Chapter 4 ยท Laws of Motion

4.1 Inertia, Momentum and Newton's First and Second Laws

Motion needs no cause โ€” only changes of motion do. The law of inertia, linear momentum $\mathbf{p}=m\mathbf{v}$, Newton's second law $\mathbf{F}=\dfrac{d\mathbf{p}}{dt}=m\mathbf{a}$, and impulse, the area under a forceโ€“time graph.

Push a heavy box across a floor and it moves; stop pushing and it stops. For almost two thousand years this everyday experience was taken as a law of nature: Aristotle taught that a body moves only while something keeps pushing it, and that its natural state is rest. The idea feels right โ€” and it is wrong. The box stops because the floor pushes back on it through friction. Polish the floor, put the box on wheels, or set it on ice, and it keeps sliding for longer and longer. In deep space, with nothing to rub against, it would never stop.

The correct picture, found by Galileo and stated sharply by Newton, is that a force is needed to change motion โ€” to start, stop, speed up, slow down or turn a body โ€” not to keep it going. This section turns that idea into two laws. The first law says what happens when no net force acts. The second tells us exactly how a net force changes a body's motion, through a quantity called momentum. Together they explain why passengers lurch forward when a bus brakes, why a cricketer pulls her hands back to catch, and why a car's crumple zone saves lives.

Aristotle's fallacy and the law of inertia

Galileo tested the question with balls rolling on smooth inclined planes. A ball rolling down a slope speeds up; one rolling up slows down. On a level plane, he reasoned, it should do neither. He then let a ball roll down one incline and up a second facing one: it rose to nearly the same height it started from, whatever the slope of the second plane. Making the second plane less and less steep, the ball had to roll further and further to reach that height. With the second plane flat, it would roll on for ever. The only reason real balls stop is friction โ€” an outside influence, not a built-in tendency to rest.

๐Ÿ“˜ Definition 4.1 โ€” Force and inertia

A force is a push or pull exerted on a body by another body (by contact, or at a distance as gravity does) that can change the body's state of rest or of uniform motion. Force is a vector; its SI unit is the newton, $\text{N}$. When several forces act, their vector sum is the net (resultant) force $\mathbf{F}_\text{net}=\mathbf{F}_1+\mathbf{F}_2+\dots$

Inertia is the tendency of a body to resist any change in its state of rest or of uniform motion in a straight line. The mass of a body is the measure of its inertia: the larger the mass, the harder it is to change its velocity.

๐Ÿ“ Law 4.1 โ€” Newton's first law of motion (law of inertia)

Every body continues in its state of rest, or of uniform motion in a straight line, unless it is compelled by a net external force to change that state:

$$\mathbf{F}_\text{net}=\mathbf{0}\iff\mathbf{a}=\mathbf{0}\iff\mathbf{v}=\text{constant}.$$

The law holds in an inertial frame of reference โ€” a frame that is not itself accelerating. The ground is very nearly an inertial frame for everyday problems; a braking bus or a car going round a bend is not.

The first law says nothing about speed: a body may move at $1000\ \text{m/s}$ with zero net force on it. Notice too that "zero net force" does not mean "no forces". A book resting on a table has two forces on it โ€” its weight down and the push of the table up โ€” which cancel. A car cruising at steady speed on a straight road has the forward push of the road on its tyres balanced by air drag and rolling resistance. In each case $\mathbf{F}_\text{net}=\mathbf{0}$ and the velocity stays constant.

Inertia explains many everyday jolts. When a bus brakes, its floor slows your feet but your upper body, which nothing has yet pushed, carries on forward. When it starts suddenly, your feet go with the floor and you seem to fall backward. A dusty carpet beaten with a stick moves, but the dust, by its inertia, stays behind and falls out.

โœ๏ธ Example 1 โ€” Zero net force or not?
Decide whether the net force is zero in each case, and explain. (a) A raindrop falling at a steady terminal speed. (b) A car moving at a constant $60\ \text{km/h}$ round a circular roundabout. (c) A stone at the very top of its flight after being thrown upward. (d) A lift moving up at a steady $2.0\ \text{m/s}$. (e) A space probe coasting far from all stars and planets at $17\ \text{km/s}$.
(a) Zero: the velocity is constant, so the weight is balanced exactly by air resistance.
(b) Not zero: the speed is constant but the direction changes, so there is a centripetal acceleration (Section 3.4) and a net force towards the centre of the roundabout.
(c) Not zero: the stone is momentarily at rest, but its acceleration is still $g$ downward, so the net force is its weight. "At rest for an instant" is not the same as "in equilibrium".
(d) Zero: constant velocity, so the cable's pull balances the weight of the lift.
(e) Zero (very nearly): no forces act, so by the first law it keeps its $17\ \text{km/s}$ without any engine.
Check The test is always the acceleration, never the velocity: $\mathbf{F}_\text{net}=\mathbf{0}$ exactly when $\mathbf{a}=\mathbf{0}$.

