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Physics · Secondary School · Dynamics

Dynamics

Why is a passenger thrown forward when the bus brakes, and why does the scale in a lift ‘lie’? It is also harder to start pushing a wardrobe than to keep it moving, something many of us have felt without ever asking why. Dynamics studies forces and what they do to motion, and with Newton's three laws, together with weight, normal force and friction, we can explain and calculate almost everything that moves around us.

  1. 1Force and net force
  2. 21st law: inertia
  3. 32nd law: F = m·a
  4. 43rd law: action and reaction
  5. 5Weight and normal force
  6. 6Friction
  7. 7Inclined plane
  8. 8Blocks and pulleys
  9. 9Work and energy
  10. 10Impulse
  11. 11Collisions
  12. ✓Challenges
STEP 1

A force has size, direction and sense

A force is a push or a pull, and to describe it we need more than a number such as ‘10 newtons’, since we also have to say which way it acts. That is why force is treated as a vector and drawn as an arrow.

When several forces act on a body, what matters is the net force \(\vec F_R\), the vector sum of all of them. Forces along the same line are added when they have the same sense and subtracted when their senses are opposite, while perpendicular forces call for Pythagoras. The polygon rule works in every case, and we apply it by placing each arrow at the tip of the previous one, so that the net force goes from the start of the first arrow to the tip of the last.

When the net force is zero, the body is in equilibrium.

\(\vec F_R = \vec F_1 + \vec F_2 + \vec F_3\)\(F_R = \sqrt{F_1^2 + F_2^2}\) (perpendicular)Equilibrium: \(\vec F_R = \vec 0\)The unit of force is the newton (N), and angles are measured anticlockwise from the right (0°). The same sense gives \(F_1 + F_2\) · opposite senses give \(|F_1 - F_2|\).

\(\vec F_1\)   \(\vec F_2\)   \(\vec F_3\)   net force \(\vec F_R\)

Let's discuss

  • Drag the arrow tips (or use the magnitude and angle controls) and set 3 N at 0° and 4 N at 90°. Is the net force 5 N, as Pythagoras would predict?
  • Can two forces of 3 N and 4 N give a net force of 8 N? Make a quick guess, then try it.
  • Turn on \(\vec F_3\) and adjust it until the net force disappears; how does it then compare with the sum of the other two?
  • Look at the dashed arrows when the body is in equilibrium; does the polygon ‘close’?
|F₁|
|F₂|
|F₃|
Net force
Angle of net force
Exercises for step 1 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    Two people push a wardrobe along the same line and in the same sense, one with \(30\ \text{N}\) and the other with \(50\ \text{N}\). What is the net force on the wardrobe?

    Show solution
    Since the forces lie on the same line and in the same sense, the magnitudes add up.
    \(F_R = 30 + 50 = 80\ \text{N}\)
    \(F_R = 80\ \text{N}\), in the sense both are pushing.
  2. basic

    A box is pulled to the right with \(50\ \text{N}\) and to the left with \(30\ \text{N}\). What is the net force (magnitude and sense)?

    Show solution
    On the same line with opposite senses, we subtract the magnitudes, and the sense is that of the larger force.
    \(F_R = 50 - 30 = 20\ \text{N}\)
    \(F_R = 20\ \text{N}\), to the right.
  3. basic

    Two perpendicular forces, of \(6\ \text{N}\) and \(8\ \text{N}\), act on the same body. What is the magnitude of the net force?

    Show solution
    Perpendicular forces form the legs of a right triangle, and the net force is its hypotenuse.
    \(F_R = \sqrt{6^2 + 8^2}\) \(= \sqrt{36 + 64}\) \(= \sqrt{100}\)
    \(F_R = 10\ \text{N}\)
  4. basic

    A lamp hangs at rest from a wire. What is the net force on it, and what does that tell you about the force of the wire?

    Show solution
    At rest (and staying at rest), the body is in equilibrium, so the net force is zero.
    Two forces act on it, the weight (down) and the tension in the wire (up), and for their sum to be zero they must have the same magnitude.
    \(F_R = 0\); the tension in the wire equals the weight of the lamp.
  5. intermediate

    Three horizontal forces act on a ring: \(10\ \text{N}\) east, \(4\ \text{N}\) west and \(8\ \text{N}\) north. Find the magnitude of the net force.

    Show solution
    We start by adding the forces along the same line (east–west), which gives \(10 - 4 = 6\ \text{N}\) east.
    That leaves two perpendicular forces, \(6\ \text{N}\) (east) and \(8\ \text{N}\) (north).
    \(F_R = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\ \text{N}\)
    \(F_R = 10\ \text{N}\), pointing between east and north.
  6. intermediate

    Two forces of \(5\ \text{N}\) and \(12\ \text{N}\) act on a body. What are the largest and smallest possible values of the net force? And what is it if the forces are perpendicular?

    Show solution
    The largest net force comes with the same sense, \(5 + 12 = 17\ \text{N}\).
    The smallest comes with opposite senses, \(12 - 5 = 7\ \text{N}\).
    For perpendicular forces, \(F_R = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\ \text{N}\).
    Maximum \(17\ \text{N}\); minimum \(7\ \text{N}\); perpendicular \(13\ \text{N}\).
  7. intermediate

    In a tug of war, team A has 3 people pulling with \(200\ \text{N}\) each. Team B has 2 people pulling with \(250\ \text{N}\) each and one pulling with \(120\ \text{N}\). What is the net force on the rope, and which way?

    Show solution
    Team A: \(3 \cdot 200 = 600\ \text{N}\).
    Team B: \(2 \cdot 250 + 120 = 620\ \text{N}\).
    The teams pull in opposite senses, so \(F_R = 620 - 600 = 20\ \text{N}\).
    \(F_R = 20\ \text{N}\), towards team B.
  8. intermediate

    Is it possible for two forces, one of \(3\ \text{N}\) and another of \(4\ \text{N}\), to have a net force of \(8\ \text{N}\)? Explain.

    Show solution
    The net force of two forces always lies between the difference and the sum of their magnitudes.
    \(4 - 3 = 1\ \text{N} \le F_R \le 4 + 3 = 7\ \text{N}\)
    No: the largest possible value is \(7\ \text{N}\) (forces in the same sense).
  9. challenge

    Two forces of \(10\ \text{N}\) each make an angle of \(120^\circ\) with each other. What is the magnitude of the net force? Use \(F_R^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta\) and \(\cos 120^\circ = -0.5\).

    Show solution
    \(F_R^2 = 10^2 + 10^2 + 2 \cdot 10 \cdot 10 \cdot (-0.5)\)
    \(F_R^2 = 100 + 100 - 100 = 100\)
    \(F_R = 10\ \text{N}\)
    By the parallelogram rule, the rhombus splits into two equilateral triangles, so the diagonal is as long as the sides.
    \(F_R = 10\ \text{N}\), along the bisector of the angle between the forces.
  10. challenge

    A block is in equilibrium under three forces. Two of them are \(\vec F_1\), \(9\ \text{N}\) east, and \(\vec F_2\), \(12\ \text{N}\) north. What is the third force (magnitude and direction)?

    Show solution
    In equilibrium, \(\vec F_1 + \vec F_2 + \vec F_3 = 0\), so \(\vec F_3 = -(\vec F_1 + \vec F_2)\).
    \(|\vec F_1 + \vec F_2| = \sqrt{9^2 + 12^2}\) \(= \sqrt{225}\) \(= 15\ \text{N}\)
    \(\vec F_3\) has the same magnitude and the opposite sense, pointing southwest, with \(\tan\alpha = \dfrac{12}{9}\), that is, about \(53^\circ\) below west.
    \(F_3 = 15\ \text{N}\), towards the southwest (opposite to the sum of the other two).
STEP 2

No net force, no change: inertia

For a long time people thought that to keep something moving you had to keep pushing it, and our own first intuition tends to agree with them. Galileo and Newton showed that a body with no net force stays as it is, either at rest or in straight-line motion at constant velocity, and we call this tendency to keep its velocity inertia.

In real life, things stop because there are forces opposing the motion, such as friction and air resistance. If we could remove friction completely, as the simulation lets us do, the puck would slide forever.

\(\vec F_R = \vec 0 \iff \vec v = \text{constant}\)Rest is the special case \(\vec v = \vec 0\). The greater the mass, the greater the inertia (the harder it is to change the velocity).

On a braking bus, nobody pushes the passenger forward; the bus slows down while the passenger keeps the velocity they had. A tablecloth pulled quickly leaves the dishes in place because friction acts on them for a very short time, and the seat belt is the force that makes you brake together with the car.

