many countries present such calculations in court. The method chains the two big ideas of this chapter: the work–energy theorem gives the speed of the wreckage just after the collision from the length of the skid, and conservation of momentum takes you back across the collision — where kinetic energy is not conserved — to the speed before it.

What to do.

A $1500\ \text{kg}$ car runs into the back of a stationary $1000\ \text{kg}$ car. The locked wreckage skids $12\ \text{m}$ before stopping; the coefficient of kinetic friction between tyres and road is $0.70$. Use $\mu_kMg\,d=\tfrac12MV^2$ to find the speed $V$ just after the collision.
Use momentum conservation to find the first car's speed before the collision, in km/h. The speed limit was $60\ \text{km/h}$ — was the driver speeding?
How much kinetic energy was lost in the collision itself?
Investigators are never sure of $\mu_k$. Repeat for $\mu_k=0.60$ and $0.80$ and give the speed as a range.
For a junction (two-dimensional) collision, describe how the direction of the skid marks would also be used.

Tools: Calculator or spreadsheet; published road-surface friction data.

Hand in: A one-page "expert report" with your calculation, the uncertainty range, and a clear statement of the assumptions (no braking before impact, constant friction, level road).

Going further. Real reconstructions also use crush depth (work done deforming the cars) and the cars' onboard event data recorders. Find out what such a recorder stores and how it could check your estimate.

Instructor note: $V=\sqrt{2\mu_kgd}=\sqrt{2\times0.70\times9.8\times12}=12.8\ \text{m/s}$; $u=\tfrac{2500}{1500}\times12.8=21.4\ \text{m/s}\approx77\ \text{km/h}$ — well over the limit. KE lost in the collision: $\tfrac12(1500)(21.4)^2-\tfrac12(2500)(12.8)^2\approx1.4\times10^5\ \text{J}$ (40%, i.e. $m_2/(m_1+m_2)$). With $\mu_k=0.60$–$0.80$ the speed is about $71$–$82\ \text{km/h}$, still clearly above $60$.

🧪 Self-check quiz

0/5 correct

Q1. The SI unit of power, the watt, is equal to

Q2. One kilowatt-hour is equal to

Q3. In a perfectly inelastic collision between two bodies (no external force),

Q4. A ball collides elastically and head-on with an identical ball at rest. After the collision,

Q5. A force acts on a body always perpendicular to its velocity. The power delivered by this force is

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