Linear momentum and the second law

Which is harder to stop: a cricket ball tossed gently, or the same ball hit hard? A bicycle or a lorry moving at the same speed? Experience says that the effect of a moving body depends both on its mass and on its velocity. Newton combined the two into one quantity.

๐Ÿ“˜ Definition 4.2 โ€” Linear momentum

The linear momentum of a body of mass $m$ moving with velocity $\mathbf{v}$ is

$$\mathbf{p}=m\mathbf{v}.$$

It is a vector in the direction of $\mathbf{v}$, with SI unit $\text{kg m/s}$ (equivalently $\text{N s}$). In components, $p_x=mv_x$ and $p_y=mv_y$.

Two simple experiments show how force and momentum are linked. A steady force applied for a longer time produces a larger change of momentum. And the same change of momentum can be produced by a large force acting briefly or a small force acting for a long time โ€” a ball can be stopped by a sharp slap or by a gentle catch. Newton's second law makes this quantitative.

๐Ÿ“ Law 4.2 โ€” Newton's second law of motion

The rate of change of momentum of a body is equal to the net external force on it, and takes place in the direction of that force:

$$\mathbf{F}_\text{net}=\frac{d\mathbf{p}}{dt}.$$

For a body of constant mass, $\dfrac{d\mathbf{p}}{dt}=m\dfrac{d\mathbf{v}}{dt}$, so

$$\mathbf{F}_\text{net}=m\mathbf{a},\qquad F_x=ma_x,\quad F_y=ma_y,\quad F_z=ma_z .$$

The unit of force follows from this law: $1\ \text{N}$ is the net force that gives a mass of $1\ \text{kg}$ an acceleration of $1\ \text{m/s}^2$, so $1\ \text{N}=1\ \text{kg m/s}^2$. (Valid in an inertial frame.)

Four points about the second law are worth fixing now.

โœ๏ธ Example 2 โ€” Braking a car
A car of mass $1200\ \text{kg}$ moving at $20\ \text{m/s}$ is brought uniformly to rest in $5.0\ \text{s}$. Find (a) its momentum before braking, (b) the braking force, (c) the stopping distance, (d) the force the seat belt must exert on a $60\ \text{kg}$ passenger to keep her moving with the car.
(a) $p=mv=1200\times20=2.4\times10^4\ \text{kg m/s}$ forward.
(b) $a=\dfrac{0-20}{5.0}=-4.0\ \text{m/s}^2$, so $F=ma=1200\times(-4.0)=-4.8\times10^3\ \text{N}$: $4.8\ \text{kN}$ backward. Equivalently $F=\dfrac{\Delta p}{\Delta t}=\dfrac{-2.4\times10^4}{5.0}$.
(c) $s=\dfrac{u+v}{2}\,t=\dfrac{20+0}{2}\times5.0=50\ \text{m}$.
(d) She needs the same acceleration: $F=60\times4.0=240\ \text{N}$ backward. Without the belt nothing provides it, and by inertia she keeps moving at $20\ \text{m/s}$ until the dashboard stops her.
Check $v^2=u^2+2as$: $0=400-2(4.0)(50)$ โœ“.
โœ๏ธ Example 3 โ€” Two forces on a puck
A $2.0\ \text{kg}$ puck rests on frictionless ice. Two ropes pull it horizontally: $6.0\ \text{N}$ due east and $8.0\ \text{N}$ due north. Find (a) the acceleration (magnitude and direction), (b) the velocity and the distance moved after $2.0\ \text{s}$.
Components With $x$ east and $y$ north: $F_x=6.0\ \text{N}$, $F_y=8.0\ \text{N}$, so $a_x=3.0\ \text{m/s}^2$, $a_y=4.0\ \text{m/s}^2$.
(a) $a=\sqrt{3.0^2+4.0^2}=5.0\ \text{m/s}^2$ at $\tan^{-1}\tfrac{4.0}{3.0}=53.1^\circ$ north of east โ€” along the net force $F_\text{net}=10\ \text{N}$.
(b) Starting from rest: $v=at=10\ \text{m/s}$ and $s=\tfrac12at^2=10\ \text{m}$, both $53.1^\circ$ north of east.
Check $F_\text{net}=ma=2.0\times5.0=10\ \text{N}=\sqrt{6.0^2+8.0^2}$ โœ“.
โœ๏ธ Example 4 โ€” A bullet stopped by a plank
A $10\ \text{g}$ bullet moving at $400\ \text{m/s}$ enters a wooden plank and comes to rest after penetrating $8.0\ \text{cm}$. Assuming a constant resisting force, find (a) the deceleration, (b) the resisting force, (c) the time taken to stop.
(a) $v^2=u^2-2as$ with $v=0$: $a=\dfrac{u^2}{2s}=\dfrac{400^2}{2\times0.080}=1.0\times10^6\ \text{m/s}^2$.
(b) $F=ma=0.010\times1.0\times10^6=1.0\times10^4\ \text{N}$ โ€” the weight of a tonne, acting on a $10\ \text{g}$ bullet.
(c) $t=\dfrac{u}{a}=\dfrac{400}{1.0\times10^6}=4.0\times10^{-4}\ \text{s}$.
Check $F\,t=1.0\times10^4\times4.0\times10^{-4}=4.0\ \text{N s}=mu=0.010\times400$ โœ“ โ€” the force times the time equals the momentum lost, which is the idea of the next subsection.
Interactive 1 โ€” Push, Then Let Go