Let's discuss

  • Push the puck with friction 0.30 and then with 0.05, and compare how far it goes in each case.
  • Set friction to zero and push; does the puck stop, and what force would be needed to stop it?
  • With the bus moving at constant speed and not braking, does the passenger need any force to keep up with the bus?
  • Without the seat belt, change the bus speed and brake again, keeping in mind that our first guess is probably that a faster bus means a harder impact on the seat; is that what we see? Then change the braking strength and the gap to the seat.
  • Brake the bus without the seat belt and look at each one's velocity. Who ‘moves forward’, the passenger or the bus that falls behind?
  • Now fasten the seat belt and brake again; which force slows the passenger down this time?
Puck velocity
Puck distance
Bus velocity
Passenger velocity
Impact on the seat (relative to the bus)
Exercises for step 2 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A bus brakes suddenly and the standing passengers are ‘thrown’ forward. Is there a force pushing them forward? Explain.

    Show solution
    There is no forward force, since the passengers were already moving with the bus's velocity.
    When the bus brakes, the floor (and the bus itself) slows down, but the passenger's body tends to keep the velocity it had, and that is what we call inertia.
    There is no forward force: the passenger keeps going by inertia while the bus stops.
  2. basic

    In the tablecloth trick, the cloth is pulled very quickly and the plates stay on the table. Why doesn't pulling slowly work?

    Show solution
    The plates are at rest and tend to stay at rest (inertia).
    If we pull quickly, friction acts for a very short time and hardly changes the plates' velocity.
    If we pull slowly, friction acts for a long time and drags the plates along.
    Quickly: friction acts for a short time and the plates gain almost no velocity.
  3. basic

    A probe in deep space, far from any planet, travels at \(2000\ \text{m/s}\) with its engines off. What will its velocity be 1 hour later, and how far will it have travelled in that time?

    Show solution
    With no forces (zero net force), the velocity stays constant, so it is still \(2000\ \text{m/s}\).
    \(d = v \cdot t = 2000 \cdot 3600 = 7\,200\,000\ \text{m}\)
    \(v = 2000\ \text{m/s}\); \(d = 7\,200\ \text{km}\).
  4. basic

    A car travels at a constant \(80\ \text{km/h}\) on a straight road. What is the net force on it?

    Show solution
    Constant velocity in a straight line means zero acceleration.
    By the 1st law, this only happens when the net force is zero, so the engine pushes and friction and the air hold back by the same amount.
    \(F_R = 0\).
  5. intermediate

    What is the headrest of a car seat for, when the car is hit from behind?

    Show solution
    In a rear-end collision, the car (and the seat) is suddenly pushed forward.
    The head, by inertia, tends to stay where it was, while the torso moves forward with the seat and the neck bends backward.
    The headrest pushes the head along with the body and so prevents the injury.
    It makes the head move with the body, which is pushed forward: by inertia, the head would be left behind.
  6. intermediate

    A hockey puck slides on almost frictionless ice at \(4\ \text{m/s}\). Ignoring friction, how far does it go in \(5\ \text{s}\)?

    Show solution
    Without friction, the horizontal net force is zero, and the puck keeps \(v = 4\ \text{m/s}\).
    \(d = v \cdot t = 4 \cdot 5\)
    \(d = 20\ \text{m}\)
  7. intermediate

    A student says: ‘If I stop pedalling, the bike stops. So you need a force to keep it moving.’ How would Newton answer?

    Show solution
    The bike stops because there are forces against the motion: friction in the axles, deformation of the tyres and air resistance.
    Without these forces, it would keep going forever at the same velocity (1st law).
    Pedalling balances the opposing forces; it does not ‘keep up’ the motion.
    A force is needed to change the velocity; the bike only stops because there are opposing forces.
  8. intermediate

    The head of a hammer is loose. Striking the end of the handle on the floor (with the head up) makes it fit tightly. Why?

    Show solution
    The whole hammer moves down together until the handle hits the floor and stops suddenly.
    The head, by inertia, keeps moving down and slides along the handle, fitting more tightly.
    The handle stops, but the head keeps moving down by inertia.
  9. challenge

    A small ball hangs from a string attached to the roof of a car. With the car stopped, the string is vertical. The car accelerates forward at \(2\ \text{m/s}^2\). Which way does the string tilt, and by what angle? Use \(g = 10\ \text{m/s}^2\) and \(\tan\alpha = a/g\).

    Show solution
    By inertia, the ball ‘lags behind’ and the string tilts backward until the horizontal component of the tension accelerates the ball along with the car.
    \(\tan\alpha = \dfrac{a}{g} = \dfrac{2}{10} = 0.2\)
    \(\alpha \approx 11.3^\circ\)
    At constant velocity, the string returns to vertical.
    It tilts backward, about \(11^\circ\) from the vertical.
  10. challenge

    A bus at \(72\ \text{km/h}\) brakes with a deceleration of \(5\ \text{m/s}^2\). A passenger without a seat belt is \(1.0\ \text{m}\) from the seat in front and slides without friction. How long after braking starts does the passenger hit the seat, and at what speed relative to the bus?

    Show solution
    \(72\ \text{km/h} = 20\ \text{m/s}\). The passenger keeps going at \(20\ \text{m/s}\) (inertia), while the bus loses \(5\ \text{m/s}\) every second.
    Seen from inside the bus, the passenger approaches the seat with a relative acceleration of \(5\ \text{m/s}^2\), starting from rest.
    \(1.0 = \dfrac{5\,t^2}{2}\) \(\Rightarrow t = \sqrt{0.4} \approx 0.63\ \text{s}\)
    \(v_{rel} = 5 \cdot 0.63 \approx 3.2\ \text{m/s}\) (the bus is still moving at \(20 - 3.2 \approx 16.8\ \text{m/s}\)).
    \(t \approx 0.63\ \text{s}\); hits at \(\approx 3.2\ \text{m/s}\) (about \(11\ \text{km/h}\)) relative to the bus.
STEP 3

A net force changes the velocity: \(F_R = m \cdot a\)

If the net force is not zero, the velocity changes, and the body gains an acceleration in the same direction as the net force; Newton's 2nd law tells us by how much, since the acceleration is proportional to the force and inversely proportional to the mass.

Doubling the force doubles the acceleration, while doubling the mass halves the acceleration. Mass is therefore a measure of inertia, and the more mass a body has, the more force we need to change its velocity.

\(\vec F_R = m\,\vec a\)\(a = \dfrac{F_R}{m}\)\(1\ \text{N} = 1\ \text{kg} \cdot \text{m/s}^2\)One newton is the force that gives 1 kg an acceleration of 1 m/s². On the \(v \times t\) graph, the acceleration is the slope of the line.

Let's discuss

  • Release the trolley and check whether the velocity always increases by the same amount every second. How does that show up on the graph?
  • Press ‘Double the force’ and release again; the new line will probably look steeper, but does it have twice the slope?
  • Reset the force and press ‘Double the mass’; what happens to the measured acceleration?
  • Find two different combinations of F and m that give the same acceleration.
  • Set the force to zero in the middle of the motion, when many of us would expect the trolley to stop; does it? (Remember step 2.)
Predicted F/m
Measured on graph
Velocity
Time
Distance
Exercises for step 3 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A net force of \(20\ \text{N}\) acts on a \(4\ \text{kg}\) block. What is the acceleration?

    Show solution
    \(a = \dfrac{F_R}{m} = \dfrac{20}{4}\)
    \(a = 5\ \text{m/s}^2\)
  2. basic

    What net force is needed to give a \(1000\ \text{kg}\) car an acceleration of \(2\ \text{m/s}^2\)?

    Show solution
    \(F_R = m \cdot a = 1000 \cdot 2\)
    \(F_R = 2000\ \text{N}\)
  3. basic

    A net force of \(30\ \text{N}\) gives a body an acceleration of \(6\ \text{m/s}^2\). What is the mass of the body?

    Show solution
    \(m = \dfrac{F_R}{a} = \dfrac{30}{6}\)
    \(m = 5\ \text{kg}\)
  4. basic

    If we double the net force on a trolley and also double its mass, what happens to the acceleration?

    Show solution
    Since \(a = \dfrac{F_R}{m}\), doubling the numerator doubles \(a\) and doubling the denominator divides \(a\) by 2.
    \(a' = \dfrac{2F_R}{2m} = \dfrac{F_R}{m} = a\)
    The acceleration does not change.
  5. intermediate

    A \(2\ \text{kg}\) block on a frictionless table is pulled with \(10\ \text{N}\) to the right and \(4\ \text{N}\) to the left. What is the acceleration?