A glider on a frictionless air track is pushed by a constant force $F$ (red arrow) for a chosen time, then released. Below, its velocityโ€“time graph is drawn as it goes. While the force acts the line slopes up with slope $a=F/m$ (second law); the moment the force stops, the line goes flat (first law). Change the force and the mass and compare the slopes; the momentum gained is always $F\,\Delta t$.

4 N2 kg3 s
acceleration $a=F/m$โ€”
speed after the pushโ€”
momentum $p=F\,\Delta t$โ€”
distance at $t=6\ \text{s}$โ€”

Impulse and impulsive forces

When a bat hits a ball, or a ball hits a wall, a very large force acts for a very short time โ€” a few milliseconds โ€” and its size changes from moment to moment. Such forces are hard to measure, but their total effect is easy to find: it is the change in momentum they produce. Multiply the second law by a short time interval $dt$: $\mathbf{F}\,dt=d\mathbf{p}$. Adding up over the whole interval gives the next result.

๐Ÿ“˜ Definition 4.3 โ€” Impulse

The impulse of a constant force $\mathbf{F}$ acting for a time $\Delta t$ is $\mathbf{J}=\mathbf{F}\,\Delta t$. For a force that varies, the impulse is the area under the forceโ€“time graph (Section 2.2), $J=\displaystyle\int_{t_1}^{t_2}F\,dt$. Its unit is $\text{N s}=\text{kg m/s}$. A large force that acts for a very short time, so that its impulse is finite, is called an impulsive force.

๐Ÿ“ Result 4.3 โ€” Impulseโ€“momentum theorem

The impulse of the net force on a body during a time interval equals the change in its momentum in that interval:

$$\mathbf{J}=\mathbf{F}_\text{avg}\,\Delta t=\Delta\mathbf{p}=m\mathbf{v}-m\mathbf{u}.$$

Here $\mathbf{F}_\text{avg}=\dfrac{\Delta\mathbf{p}}{\Delta t}$ is the average force โ€” the constant force that would give the same impulse in the same time. For a given change of momentum, the longer the time, the smaller the average force.

This is why a cricketer draws her hands back while catching, why high jumpers land on thick foam, why cars have crumple zones and airbags, and why glassware is packed in straw or bubble wrap: each makes $\Delta t$ longer for the same $\Delta p$, and so makes the force smaller. A martial-arts expert breaking a tile does the opposite: a fast, stiff strike delivers its momentum in a very short time.

โœ๏ธ Example 5 โ€” Catching a cricket ball
A $0.16\ \text{kg}$ cricket ball arrives at $25\ \text{m/s}$. Find the average force on the fielder's hands if she stops it (a) in $0.10\ \text{s}$ by drawing her hands back, (b) in $0.020\ \text{s}$ with stiff hands. (c) Compare each with the weight of the ball.
Impulse $|\Delta p|=mu=0.16\times25=4.0\ \text{kg m/s}$ in both cases.
(a) $F_\text{avg}=\dfrac{4.0}{0.10}=40\ \text{N}$.
(b) $F_\text{avg}=\dfrac{4.0}{0.020}=200\ \text{N}$ โ€” five times as much, for the same catch.
(c) The weight is $mg=0.16\times9.8=1.57\ \text{N}$, so the forces are about $25$ and $130$ times the weight. During impacts the weight can usually be neglected.
Check These are the default settings of Interactive 2: drag the stopping time from $100\ \text{ms}$ to $20\ \text{ms}$ and watch the area stay fixed while the peak shoots up.
โœ๏ธ Example 6 โ€” A ball rebounds from a wall
A $0.060\ \text{kg}$ tennis ball hits a wall head-on at $20\ \text{m/s}$ and rebounds along the same line at $15\ \text{m/s}$. The contact lasts $5.0\ \text{ms}$. Find (a) the impulse on the ball, (b) the average force of the wall on the ball, (c) the average force of the ball on the wall.
Signs Take $+$ away from the wall. Then $u=-20\ \text{m/s}$ and $v=+15\ \text{m/s}$.
(a) $J=m(v-u)=0.060\,(15-(-20))=0.060\times35=2.1\ \text{N s}$, away from the wall.
(b) $F_\text{avg}=\dfrac{2.1}{5.0\times10^{-3}}=4.2\times10^2\ \text{N}$, away from the wall.
(c) By Newton's third law (Section 4.2), $4.2\times10^2\ \text{N}$ into the wall.
Check Forgetting the sign change gives $0.060\times(15-20)$, only $0.30\ \text{N s}$. A bounce always needs more impulse than a dead stop.
Interactive 2 โ€” Catch It Softly: the Forceโ€“Time Graph

A ball is caught and brought to rest. The force on it rises and falls smoothly (modelled as a half sine wave) over the stopping time. The shaded area is the impulse, which must equal the ball's momentum โ€” so it is the same whatever the stopping time. The dashed curve is a stiff $20\ \text{ms}$ catch with the same area. Make the catch softer and watch the peak force fall. Press Play to see the area fill up as the ball slows.

25 m/s100 ms
impulse $=\Delta p$โ€”
average forceโ€”
peak forceโ€”
ball's speed nowโ€”
โš ๏ธ Common mistakes

1. "A moving body needs a force to keep it moving." This is exactly Aristotle's error. A body at constant velocity has zero net force on it; forces you see (an engine's push) are only balancing others (friction, drag).

2. Using one force instead of the net force. $\mathbf{F}=m\mathbf{a}$ needs the vector sum of all forces on the body. Draw every force first, then add.

3. Forgetting that momentum and impulse are vectors. For a rebound, $|\Delta\mathbf{p}|=m(u+v)$, not $m(u-v)$ (Example 6). Choose a positive direction and keep to it.

4. "At rest means no force." At the top of a vertical throw the velocity is zero but the net force is still the weight (Example 1(c)).

5. Mixing up mass and weight. Mass ($\text{kg}$) measures inertia and is the same everywhere; weight $mg$ ($\text{N}$) is a force and depends on $g$.

Exercises

Take $g=9.8\ \text{m/s}^2$ unless told otherwise.