    Show solution
    \(F_R = 10 - 4 = 6\ \text{N}\), to the right.
    \(a = \dfrac{6}{2} = 3\ \text{m/s}^2\)
    \(a = 3\ \text{m/s}^2\), to the right.
  6. intermediate

    A \(5\ \text{kg}\) block starts from rest, without friction, pushed by a constant force of \(10\ \text{N}\). What is its velocity after \(4\ \text{s}\), and how far has it gone?

    Show solution
    \(a = \dfrac{10}{5} = 2\ \text{m/s}^2\)
    \(v = a\,t = 2 \cdot 4 = 8\ \text{m/s}\)
    \(d = \dfrac{a\,t^2}{2} = \dfrac{2 \cdot 16}{2} = 16\ \text{m}\)
    \(v = 8\ \text{m/s}\); \(d = 16\ \text{m}\).
  7. intermediate

    A \(1200\ \text{kg}\) car goes from 0 to \(108\ \text{km/h}\) in \(10\ \text{s}\). What is the average net force on it?

    Show solution
    \(108\ \text{km/h} = 30\ \text{m/s}\)
    \(a = \dfrac{\Delta v}{\Delta t} = \dfrac{30}{10} = 3\ \text{m/s}^2\)
    \(F_R = m\,a = 1200 \cdot 3\)
    \(F_R = 3600\ \text{N}\)
  8. intermediate

    On the \(v \times t\) graph of a \(2\ \text{kg}\) trolley, the velocity goes from 0 to \(12\ \text{m/s}\) in \(4\ \text{s}\), along a straight line. What is the net force?

    Show solution
    The slope of the \(v \times t\) graph is the acceleration, so \(a = \dfrac{12}{4} = 3\ \text{m/s}^2\).
    \(F_R = m\,a = 2 \cdot 3\)
    \(F_R = 6\ \text{N}\)
  9. challenge

    An \(800\ \text{kg}\) car moving at \(20\ \text{m/s}\) brakes and stops in \(40\ \text{m}\), with constant deceleration. What is the braking force? Use \(v^2 = v_0^2 - 2\,a\,d\).

    Show solution
    \(0 = 20^2 - 2 \cdot a \cdot 40\) \(\Rightarrow a = \dfrac{400}{80} = 5\ \text{m/s}^2\)
    \(F = m\,a = 800 \cdot 5\)
    \(F = 4000\ \text{N}\), against the motion.
  10. challenge

    The same force gives an acceleration of \(6\ \text{m/s}^2\) to a body of mass \(m_1\) and \(3\ \text{m/s}^2\) to another of mass \(m_2\). What acceleration does this force give to the two together (attached to each other)?

    Show solution
    \(m_1 = \dfrac{F}{6}\) and \(m_2 = \dfrac{F}{3}\)
    \(m_1 + m_2 = \dfrac{F}{6} + \dfrac{2F}{6} = \dfrac{F}{2}\)
    \(a = \dfrac{F}{m_1 + m_2} = \dfrac{F}{F/2} = 2\ \text{m/s}^2\)
    \(a = 2\ \text{m/s}^2\)
STEP 4

Every force has a partner: action and reaction

Forces always come in pairs, and if A pushes B, then B pushes A with a force of the same magnitude, along the same line and in the opposite sense, which is what Newton's 3rd law states.

What often confuses us is that the two forces act on different bodies, and that is why they do not cancel out; when two skaters push each other, the forces are equal, yet the lighter skater gets more acceleration (\(a = F/m\)).

You walk because you push the ground backward and the ground pushes you forward. A rocket works the same way, pushing the gases backward while the gases push it forward, even in a vacuum.

\(\vec F_{A \to B} = -\vec F_{B \to A}\)\(a_A = \dfrac{F}{m_A},\quad a_B = \dfrac{F}{m_B}\)With the same push time, \(m_A\,v_A = m_B\,v_B\), so whoever has half the mass leaves with twice the speed.

action–reaction pair   velocity

Let's discuss

  • Push with equal masses and compare the two velocities; are they equal?
  • In ‘Adult and child’, who feels the larger force, and who moves off faster? Intuition may point to the child for the first question, so check the numbers.
  • Pause during the push and compare the two red arrows; are they the same size?
  • If the forces are equal and opposite, why don't the two stay still?
Force on A
Force on B
Acceleration of A
Acceleration of B
Velocity of A
Velocity of B
Exercises for step 4 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    If action and reaction have the same magnitude and opposite senses, why don't they cancel out?

    Show solution
    Only forces acting on the SAME body cancel (add up).
    Action and reaction act on different bodies, one on A and the other on B.
    They don't cancel because they act on different bodies.
  2. basic

    When you walk, what force makes you move forward?

    Show solution
    Your foot pushes the ground backward (through friction).
    By the 3rd law, the ground pushes your foot forward with the same magnitude.
    The force of the ground on your foot, forward (the reaction to your backward push).
  3. basic

    Two skaters, a \(40\ \text{kg}\) girl and a \(60\ \text{kg}\) boy, push each other. The force on the girl is \(120\ \text{N}\). What is the force on the boy, and what is the acceleration of each?

    Show solution
    By the 3rd law, the force on the boy is also \(120\ \text{N}\), in the opposite sense.
    Girl: \(a = \dfrac{120}{40} = 3\ \text{m/s}^2\)
    Boy: \(a = \dfrac{120}{60} = 2\ \text{m/s}^2\)
    \(120\ \text{N}\) on each; girl \(3\ \text{m/s}^2\), boy \(2\ \text{m/s}^2\).
  4. basic

    A book rests on a table. The Earth pulls the book (weight). What is the reaction to this force, and is it the normal force?

    Show solution
    The reaction to the weight is the force with which the BOOK pulls the EARTH, upward, applied at the Earth's centre.
    The normal force (table pushing the book) also equals the weight here, but it belongs to a different interaction, and the reaction to the normal force is the book pushing the table down.
    The reaction to the weight is the book pulling the Earth; it is not the normal force.
  5. intermediate

    Skaters of \(50\ \text{kg}\) and \(75\ \text{kg}\), at rest on the ice, push each other with \(150\ \text{N}\) for \(0.5\ \text{s}\). With what speed does each one move off?

    Show solution
    50 kg skater: \(a = \dfrac{150}{50} = 3\ \text{m/s}^2\), \(v = 3 \cdot 0.5 = 1.5\ \text{m/s}\)
    75 kg skater: \(a = \dfrac{150}{75} = 2\ \text{m/s}^2\), \(v = 2 \cdot 0.5 = 1.0\ \text{m/s}\)
    \(1.5\ \text{m/s}\) and \(1.0\ \text{m/s}\), in opposite senses.
  6. intermediate

    In space there is no air to ‘push against’. How can a rocket accelerate?

    Show solution
    The rocket pushes the gases backward with a large force as it expels them.
    By the 3rd law, the gases push the rocket forward with the same magnitude.
    No air is needed, because the force pair is between the rocket and the gases.
    The rocket pushes the gases backward and the gases push the rocket forward.
  7. intermediate

    A \(10\,000\ \text{kg}\) lorry hits a \(1000\ \text{kg}\) car. At one instant, the force of the lorry on the car is \(50\,000\ \text{N}\). What is the force of the car on the lorry, and what are the accelerations?

    Show solution
    By the 3rd law, the car pushes the lorry with the same \(50\,000\ \text{N}\).
    Car: \(a = \dfrac{50\,000}{1000} = 50\ \text{m/s}^2\)
    Truck: \(a = \dfrac{50\,000}{10\,000} = 5\ \text{m/s}^2\)
    \(50\,000\ \text{N}\) on both; the car's acceleration is 10 times larger (\(50\) versus \(5\ \text{m/s}^2\)).
  8. intermediate

    A \(100\ \text{g}\) apple falls, pulled by the Earth with \(1\ \text{N}\). With what force does the apple pull the Earth, and what is the Earth's acceleration? (mass of the Earth \(\approx 6 \cdot 10^{24}\ \text{kg}\))

    Show solution
    By the 3rd law, the apple pulls the Earth with \(1\ \text{N}\), upward.
    \(a_{Earth} = \dfrac{1}{6 \cdot 10^{24}} \approx 1.7 \cdot 10^{-25}\ \text{m/s}^2\)
    This acceleration is so small that it cannot be noticed.
    \(1\ \text{N}\); \(a \approx 1.7 \cdot 10^{-25}\ \text{m/s}^2\).
  9. challenge

    An \(80\ \text{kg}\) astronaut (including the suit), at rest in space, throws a \(2\ \text{kg}\) tool at \(8\ \text{m/s}\). With what speed does the astronaut recoil?