I
Inertia and the first law.
  • 1 Why does a standing passenger fall forward when a bus brakes suddenly, and backward when it starts suddenly?
  • 2 Explain why beating a hanging carpet with a stick removes dust from it.
  • 3 A space probe far from all stars and planets moves at $5.0\ \text{km/s}$ with its engines off. What force is needed to keep it moving at this speed? Explain.
  • 4 A small ball hangs from the ceiling of a train on a string. As the train starts, the string swings back. Explain this using the first law, as seen by someone standing on the platform.
  • 5 In which cases is the net force zero? (a) A car at a steady $60\ \text{km/h}$ on a straight level road. (b) The same car taking a bend at $60\ \text{km/h}$. (c) A parachutist falling at a constant speed. (d) A ball at the top of its flight.
  • 6 A $3.0\ \text{kg}$ body moves at a constant $4.0\ \text{m/s}$ due east. Three forces act on it: $12\ \text{N}$ due east, $5.0\ \text{N}$ due north, and an unknown third force. Find the third force (magnitude and direction).
1 Braking: friction from the floor slows the feet, but the upper body, by inertia, keeps moving forward. Starting: the feet are carried forward while the upper body tends to stay at rest, so it is left behind.
2 The stick suddenly moves the carpet; the dust particles, by inertia, stay where they were, so they separate from the carpet and fall.
3 None. With no net force the velocity stays constant (first law); a force would be needed only to change the speed or direction.
4 The train's floor and ceiling accelerate forward; the ball, by inertia, at first stays at rest, so the top of the string moves ahead of it and the string tilts back. Then the tilted string pulls the ball forward and it accelerates with the train.
5 (a) and (c) only: constant velocity. In (b) the direction changes; in (d) the acceleration is $g$.
6 Constant velocity, so $\mathbf{F}_\text{net}=\mathbf{0}$: $\mathbf{F}_3=-(12\,\hat{\mathbf{i}}+5.0\,\hat{\mathbf{j}})\ \text{N}$. Magnitude $\sqrt{144+25}=13\ \text{N}$, directed $\tan^{-1}\tfrac{5.0}{12}=22.6^\circ$ south of west. (The speed and mass are not needed.)
II
The second law, $\mathbf{F}=m\mathbf{a}$.
  • 1 A net force of $15\ \text{N}$ acts on a $3.0\ \text{kg}$ body. Find its acceleration.
  • 2 What net force gives a $1500\ \text{kg}$ car an acceleration of $2.4\ \text{m/s}^2$?
  • 3 A $0.50\ \text{kg}$ trolley speeds up uniformly from $2.0\ \text{m/s}$ to $8.0\ \text{m/s}$ in $3.0\ \text{s}$. Find the net force on it.
  • 4 The momentum of a body is $p=(4.0\,t^2+2.0)\ \text{kg m/s}$, with $t$ in seconds. Find the force on it at $t=1.5\ \text{s}$.
  • 5 A constant force changes the speed of a $5.0\ \text{kg}$ body from $3.0\ \text{m/s}$ to $7.0\ \text{m/s}$ in $2.0\ \text{s}$ (no change of direction). Find the force and the distance covered meanwhile.
  • 6 A $2.0\ \text{kg}$ body is acted on by a net force $\mathbf{F}=(4.0\,\hat{\mathbf{i}}-6.0\,\hat{\mathbf{j}})\ \text{N}$. Find its acceleration as a vector and its magnitude.
  • 7 A $6000\ \text{kg}$ truck moving at $54\ \text{km/h}$ is stopped by its brakes in $15\ \text{m}$. Find the (constant) braking force.
  • 8 A certain force gives a mass $m_1$ an acceleration of $6.0\ \text{m/s}^2$ and a mass $m_2$ an acceleration of $3.0\ \text{m/s}^2$. What acceleration would it give the two masses fastened together?
1 $a=\dfrac{15}{3.0}=5.0\ \text{m/s}^2$ along the force.
2 $F=1500\times2.4=3.6\times10^3\ \text{N}$.
3 $a=\dfrac{8.0-2.0}{3.0}=2.0\ \text{m/s}^2$; $F=0.50\times2.0=1.0\ \text{N}$.
4 $F=\dfrac{dp}{dt}=8.0\,t$; at $t=1.5\ \text{s}$, $F=12\ \text{N}$.
5 $a=\dfrac{7.0-3.0}{2.0}=2.0\ \text{m/s}^2$, $F=10\ \text{N}$; $s=\dfrac{3.0+7.0}{2}\times2.0=10\ \text{m}$.
6 $\mathbf{a}=\dfrac{\mathbf{F}}{m}=(2.0\,\hat{\mathbf{i}}-3.0\,\hat{\mathbf{j}})\ \text{m/s}^2$; $a=\sqrt{4.0+9.0}=3.6\ \text{m/s}^2$.
7 $u=15\ \text{m/s}$; $a=\dfrac{u^2}{2s}=\dfrac{225}{30}=7.5\ \text{m/s}^2$; $F=6000\times7.5=4.5\times10^4\ \text{N}$ (opposite to the motion).
8 $m_1=\dfrac F{6.0}$, $m_2=\dfrac F{3.0}$, so $m_1+m_2=\dfrac F{2.0}$ and $a=\dfrac{F}{m_1+m_2}=2.0\ \text{m/s}^2$.
III
Momentum and impulse.
  • 1 Find the momentum of a $0.145\ \text{kg}$ ball moving at $40\ \text{m/s}$.
  • 2 Which has more momentum: a $60\ \text{kg}$ sprinter at $8.0\ \text{m/s}$ or a $10\ \text{g}$ bullet at $500\ \text{m/s}$? By what factor?
  • 3 A force of $50\ \text{N}$ acts for $0.20\ \text{s}$ on a $2.0\ \text{kg}$ body initially at rest. Find the impulse and the final speed.
  • 4 A $0.40\ \text{kg}$ football at rest is kicked and leaves the foot at $25\ \text{m/s}$. The contact lasts $10\ \text{ms}$. Find the average force of the kick.
  • 5 A $0.25\ \text{kg}$ ball hits a wall perpendicularly at $12\ \text{m/s}$ and rebounds at $10\ \text{m/s}$. Find the impulse on the ball.
  • 6 The force on a ball rises steadily from $0$ to $800\ \text{N}$ in $2.0\ \text{ms}$ and then falls steadily to $0$ in a further $3.0\ \text{ms}$. Find the impulse. If the $50\ \text{g}$ ball was at rest, how fast does it leave?
  • 7 Why does a high jumper land on a thick foam mattress rather than on hard ground?
  • 8 A $70\ \text{kg}$ person jumps from a wall $1.25\ \text{m}$ high. Find the speed on landing, and the average net force that stops him if he lands (a) stiff-legged, stopping in $0.010\ \text{s}$, (b) bending his knees, stopping in $0.20\ \text{s}$.
1 $p=0.145\times40=5.8\ \text{kg m/s}$.
2 Sprinter: $480\ \text{kg m/s}$; bullet: $5.0\ \text{kg m/s}$. The sprinter has $96$ times as much.
3 $J=50\times0.20=10\ \text{N s}$; $v=\dfrac{J}{m}=5.0\ \text{m/s}$.
4 $\Delta p=0.40\times25=10\ \text{kg m/s}$; $F_\text{avg}=\dfrac{10}{0.010}=1.0\times10^3\ \text{N}$.
5 $J=0.25\,(10+12)=5.5\ \text{N s}$, directed away from the wall.
6 Area of the triangle: $J=\tfrac12\times800\times5.0\times10^{-3}=2.0\ \text{N s}$; $v=\dfrac{2.0}{0.050}=40\ \text{m/s}$.
7 The foam stops him over a longer time (and distance), so for the same change of momentum the average force on his body is much smaller.
8 $v=\sqrt{2gh}=\sqrt{2\times9.8\times1.25}=4.95\ \text{m/s}$; $\Delta p=70\times4.95=347\ \text{kg m/s}$. (a) $F=\dfrac{347}{0.010}=3.5\times10^4\ \text{N}$; (b) $F=\dfrac{347}{0.20}=1.7\times10^3\ \text{N}$ โ€” twenty times less. (The ground must push with this plus his weight, $686\ \text{N}$.)
IV
Applications and harder problems.
  • 1 A $1.0\ \text{kg}$ block slides at $4.0\ \text{m/s}$ along $+x$ on a smooth floor. A constant force of $2.0\ \text{N}$ along $-x$ then acts for $3.0\ \text{s}$. Find its final velocity and its displacement during those $3.0\ \text{s}$.
  • 2 A machine gun fires $10$ bullets per second, each of mass $40\ \text{g}$, at $500\ \text{m/s}$. What average force must the gunner exert to hold the gun still?
  • 3 A hose delivers $2.0\ \text{kg}$ of water per second at $15\ \text{m/s}$ horizontally against a wall; the water stops dead and runs down the wall. Find the force on the wall.
  • 4 The velocity of a $2.0\ \text{kg}$ body along a straight line is $v=3.0\,t^2\ \text{m/s}$. Find the net force on it at $t=2.0\ \text{s}$.
  • 5 (Modelling) In a crash at $15\ \text{m/s}$, a $60\ \text{kg}$ driver held by a seat belt and airbag comes to rest over $0.50\ \text{m}$; without them, her head hits the steering wheel and stops in $0.050\ \text{m}$. Assuming constant deceleration, find the average force in each case and express each deceleration in multiples of $g$.
  • 6 A $0.50\ \text{kg}$ ball moving at $6.0\ \text{m/s}$ along $+x$ is struck so that it moves off at $8.0\ \text{m/s}$ along $+y$. Find the impulse given to it (magnitude and direction).
1 $a=-2.0\ \text{m/s}^2$: $v=4.0-2.0\times3.0=-2.0\ \text{m/s}$ (now moving along $-x$); $x=4.0\times3.0-\tfrac12(2.0)(9.0)=3.0\ \text{m}$ (it went $4.0\ \text{m}$ forward and came $1.0\ \text{m}$ back).
2 Momentum given to bullets per second: $10\times0.040\times500=200\ \text{kg m/s}$ each second, so $F=200\ \text{N}$.
3 $F=\dfrac{\Delta p}{\Delta t}=2.0\times15=30\ \text{N}$, towards the wall.
4 $a=\dfrac{dv}{dt}=6.0\,t=12\ \text{m/s}^2$ at $t=2.0\ \text{s}$; $F=2.0\times12=24\ \text{N}$.
5 $a=\dfrac{v^2}{2s}$: $\dfrac{225}{1.0}=225\ \text{m/s}^2\approx23g$, $F=60\times225=1.4\times10^4\ \text{N}$; without restraint $a=2250\ \text{m/s}^2\approx2.3\times10^2g$, $F=1.4\times10^5\ \text{N}$. Ten times the stopping distance means a tenth of the force.
6 $\Delta\mathbf{p}=0.50\,(8.0\,\hat{\mathbf{j}}-6.0\,\hat{\mathbf{i}})=(-3.0\,\hat{\mathbf{i}}+4.0\,\hat{\mathbf{j}})\ \text{N s}$: magnitude $5.0\ \text{N s}$, at $\tan^{-1}\tfrac{4.0}{3.0}=53.1^\circ$ from $-x$ towards $+y$ (i.e. $126.9^\circ$ from $+x$).
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E1. Board ยท 3 marks State Newton's second law of motion in terms of momentum. Deduce from it the relation $\mathbf{F}=m\mathbf{a}$, stating the condition needed, and define the SI unit of force.