    Show solution
    The forces between astronaut and tool are equal and opposite and act for the same time \(\Delta t\).
    Since \(F = m\,a = m\,\dfrac{\Delta v}{\Delta t}\), the product \(m \cdot \Delta v\) is the same for both.
    \(80 \cdot v = 2 \cdot 8\) \(\Rightarrow v = \dfrac{16}{80} = 0.2\ \text{m/s}\)
    \(v = 0.2\ \text{m/s}\), in the opposite sense to the tool.
  10. challenge

    Two skaters, A (\(50\ \text{kg}\)) and B (\(70\ \text{kg}\)), at rest \(10\ \text{m}\) apart, pull on a rope between them. Ignoring friction, where do they meet?

    Show solution
    The rope pulls both with the same force \(F\) the whole time.
    \(a_A = F/50\) and \(a_B = F/70\); starting from rest, the distances are proportional to the accelerations, so \(\dfrac{d_A}{d_B} = \dfrac{70}{50}\).
    \(d_A + d_B = 10\) \(\Rightarrow d_A = 10 \cdot \dfrac{70}{120} \approx 5.83\ \text{m}\), \(d_B \approx 4.17\ \text{m}\)
    A moves \(\approx 5.83\ \text{m}\) and B \(\approx 4.17\ \text{m}\): they meet closer to B's starting position.
STEP 5

Does the scale measure your weight? Not exactly

Weight is the force with which the Earth pulls a body, \(W = m\,g\). Mass (in kg) is the amount of matter and does not change from place to place, whereas weight (in N) depends on gravity; on the Moon, where \(g\) is about 1.6 m/s², your weight drops to one sixth while your mass stays the same.

The normal force \(N\) is the push of a surface, perpendicular to it, and something that may surprise us is that the scale measures this normal force rather than our weight. In a lift accelerating upward, the normal force has to be greater than the weight, and in one accelerating downward it has to be smaller. In free fall, the floor ‘falls away’ together with you, so \(N = 0\) and you feel ‘weightless’.

\(W = m\,g\)\(N - W = m\,a\)\(N = m\,(g + a)\)We take \(a\) as positive upward, with \(g = 10\ \text{m/s}^2\) on Earth and \(1.6\ \text{m/s}^2\) on the Moon. The scale is calibrated on Earth, so it shows \(N/10\) in ‘kg’.

weight \(W\)   normal \(N\)   acceleration   velocity

Let's discuss

  • Go up at constant velocity and watch the scale; does the reading change, and why?
  • In both ‘Start going up’ and ‘Brake going down’, the scale reads more. What do these two motions have in common, even though they seem so different?
  • Cut the cable and read N; has the person stopped having weight?
  • Take the lift to the Moon; does the mass change, and does the scale reading?
Weight W
Normal N
Scale reading
Acceleration
Velocity
Exercises for step 5 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A person has a mass of \(60\ \text{kg}\). What is their weight on Earth (\(g = 10\ \text{m/s}^2\)) and on the Moon (\(g = 1.6\ \text{m/s}^2\))? And their mass on the Moon?

    Show solution
    Earth: \(W = m\,g = 60 \cdot 10 = 600\ \text{N}\)
    Moon: \(W = 60 \cdot 1.6 = 96\ \text{N}\)
    The mass (amount of matter, inertia) does not change.
    \(600\ \text{N}\) on Earth, \(96\ \text{N}\) on the Moon; mass \(60\ \text{kg}\) in both.
  2. basic

    What is wrong with the sentence ‘my weight is 60 kilograms’?

    Show solution
    The kilogram is a unit of mass, while weight is a force, measured in newtons.
    With \(g = 10\ \text{m/s}^2\), a mass of \(60\ \text{kg}\) has a weight of \(600\ \text{N}\).
    The correct form is ‘my mass is 60 kg’ (or ‘my weight is 600 N’).
  3. basic

    A \(2\ \text{kg}\) book rests on a horizontal table. What is the normal force? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    At rest, \(F_R = 0\), so \(N = W\).
    \(N = m\,g = 2 \cdot 10\)
    \(N = 20\ \text{N}\)
  4. basic

    A lift goes up at constant velocity. A \(70\ \text{kg}\) person stands on a scale inside it. What does the scale read? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    At constant velocity, \(a = 0\), so \(N = W = 70 \cdot 10 = 700\ \text{N}\).
    The scale (calibrated for \(g = 10\)) shows \(700/10 = 70\ \text{kg}\).
    It reads \(70\ \text{kg}\) (\(N = 700\ \text{N}\)), the same as with the lift at rest.
  5. intermediate

    A lift accelerates upward at \(2\ \text{m/s}^2\). A \(50\ \text{kg}\) person stands on a scale inside it. What is the normal force and what does the scale read? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(N - W = m\,a\) \(\Rightarrow N = m\,(g + a)\)
    \(N = 50 \cdot (10 + 2) = 600\ \text{N}\)
    Reading: \(600/10 = 60\ \text{kg}\)
    \(N = 600\ \text{N}\); the scale reads \(60\ \text{kg}\).
  6. intermediate

    A lift accelerates downward at \(3\ \text{m/s}^2\). What does the scale read for an \(80\ \text{kg}\) person? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    With a downward acceleration, \(W - N = m\,a\) \(\Rightarrow N = m\,(g - a)\)
    \(N = 80 \cdot (10 - 3) = 560\ \text{N}\)
    Reading: \(560/10 = 56\ \text{kg}\)
    \(N = 560\ \text{N}\); the scale reads \(56\ \text{kg}\).
  7. intermediate

    A \(5\ \text{kg}\) box sits on a table, and a person presses down on it with \(30\ \text{N}\). What is the normal force? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    At rest, the normal force (up) balances the weight and the push (down).
    \(N = W + F = 50 + 30\)
    \(N = 80\ \text{N}\): the normal force is not always equal to the weight.
  8. intermediate

    On the Space Station, about \(400\ \text{km}\) up, \(g\) is still about \(8.7\ \text{m/s}^2\). Why do astronauts float ‘weightless’?

    Show solution
    They DO have weight, since the Earth pulls them with almost 90% of the force they would feel on the ground.
    The station and the astronauts are in free fall together (they fall around the Earth, in orbit).
    Since everything falls with the same acceleration, the floor does not push on their feet, and \(N = 0\).
    They are in free fall together with the station: the normal force is zero, hence the feeling of ‘weightlessness’.
  9. challenge

    On a scale inside a lift, a \(70\ \text{kg}\) person reads \(84\ \text{kg}\) (scale calibrated with \(g = 10\ \text{m/s}^2\)). What is the lift's acceleration? Is it necessarily going up?

    Show solution
    \(N = 84 \cdot 10 = 840\ \text{N}\)
    \(N = m\,(g + a)\) \(\Rightarrow 840 = 70\,(10 + a)\) \(\Rightarrow a = 12 - 10 = 2\ \text{m/s}^2\), upward.
    An upward acceleration happens when going up faster and faster OR when going down and braking.
    \(a = 2\ \text{m/s}^2\) upward; it may be going up and speeding up, or going down and braking.
  10. challenge

    A \(1000\ \text{kg}\) lift carries a \(60\ \text{kg}\) person and accelerates upward at \(1.5\ \text{m/s}^2\). Find the tension in the cable and the normal force on the person. Use \(g = 10\ \text{m/s}^2\).

    Show solution
    For the cable, we take the lift + person system, \(1060\ \text{kg}\), and write \(T - (m_{tot})\,g = m_{tot}\,a\)
    \(T = 1060 \cdot (10 + 1.5) = 1060 \cdot 11.5 = 12\,190\ \text{N}\)
    For the person, \(N = 60 \cdot 11.5 = 690\ \text{N}\)
    \(T = 12\,190\ \text{N}\); \(N = 690\ \text{N}\).
STEP 6

Friction: first it holds, then it slows

Give a heavy piece of furniture a gentle push and it does not move, because static friction grows along with your push and balances it, up to a maximum value \(\mu_s\,N\), beyond which the furniture slides.

While the furniture slides, kinetic friction acts, with a fixed value \(\mu_k\,N\) that is usually smaller than the static maximum. That is why pushing gets easier once it ‘breaks free’.

The coefficients \(\mu\) depend on the surfaces in contact (rubber and asphalt, wood and wood...), and, perhaps against our intuition, not on the contact area.

\(f_s \le \mu_s\,N\)\(f_k = \mu_k\,N\)\(\mu_k < \mu_s\)On a horizontal floor, \(N = m\,g\). The coefficient \(\mu\) has no unit, and the values in the simulation are typical and approximate.

applied force \(F\)   friction \(f\)   normal   weight

Let's discuss

  • Press ‘Pull harder and harder’ and watch the graph; while the block is at rest, why does it rise along a sloped line?
  • At the instant the block starts to slide, what happens to the friction and to the acceleration?
  • Stand the block ‘on end’, with less area in contact. We might expect the peak of the graph to change; does it?
  • Double the mass; what happens to the maximum friction, and why?
  • With the block sliding, lower the force until it is less than the kinetic friction; what does the block do then?
Applied force
Friction now
Static max (μs·N)
Kinetic (μk·N)
Acceleration
State
Exercises for step 6 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A \(10\ \text{kg}\) block slides on a horizontal floor with \(\mu_k = 0.3\). What is the kinetic friction force? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    On a horizontal floor, \(N = W = 10 \cdot 10 = 100\ \text{N}\).
    \(f_k = \mu_k\,N = 0.3 \cdot 100\)
    \(f_k = 30\ \text{N}\)
  2. basic

    A \(20\ \text{kg}\) box is on the floor, with \(\mu_s = 0.5\). What is the smallest horizontal force that can get it moving? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(N = 20 \cdot 10 = 200\ \text{N}\)
    \(f_{s,\text{max}} = \mu_s\,N = 0.5 \cdot 200 = 100\ \text{N}\)
    You have to exceed this value.
    A little more than \(100\ \text{N}\).
  3. basic

    Why is it harder to start pushing a wardrobe than to keep it sliding?

    Show solution
    At rest, static friction acts, and it can reach \(\mu_s\,N\).
    While sliding, kinetic friction \(\mu_k\,N\) acts, and usually \(\mu_k < \mu_s\).
    The maximum static friction is greater than the kinetic friction.
  4. basic

    The same \(20\ \text{kg}\) box (\(\mu_s = 0.5\)) is pushed with \(60\ \text{N}\) and does not move. What is the friction?

    Show solution
    The box stays at rest, so \(F_R = 0\).
    Static friction therefore balances the push exactly, \(f_s = 60\ \text{N}\).
    The value \(\mu_s\,N = 100\ \text{N}\) is only the MAXIMUM that static friction can reach.
    \(f_s = 60\ \text{N}\) (not \(100\ \text{N}\)).
  5. intermediate

    A \(5\ \text{kg}\) block is pulled by a horizontal force of \(25\ \text{N}\) and slides with \(\mu_k = 0.2\). What is the acceleration? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(N = 50\ \text{N}\); \(f_k = 0.2 \cdot 50 = 10\ \text{N}\)
    \(F_R = 25 - 10 = 15\ \text{N}\)
    \(a = \dfrac{15}{5}\)
    \(a = 3\ \text{m/s}^2\)
  6. intermediate

    A brick is dragged lying flat (larger contact area) and then standing on end (smaller area). Does the friction change?

    Show solution
    Friction depends on \(\mu\) (the surfaces) and on the normal force \(N\); the normal force equals the weight in both cases.
    With less area, the pressure at each point increases and makes up for the smaller area.
    It does not change: friction does not depend on the contact area.
  7. intermediate

    A \(1000\ \text{kg}\) car at \(20\ \text{m/s}\) locks its wheels and skids to a stop, with \(\mu_k = 0.5\). What are the braking distance and the time? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(f_k = 0.5 \cdot 1000 \cdot 10 = 5000\ \text{N}\); \(a = \dfrac{5000}{1000} = 5\ \text{m/s}^2\)
    \(d = \dfrac{v_0^2}{2a} = \dfrac{400}{10} = 40\ \text{m}\)
    \(t = \dfrac{v_0}{a} = \dfrac{20}{5} = 4\ \text{s}\)
    \(d = 40\ \text{m}\); \(t = 4\ \text{s}\).
  8. intermediate

    A block is launched across the floor at \(6\ \text{m/s}\) and stops after \(9\ \text{m}\). What is the coefficient of kinetic friction? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(0 = 6^2 - 2\,a \cdot 9\) \(\Rightarrow a = \dfrac{36}{18} = 2\ \text{m/s}^2\)
    Only friction slows it down, so \(\mu_k\,m\,g = m\,a\) \(\Rightarrow \mu_k = \dfrac{a}{g} = \dfrac{2}{10}\)
    \(\mu_k = 0.2\)
  9. challenge

    A \(40\ \text{kg}\) box has \(\mu_s = 0.6\) and \(\mu_k = 0.4\) with the floor. (a) What is the friction if it is pulled with \(200\ \text{N}\)? (b) What is the minimum force to get it moving? (c) Once moving, what is the acceleration with \(260\ \text{N}\)? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(N = 400\ \text{N}\); \(f_{s,\text{max}} = 0.6 \cdot 400 = 240\ \text{N}\); \(f_k = 0.4 \cdot 400 = 160\ \text{N}\)
    (a) Since \(200 < 240\), it stays at rest and \(f_s = 200\ \text{N}\).
    (b) You must exceed \(240\ \text{N}\).
    (c) \(a = \dfrac{260 - 160}{40} = 2.5\ \text{m/s}^2\)
    (a) \(200\ \text{N}\); (b) more than \(240\ \text{N}\); (c) \(2.5\ \text{m/s}^2\).
  10. challenge

    A \(1000\ \text{kg}\) car at \(30\ \text{m/s}\) brakes. With ABS brakes, the tyres don't skid and the friction is static (\(\mu_s = 0.8\)); with locked wheels, it is kinetic (\(\mu_k = 0.6\)). Compare the braking distances. Use \(g = 10\ \text{m/s}^2\).

    Show solution
    With ABS: \(a = \mu_s\,g = 8\ \text{m/s}^2\); \(d = \dfrac{30^2}{2 \cdot 8} = \dfrac{900}{16} \approx 56.3\ \text{m}\)
    Locked: \(a = \mu_k\,g = 6\ \text{m/s}^2\); \(d = \dfrac{900}{12} = 75\ \text{m}\)
    ABS: \(\approx 56\ \text{m}\); locked: \(75\ \text{m}\). Not skidding makes use of static friction, which is larger.
STEP 7

On a ramp, the weight splits in two

On an inclined plane at angle \(\theta\), it helps to split the weight into two parts. One lies along the ramp, \(W\sin\theta\), and pulls the block down, while the other is perpendicular to the ramp, \(W\cos\theta\), pressing the block against the ramp and balanced by the normal force.

If we neglect friction, the block slides down with \(a = g\sin\theta\), whatever its mass. With friction, it only starts to slip when \(W\sin\theta\) exceeds \(\mu_s\,W\cos\theta\), that is, when \(\tan\theta > \mu_s\).

\(W_x = W\sin\theta\)\(N = W\cos\theta\)\(a = g\,(\sin\theta - \mu_k\cos\theta)\)\(\tan\theta_{\text{critical}} = \mu_s\)Without friction, \(a = g\sin\theta\). The angle at which the block starts to slip does not depend on the mass.

weight and components   normal   friction

Let's discuss

  • With wood, increase the angle slowly, one degree at a time, and note the angle at which the block starts to slip. How does it compare with the ‘critical angle’?
  • Change the mass and repeat; does the critical angle change?
  • Without friction, is the acceleration 5 m/s² at 30°, and what does it become at 90°?
  • Once the block starts sliding down, reduce the angle a little; does it stop, and why do we have to reduce it quite a lot?
W·sin θ (along)
N = W·cos θ
Friction
Acceleration
Critical angle
Velocity
Exercises for step 7 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A \(4\ \text{kg}\) block is on a frictionless \(30^\circ\) inclined plane. Find the weight, the component along the plane and the normal force. Use \(g = 10\ \text{m/s}^2\), \(\sin 30^\circ = 0.5\) and \(\cos 30^\circ \approx 0.87\).

    Show solution
    \(W = 4 \cdot 10 = 40\ \text{N}\)
    \(W_x = W\sin 30^\circ = 40 \cdot 0.5 = 20\ \text{N}\)
    \(N = W\,\cos 30^\circ \approx 40 \cdot 0.87 \approx 34.6\ \text{N}\)
    \(W = 40\ \text{N}\); \(W_x = 20\ \text{N}\); \(N \approx 34.6\ \text{N}\).
  2. basic

    What is the acceleration of a block sliding, without friction, down a \(30^\circ\) ramp? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(a = g\sin\theta = 10 \cdot 0.5\)
    \(a = 5\ \text{m/s}^2\) (it does not depend on the mass).
  3. basic

    When we increase the angle of a ramp, does the normal force on the block increase or decrease? And the component of the weight along the ramp?