Law The rate of change of momentum of a body equals the net external force on it and is in the direction of that force: $\mathbf{F}=\dfrac{d\mathbf{p}}{dt}$. [1]
Deduction $\mathbf{p}=m\mathbf{v}$. If $m$ is constant, $\dfrac{d\mathbf{p}}{dt}=m\dfrac{d\mathbf{v}}{dt}=m\mathbf{a}$, so $\mathbf{F}=m\mathbf{a}$. [1]
Unit One newton is the force that gives a $1\ \text{kg}$ mass an acceleration of $1\ \text{m/s}^2$: $1\ \text{N}=1\ \text{kg m/s}^2$. [1]
Exam tip Say "net external force" and "constant mass" โ€” examiners look for both conditions.

E2. NEET A body of mass $2.0\ \text{kg}$ at rest is acted on by a force along a straight line that rises uniformly from $0$ to $6.0\ \text{N}$ in the first $2.0\ \text{s}$ and then stays at $6.0\ \text{N}$ until $t=4.0\ \text{s}$. Its speed at $t=4.0\ \text{s}$ is (a) $6.0\ \text{m/s}$ (b) $9.0\ \text{m/s}$ (c) $12\ \text{m/s}$ (d) $18\ \text{m/s}$

Idea Impulse = area under the $F$โ€“$t$ graph = $\Delta p$.
Area Triangle $\tfrac12\times2.0\times6.0=6.0$ plus rectangle $2.0\times6.0=12$: $J=18\ \text{N s}$.
Answer $v=\dfrac{18}{2.0}=9.0\ \text{m/s}$ โ€” (b).
Exam tip (d) forgets to divide by the mass; (c) treats the force as $6.0\ \text{N}$ for all $4.0\ \text{s}$.

E3. JEE Main A $0.20\ \text{kg}$ ball strikes a smooth wall at $10\ \text{m/s}$, its path making an angle of $30^\circ$ with the wall, and rebounds with the same speed at the same angle. The contact lasts $0.010\ \text{s}$. The average force exerted by the ball on the wall is (a) $100\ \text{N}$ (b) $200\ \text{N}$ (c) $346\ \text{N}$ (d) $400\ \text{N}$

Components Along the wall the velocity $10\cos30^\circ$ is unchanged (smooth wall). Perpendicular to the wall the component $10\sin30^\circ=5.0\ \text{m/s}$ reverses.
Impulse $|\Delta p|=2m\,(5.0)=2\times0.20\times5.0=2.0\ \text{N s}$, perpendicular to the wall.
Answer $F=\dfrac{2.0}{0.010}=200\ \text{N}$ โ€” (b).
Exam tip The angle is with the wall, so the normal component uses $\sin30^\circ$; using $\cos30^\circ$ gives (c), and using the full speed gives (d).

E4. JEE Main (Numerical value) A horizontal jet of water of cross-sectional area $2.0\ \text{cm}^2$ leaves a nozzle at $10\ \text{m/s}$ and strikes a vertical wall normally. The water then flows down the wall without rebounding. Taking the density of water as $1.0\times10^3\ \text{kg/m}^3$, find the force (in N) on the wall.

Mass per second In $1\ \text{s}$ a column $10\ \text{m}$ long reaches the wall: $\dfrac{dm}{dt}=\rho Av=1.0\times10^3\times2.0\times10^{-4}\times10=2.0\ \text{kg/s}$.
Momentum lost per second Each kilogram loses $10\ \text{kg m/s}$ of horizontal momentum: $F=\dfrac{dp}{dt}=\rho Av^2=2.0\times10=20\ \text{N}$.
Answer $20$.
Exam tip If the water bounced straight back with the same speed the force would double, to $2\rho Av^2$. Check dimensions: $[\rho Av^2]=\text{kg m}^{-3}\cdot\text{m}^2\cdot\text{m}^2\text{s}^{-2}=\text{kg m s}^{-2}$ โœ“.

E5. JEE Advanced (One or more correct) A particle of mass $0.50\ \text{kg}$ moves along the $x$-axis with $x=2.0\,t^3-6.0\,t^2+3.0$ (SI units). Which statements are correct? (A) The net force on it is zero at $t=1.0\ \text{s}$. (B) Its momentum is zero at $t=2.0\ \text{s}$. (C) The impulse of the net force from $t=0$ to $t=3.0\ \text{s}$ is $9.0\ \text{N s}$. (D) The net force at $t=0$ is $+6.0\ \text{N}$.

Rates $v=\dfrac{dx}{dt}=6.0\,t^2-12\,t$, $\ a=\dfrac{dv}{dt}=12\,t-12$, $\ F=ma=6.0\,t-6.0$ (in N).
(A), (B) $F(1.0)=0$ โœ“. $v(2.0)=24-24=0$, so $p=0$ โœ“.
(C), (D) $v(0)=0$, $v(3.0)=54-36=18\ \text{m/s}$: $J=\Delta p=0.50\times18=9.0\ \text{N s}$ โœ“. $F(0)=-6.0\ \text{N}$, not $+6.0$ โœ—.
Answer (A), (B), (C).
Exam tip Check (C) another way: the area under $F=6.0t-6.0$ from $0$ to $3.0$ is $-3.0+12=9.0\ \text{N s}$ โœ“ (the negative triangle counts).