    Show solution
    In \(N = W\cos\theta\), the cosine decreases as \(\theta\) increases, so the normal force decreases (it is zero at \(90^\circ\)).
    In \(W_x = W\sin\theta\), the sine increases, so the component along the ramp increases.
    The normal force decreases and the component along the ramp increases.
  4. basic

    A block slides without friction down a ramp with \(\sin\theta = 0.6\) (\(\theta \approx 37^\circ\)). What is the acceleration? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(a = g\sin\theta = 10 \cdot 0.6\)
    \(a = 6\ \text{m/s}^2\)
  5. intermediate

    A \(10\ \text{kg}\) block slides down a ramp with \(\sin\theta = 0.6\) and \(\cos\theta = 0.8\), with \(\mu_k = 0.25\). What is the acceleration? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(W_x = 100 \cdot 0.6 = 60\ \text{N}\); \(N = 100 \cdot 0.8 = 80\ \text{N}\)
    \(f_k = 0.25 \cdot 80 = 20\ \text{N}\), up the ramp.
    \(a = \dfrac{60 - 20}{10}\)
    \(a = 4\ \text{m/s}^2\)
  6. intermediate

    A \(2\ \text{kg}\) block is at rest on a \(30^\circ\) ramp. What is the friction on it, and in which direction? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    At rest, the net force along the ramp is zero.
    Static friction balances \(W_x = 20 \cdot 0.5 = 10\ \text{N}\).
    \(f_s = 10\ \text{N}\), up along the ramp.
  7. intermediate

    A box on a board starts to slip when the board reaches \(37^\circ\) (\(\tan 37^\circ \approx 0.75\)). What is the coefficient of static friction?

    Show solution
    At the angle where it starts to slip, \(W\sin\theta = \mu_s\,W\cos\theta\).
    \(\mu_s = \tan\theta \approx 0.75\)
    \(\mu_s \approx 0.75\)
  8. intermediate

    Two boxes of the same material, one of \(1\ \text{kg}\) and one of \(10\ \text{kg}\), sit on the same board, which is gradually tilted. Which one slips first?

    Show solution
    The condition for slipping is \(\tan\theta > \mu_s\), in which the mass cancels out (it appears in \(W_x\) and in \(N\)).
    Both slip at the same angle.
  9. challenge

    A block starts from rest at the top of a frictionless \(30^\circ\) ramp, \(10\ \text{m}\) long. With what speed does it reach the bottom, and how long does it take? Compare with free fall from the same height. Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(a = 10 \cdot 0.5 = 5\ \text{m/s}^2\)
    \(v = \sqrt{2\,a\,L} = \sqrt{2 \cdot 5 \cdot 10} = 10\ \text{m/s}\); \(t = \dfrac{v}{a} = 2\ \text{s}\)
    The height is \(h = 10 \cdot 0.5 = 5\ \text{m}\), and in free fall \(v = \sqrt{2 \cdot 10 \cdot 5} = 10\ \text{m/s}\), in \(1\ \text{s}\).
    \(v = 10\ \text{m/s}\) in \(2\ \text{s}\): the same final speed as free fall, but it takes twice as long.
  10. challenge

    You want to push a \(50\ \text{kg}\) box at constant velocity up a ramp with \(\sin\theta = 0.6\), \(\cos\theta = 0.8\) and \(\mu_k = 0.5\). What force parallel to the ramp is needed? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(W_x = 500 \cdot 0.6 = 300\ \text{N}\) (down the ramp)
    \(N = 500 \cdot 0.8 = 400\ \text{N}\); \(f_k = 0.5 \cdot 400 = 200\ \text{N}\) (going up, friction points down)
    At constant velocity, \(F = W_x + f_k = 300 + 200\)
    \(F = 500\ \text{N}\)
STEP 8

Several blocks, one acceleration

When blocks move together (touching, or connected by a taut string), they all have the same acceleration. The trick is to do two calculations, one for the whole system, to find \(a\), and another for a single block, to find the force between them.

With blocks in contact, the force \(F\) accelerates both, and the rear block pushes the front one with the contact force; in the block on a table + hanging block system, only the weight of the hanging block pulls the system. The Atwood machine hangs two blocks on a pulley, and the difference between their weights accelerates both masses.

In contact: \(a = \dfrac{F}{m_1 + m_2}\), \(F_c = m_2\,a\)Table: \(a = \dfrac{m_2\,g}{m_1 + m_2}\), \(T = m_1\,a\)Atwood: \(a = \dfrac{(m_1 - m_2)\,g}{m_1 + m_2}\)\(T = \dfrac{2\,m_1 m_2\,g}{m_1 + m_2}\)If we assume an ideal string (massless, does not stretch) and an ideal pulley, the tension is the same at both ends. We also neglect friction and take \(g = 10\ \text{m/s}^2\).

weight   normal   tension   applied force   contact

Let's discuss

  • In ‘table’ mode, is the tension greater or smaller than the weight of the hanging block? Check the diagrams before answering.
  • In the Atwood machine, make the masses equal and see what happens, then try 3 kg and 2 kg and check whether the acceleration is 2 m/s².
  • With the blocks in contact, swap the masses; does the acceleration change, and does the contact force?
  • In each diagram, do the forces add up to exactly \(m \cdot a\) for that block?
Acceleration
Tension
Weight of block 1
Weight of block 2
Velocity
Exercises for step 8 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    Two blocks in contact, A (\(3\ \text{kg}\)) and B (\(2\ \text{kg}\)), are on a frictionless table. A force of \(20\ \text{N}\) pushes A, which pushes B. What are the acceleration and the contact force between them?

    Show solution
    For the system, \(a = \dfrac{F}{m_A + m_B} = \dfrac{20}{5} = 4\ \text{m/s}^2\)
    The only horizontal force on B is the contact force, so \(F_c = m_B\,a = 2 \cdot 4 = 8\ \text{N}\)
    \(a = 4\ \text{m/s}^2\); contact force \(8\ \text{N}\).
  2. basic

    In the previous situation, the same \(20\ \text{N}\) force now pushes block B (\(2\ \text{kg}\)), which pushes A. What is the contact force now?

    Show solution
    The acceleration does not change, \(a = \dfrac{20}{5} = 4\ \text{m/s}^2\).
    Now the block pushed by contact is A, and \(F_c = m_A\,a = 3 \cdot 4 = 12\ \text{N}\).
    \(F_c = 12\ \text{N}\): it is larger when you push from the side of the lighter block.
  3. basic

    In an Atwood machine, blocks of \(3\ \text{kg}\) and \(2\ \text{kg}\) are hung. What is the acceleration? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(a = \dfrac{(m_1 - m_2)\,g}{m_1 + m_2} = \dfrac{(3 - 2) \cdot 10}{5}\)
    \(a = 2\ \text{m/s}^2\) (the \(3\ \text{kg}\) block goes down).
  4. basic

    In an Atwood machine with equal masses, one of the blocks is given a small push. What happens next?

    Show solution
    With equal masses, \(a = \dfrac{(m - m)\,g}{2m} = 0\).
    With zero acceleration, the velocity stays constant (1st law).
    The blocks keep moving at the velocity they gained, constant.
  5. intermediate

    A \(6\ \text{kg}\) block is on a frictionless table, connected by a string (passing over a pulley at the edge) to a hanging \(2\ \text{kg}\) block. Find the acceleration and the tension. Use \(g = 10\ \text{m/s}^2\).

    Show solution
    Only the weight of the hanging block moves the system, so \(a = \dfrac{m_2\,g}{m_1 + m_2} = \dfrac{20}{8} = 2.5\ \text{m/s}^2\)
    For the block on the table, \(T = m_1\,a = 6 \cdot 2.5 = 15\ \text{N}\)
    As a check on the hanging block, \(20 - 15 = 2 \cdot 2.5\) ✓
    \(a = 2.5\ \text{m/s}^2\); \(T = 15\ \text{N}\).
  6. intermediate

    In the Atwood machine with \(3\ \text{kg}\) and \(2\ \text{kg}\) (\(a = 2\ \text{m/s}^2\)), what is the tension in the string? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    2 kg block (goes up): \(T - 20 = 2 \cdot 2\) \(\Rightarrow T = 24\ \text{N}\)
    Checking on the 3 kg block (goes down): \(30 - T = 3 \cdot 2\) \(\Rightarrow T = 24\ \text{N}\) ✓
    By the formula: \(T = \dfrac{2\,m_1 m_2\,g}{m_1 + m_2} = \dfrac{2 \cdot 3 \cdot 2 \cdot 10}{5} = 24\ \text{N}\)
    \(T = 24\ \text{N}\)
  7. intermediate

    A locomotive pulls two wagons of \(1000\ \text{kg}\) each, joined by couplings, with an acceleration of \(0.5\ \text{m/s}^2\) and no friction. What is the force in the locomotive–wagon 1 coupling and in the wagon 1–wagon 2 coupling?