E6. Board ยท Case study In a crash test, a car carrying a $50\ \text{kg}$ dummy hits a rigid barrier at $14\ \text{m/s}$. (i) The dummy, held by a seat belt and airbag, comes to rest in $0.20\ \text{s}$. Find the average force on it. (2 marks) (ii) Explain, using the impulseโ€“momentum theorem, how the airbag protects the dummy. (1 mark) (iii) Find the average force if instead the dummy's chest strikes a rigid dashboard and stops in $0.010\ \text{s}$. (1 mark)

(i) $\Delta p=50\times14=700\ \text{kg m/s}$ [1]; $F=\dfrac{700}{0.20}=3.5\times10^3\ \text{N}$ [1].
(ii) The change in momentum is fixed by the crash; the airbag increases the time over which it happens, so $F=\Delta p/\Delta t$ is smaller (and it spreads the force over a larger area). [1]
(iii) $F=\dfrac{700}{0.010}=7.0\times10^4\ \text{N}$ โ€” twenty times larger. [1]
Exam tip Always link the explanation to the formula: "same $\Delta p$, longer $\Delta t$, smaller $F$".
๐Ÿ”ฌ Researchโ€“Teaching NexusResearch-tutored ยท optional, about 40 minutes

Research question. How thick must the padding be to save a falling egg โ€” and what does that tell engineers who design cycle helmets and parcel packaging?

Why it matters. Packaging engineers, helmet designers and car-safety teams all work with the same trade-off: a falling or crashing object arrives with a fixed momentum, and the only way to reduce the force is to stretch the stopping time, which needs stopping distance. International helmet standards limit the peak deceleration of a test head-form dropped onto an anvil; courier companies drop-test boxes from about $1\ \text{m}$. An egg is a cheap, fragile stand-in: it cracks when the force on it exceeds roughly $25\ \text{N}$.

What to do.

  1. An egg of mass $60\ \text{g}$ is dropped from $1.0\ \text{m}$. Find its speed and momentum on impact.
  2. Using $F_\text{avg}=\Delta p/\Delta t$, find the shortest stopping time that keeps the average force below $25\ \text{N}$. Assuming uniform deceleration, convert this into a minimum stopping distance ($s=\tfrac12v\,\Delta t$).
  3. Repeat for drop heights of $0.5$, $2.0$ and $4.0\ \text{m}$ and plot minimum padding thickness against height. Is the relation linear?
  4. Test it: drop eggs (in sealed plastic bags!) onto layers of foam, towels or crumpled paper of measured thickness. Record which survive. How well does your model predict the result? Why might the peak force (Interactive 2) matter more than the average?

Tools: Eggs, plastic bags, metre stick, padding materials, spreadsheet; optionally a phone accelerometer app taped to a padded box.

Hand in: Your calculations, the graph of padding against height, a results table, and a short recommendation for a parcel company shipping eggs.

Going further. Real padding does not decelerate uniformly: it is soft at first and stiffens as it compresses. Sketch how the forceโ€“time curve might look, and explain why helmet foams are designed to crush at a nearly constant force.

Instructor note: from $1.0\ \text{m}$, $v=\sqrt{2\times9.8\times1.0}=4.43\ \text{m/s}$ and $p=0.266\ \text{kg m/s}$; $\Delta t\ge0.266/25=10.6\ \text{ms}$ and $s\ge\tfrac12(4.43)(0.0106)=2.4\ \text{cm}$. Because $s=\dfrac{mv^2}{2F}=\dfrac{mgh}{F}$, the minimum thickness is proportional to the height: $1.2$, $4.7$ and $9.4\ \text{cm}$ for $0.5$, $2.0$, $4.0\ \text{m}$. With a half-sine pulse the peak is $\pi/2$ times the average, so real padding must be thicker than this estimate.

๐Ÿงช Self-check quiz

0/5 correct

Q1. A body moves with constant velocity along a straight line. The net force on it is

Q2. Which unit is equivalent to the unit of linear momentum?

Q3. The same force acts for the same time on a $1\ \text{kg}$ body and a $4\ \text{kg}$ body, both initially at rest. Afterwards,

Q4. A $0.10\ \text{kg}$ ball hits a wall at $20\ \text{m/s}$ and rebounds along the same line at $20\ \text{m/s}$. The magnitude of its change of momentum is

Q5. A fielder pulls her hands back while catching a fast ball in order to

โ† Chapter 4 โ€” Laws of Motion4.2 Newton's Third Law and Conservation of Momentum โ†’