    Show solution
    The locomotive–wagon 1 coupling pulls BOTH wagons, so \(T_1 = 2000 \cdot 0.5 = 1000\ \text{N}\)
    The wagon 1–wagon 2 coupling pulls only the last one, so \(T_2 = 1000 \cdot 0.5 = 500\ \text{N}\)
    \(T_1 = 1000\ \text{N}\); \(T_2 = 500\ \text{N}\).
  8. intermediate

    In the ‘block on a table + hanging block’ system, is the tension in the string greater than, smaller than or equal to the weight of the hanging block while it accelerates downward? Why?

    Show solution
    The hanging block accelerates downward, so the net force on it points down, \(W_2 - T = m_2\,a > 0\).
    Hence \(T < W_2\); if they were equal, it would not accelerate.
    Smaller: \(T = W_2 - m_2\,a\).
  9. challenge

    A \(4\ \text{kg}\) block on a table with \(\mu_k = 0.25\) is connected by a string to a hanging \(2\ \text{kg}\) block. Find the acceleration and the tension. Use \(g = 10\ \text{m/s}^2\).

    Show solution
    The friction on the block on the table is \(f_k = 0.25 \cdot 40 = 10\ \text{N}\)
    System: \(a = \dfrac{m_2\,g - f_k}{m_1 + m_2}\) \(= \dfrac{20 - 10}{6} \approx 1.67\ \text{m/s}^2\)
    Hanging block: \(T = 20 - 2 \cdot 1.67 \approx 16.7\ \text{N}\)
    Checking on the table: \(T - 10 = 4 \cdot 1.67 \approx 6.7\) ✓
    \(a \approx 1.67\ \text{m/s}^2\); \(T \approx 16.7\ \text{N}\).
  10. challenge

    In an Atwood machine with \(5\ \text{kg}\) and \(3\ \text{kg}\), released from rest, the heavier block is \(1.0\ \text{m}\) above the floor. Find the acceleration, the tension, the time until it reaches the floor and its speed at that instant. Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(a = \dfrac{(5 - 3) \cdot 10}{8} = 2.5\ \text{m/s}^2\)
    \(T = \dfrac{2 \cdot 5 \cdot 3 \cdot 10}{8} = 37.5\ \text{N}\)
    \(1.0 = \dfrac{2.5\,t^2}{2}\) \(\Rightarrow t = \sqrt{0.8} \approx 0.89\ \text{s}\)
    \(v = a\,t \approx 2.5 \cdot 0.894 \approx 2.24\ \text{m/s}\)
    \(a = 2.5\ \text{m/s}^2\); \(T = 37.5\ \text{N}\); \(t \approx 0.89\ \text{s}\); \(v \approx 2.24\ \text{m/s}\).
STEP 9

Work and energy: what is conserved on the roller coaster

A force does work when it moves a body, and we compute it as \(\tau = F\,d\cos\theta\), where \(\theta\) is the angle between the force and the displacement. A force perpendicular to the motion does no work.

Work transfers energy. A moving body has kinetic energy \(E_k\), while a body up high has gravitational potential energy \(E_p\), and the work–energy theorem tells us that the work done by the net force equals the change in \(E_k\).

If we neglect friction, the mechanical energy \(E_k + E_p\) is conserved, and the cart trades height for speed and back again without ever going higher than where it started. With friction, part of the energy turns into heat (dissipated energy), and it is the sum of the three that stays equal to the initial energy.

\(\tau = F\,d\cos\theta\)\(E_k = \dfrac{m\,v^2}{2}\)\(E_p = m\,g\,h\)\(\tau_R = \Delta E_k\)\(E_k + E_p = \text{constant}\) (no friction)We measure energy and work in joules, \(1\ \text{J} = 1\ \text{N} \cdot \text{m}\). Without friction, \(v = \sqrt{2\,g\,(h_0 - h)}\), regardless of the mass.

potential \(E_p\)   kinetic \(E_k\)   dissipated (heat)

Let's discuss

  • Release from 8 m without friction; does the total bar change while the cart moves, and where is the speed greatest?
  • The hump is 5 m high, so what is the smallest starting height that gets over it without friction? Try 4.5 m and 5.5 m.
  • Turn on friction and watch the bars; where does the energy that seems to ‘disappear’ from the \(E_k\) and \(E_p\) bars go?
  • With friction, is 6 m still enough to get over the hump, and what height is needed now?
  • Change the mass and the energies change with it, but does the speed at each point change too?
Height
Speed
Kinetic Ek
Potential Ep
Dissipated
Total
Exercises for step 9 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A person pushes a box with a horizontal force of \(50\ \text{N}\) over \(4\ \text{m}\), in the direction of motion. How much work does this force do?

    Show solution
    Force and displacement point in the same direction, so \(\cos 0^\circ = 1\).
    \(\tau = F\,d\cos\theta = 50 \cdot 4 \cdot 1\)
    \(\tau = 200\ \text{J}\)
  2. basic

    What is the kinetic energy of a \(1000\ \text{kg}\) car at \(20\ \text{m/s}\)?

    Show solution
    \(E_k = \dfrac{m\,v^2}{2} = \dfrac{1000 \cdot 400}{2}\)
    \(E_k = 200\,000\ \text{J} = 200\ \text{kJ}\)
  3. basic

    What is the gravitational potential energy of a \(2\ \text{kg}\) flowerpot \(5\ \text{m}\) above the ground? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(E_p = m\,g\,h = 2 \cdot 10 \cdot 5\)
    \(E_p = 100\ \text{J}\)
  4. basic

    A person carries a rucksack while walking in a straight horizontal line at constant velocity. Does the force they apply to hold the rucksack (vertical, upward) do any work?

    Show solution
    The force is vertical and the displacement is horizontal, so \(\theta = 90^\circ\) and \(\cos 90^\circ = 0\).
    \(\tau = F\,d\cos 90^\circ = 0\)
    No: the work done by this force is zero (even though the person gets tired).
  5. intermediate

    A sledge is pulled \(10\ \text{m}\) with a rope at \(60^\circ\) to the ground, with a tension of \(100\ \text{N}\). How much work does the tension do? (\(\cos 60^\circ = 0.5\))

    Show solution
    \(\tau = F\,d\cos\theta = 100 \cdot 10 \cdot 0.5\)
    Only the horizontal component of the force (\(50\ \text{N}\)) does work.
    \(\tau = 500\ \text{J}\)
  6. intermediate

    A ball is dropped from a height of \(20\ \text{m}\). Ignoring air resistance, with what speed does it hit the ground? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    By conservation of energy, \(m\,g\,h = \dfrac{m\,v^2}{2}\) (the mass cancels).
    \(v = \sqrt{2\,g\,h} = \sqrt{2 \cdot 10 \cdot 20} = \sqrt{400}\)
    \(v = 20\ \text{m/s}\)
  7. intermediate

    A roller-coaster car starts from rest at a height of \(45\ \text{m}\). Without friction, what is its speed at a point \(25\ \text{m}\) high? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(m\,g\,h_1 = m\,g\,h_2 + \dfrac{m\,v^2}{2}\)
    \(v = \sqrt{2\,g\,(h_1 - h_2)}\) \(= \sqrt{2 \cdot 10 \cdot 20}\) \(= \sqrt{400}\)
    \(v = 20\ \text{m/s}\)
  8. intermediate

    A \(2\ \text{kg}\) block goes from \(3\ \text{m/s}\) to \(7\ \text{m/s}\). How much work does the net force do? If it acted over \(8\ \text{m}\), what is its value?

    Show solution
    By the work–energy theorem, \(\tau_R = \Delta E_k\) \(= \dfrac{2 \cdot 7^2}{2} - \dfrac{2 \cdot 3^2}{2}\) \(= 49 - 9\) \(= 40\ \text{J}\)
    \(F_R = \dfrac{\tau}{d} = \dfrac{40}{8} = 5\ \text{N}\)
    \(\tau = 40\ \text{J}\); \(F_R = 5\ \text{N}\).
  9. challenge

    A \(50\ \text{kg}\) skateboarder starts from rest at the top of a ramp \(5\ \text{m}\) high and reaches the bottom at \(8\ \text{m/s}\). How much energy was dissipated? What fraction of the initial energy is that? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    Initial: \(E_p = 50 \cdot 10 \cdot 5 = 2500\ \text{J}\)
    Final: \(E_k = \dfrac{50 \cdot 64}{2} = 1600\ \text{J}\)
    Dissipated: \(2500 - 1600 = 900\ \text{J}\); fraction \(\dfrac{900}{2500} = 0.36\)
    \(900\ \text{J}\), or \(36\%\) of the initial energy (it became heat and sound).
  10. challenge

    A \(2\ \text{kg}\) block is launched at \(6\ \text{m/s}\) across a horizontal floor with \(\mu_k = 0.3\). Using energy, find how far it travels before stopping, and how far it goes if the initial speed doubles. Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(E_k = \dfrac{2 \cdot 36}{2} = 36\ \text{J}\); \(f_k = 0.3 \cdot 20 = 6\ \text{N}\)
    The work done by friction uses up all of \(E_k\), so \(f_k\,d = 36\) \(\Rightarrow d = 6\ \text{m}\)
    With \(12\ \text{m/s}\), \(E_k = 144\ \text{J}\) (4 times) and \(d = 24\ \text{m}\)
    \(d = 6\ \text{m}\); doubling the speed, \(d = 24\ \text{m}\) (4 times farther).
STEP 10

Stopping slowly hurts less: impulse and momentum

We call the product of mass and velocity momentum, \(\vec p = m\,\vec v\). It is a vector pointing the same way as the velocity, which is why a ball that bounces back off a wall ends up with \(p\) of the opposite sign.

To change a body's momentum we have to apply a force for some time. The product \(\vec F\,\Delta t\) is called the impulse, and when the force varies, the impulse is the area under the \(F \times t\) graph. From the 2nd law, \(F = m\,\Delta v/\Delta t\), we obtain the impulse–momentum theorem, which states that the impulse of the net force equals the change in momentum.

Consider the egg in the simulation, which falls from a given height and reaches the ground with a well-defined momentum that it has to lose in order to stop, whether it lands on concrete or on a mattress. The \(\Delta p\) is the same in both cases, and what changes is how long the stop takes. Since \(F_{\text{mean}} = \Delta p/\Delta t\), an impact that lasts ten times longer needs a force ten times smaller.

\(\vec p = m\,\vec v\)\(\vec I = \vec F\,\Delta t\)\(I = \text{area under the } F \times t \text{ graph}\)\(\vec I_R = \Delta \vec p = m\,\vec v_f - m\,\vec v_i\)\(F_{\text{mean}} = \dfrac{\Delta p}{\Delta t}\)We measure \(p\) in kg·m/s and \(I\) in N·s, which are the same unit. In the simulation the egg has a mass of 60 g and stops without bouncing; since its weight (0.6 N) is small compared with the force from the ground during the impact, we treat the force from the ground as the net force.

This is the principle behind the airbag, the gym crash mat, the goalkeeper's glove and bending your knees when you land from a jump. In each of these cases the momentum to be lost stays the same, and what we gain is stopping time, with a much smaller peak force. Our intuition tends to credit the protection to how soft the material is, and it may be more useful to think of softness as a way of stretching the impact, since it is this time that divides the \(\Delta p\) and sets the size of the force.

concrete   grass   mattress

Let's discuss

  • Drop the egg from 1.8 m onto concrete, then onto grass and onto the mattress. Do the three areas under the curves look equal? Check the values written on the graph.
  • Compare the peak forces. How many times larger is the peak on concrete than on the mattress, and how many times longer is the impact on the mattress? Do the two ratios match?
  • On grass, look for the greatest height from which the egg still survives, lowering the height a little at a time. We assume the shell cracks at about 25 N, a value that only gives the order of magnitude.
  • Change the height from 0.8 m to 3.2 m, four times higher. What happens to the landing speed, the area and the peak force? The answer may not be ‘four times’.
Landing speed
Momentum
Impulse (area)
Impact duration
Peak force
Mean force
Exercises for step 10 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    What is the momentum of a \(0.4\ \text{kg}\) ball moving at \(15\ \text{m/s}\)? And that of a \(1000\ \text{kg}\) car at \(20\ \text{m/s}\)?

    Show solution
    In each case we multiply the mass by the velocity.
    Ball: \(p = m\,v = 0.4 \cdot 15 = 6\ \text{kg·m/s}\)
    Car: \(p = 1000 \cdot 20 = 20\,000\ \text{kg·m/s}\)
    \(6\ \text{kg·m/s}\) for the ball and \(20\,000\ \text{kg·m/s}\) for the car, each in its direction of motion.
  2. basic

    A constant force of \(50\ \text{N}\) acts on a body for \(0.2\ \text{s}\). What is the impulse of this force?

    Show solution
    With a constant force, the impulse is the force multiplied by the time.
    \(I = F\,\Delta t = 50 \cdot 0.2\)
    \(I = 10\ \text{N·s}\), in the direction of the force.
  3. basic

    When catching a hard shot, a goalkeeper draws the hands back with the ball, and the glove has thick foam. Why does this protect the hands?

    Show solution
    The ball has to lose all of its momentum, and that \(\Delta p\) does not depend on how the goalkeeper holds it.
    By drawing the hands back and letting the foam compress, the goalkeeper increases the stopping time \(\Delta t\).
    Since \(F_{\text{mean}} = \Delta p/\Delta t\), a longer time means a smaller force on the hands.
    The \(\Delta p\) is the same; with a longer stopping time, the force on the hands is smaller.
  4. basic

    A \(2\ \text{kg}\) trolley, initially at rest, receives an impulse of \(8\ \text{N·s}\). At what speed does it move off?

    Show solution
    By the impulse–momentum theorem, \(I = \Delta p = m\,v - 0\).
    \(v = \dfrac{I}{m} = \dfrac{8}{2}\)
    \(v = 4\ \text{m/s}\), in the direction of the impulse.
  5. intermediate

    A \(0.5\ \text{kg}\) ball hits a wall head-on at \(10\ \text{m/s}\) and bounces back at \(8\ \text{m/s}\) along the same line. What is the magnitude of the change in momentum? If the contact lasted \(0.03\ \text{s}\), what was the mean force from the wall?

    Show solution
    We take the direction in which the ball arrives as positive.
    \(p_i = 0.5 \cdot 10 = 5\ \text{kg·m/s}\) and \(p_f = 0.5 \cdot (-8) = -4\ \text{kg·m/s}\)
    \(\Delta p = -4 - 5 = -9\ \text{kg·m/s}\), that is, \(9\ \text{kg·m/s}\) pointing away from the wall.
    \(F_{\text{mean}} = \dfrac{9}{0.03} = 300\ \text{N}\)
    Because the ball bounces back, the two speeds add up in \(\Delta p\).
    \(|\Delta p| = 9\ \text{kg·m/s}\); \(F_{\text{mean}} = 300\ \text{N}\).
  6. intermediate

    In a kick, the force of the foot on a \(0.2\ \text{kg}\) ball at rest rises from zero to \(400\ \text{N}\) in \(0.01\ \text{s}\) and falls back to zero in another \(0.01\ \text{s}\), forming a triangle on the \(F \times t\) graph. What is the impulse, and at what speed does the ball leave the foot?

    Show solution
    The impulse is the area of the triangle, with base \(0.02\ \text{s}\) and height \(400\ \text{N}\).
    \(I = \dfrac{0.02 \cdot 400}{2} = 4\ \text{N·s}\)
    \(v = \dfrac{I}{m} = \dfrac{4}{0.2} = 20\ \text{m/s}\)
    \(I = 4\ \text{N·s}\); the ball leaves at \(20\ \text{m/s}\).
  7. intermediate

    A \(70\ \text{kg}\) person jumps off a wall and reaches the ground at \(4\ \text{m/s}\). With straight legs they stop in \(0.05\ \text{s}\); bending the knees, in \(0.4\ \text{s}\). What is the mean net force in each case?

    Show solution
    In both cases the person loses \(\Delta p = 70 \cdot 4 = 280\ \text{kg·m/s}\).
    Straight legs: \(F = \dfrac{280}{0.05} = 5600\ \text{N}\)
    Bent knees: \(F = \dfrac{280}{0.4} = 700\ \text{N}\)
    The time became 8 times longer, and the force 8 times smaller.
    \(5600\ \text{N}\) with straight legs and \(700\ \text{N}\) with bent knees.
  8. intermediate

    In a crash at \(20\ \text{m/s}\), a \(60\ \text{kg}\) passenger is stopped by the airbag in \(0.2\ \text{s}\). Without the airbag, they would hit the dashboard and stop in \(0.02\ \text{s}\). Compare the mean forces on the passenger.

    Show solution
    In either case, \(\Delta p = 60 \cdot 20 = 1200\ \text{kg·m/s}\).
    With the airbag: \(F = \dfrac{1200}{0.2} = 6000\ \text{N}\)
    Without it: \(F = \dfrac{1200}{0.02} = 60\,000\ \text{N}\)
    \(6000\ \text{N}\) with the airbag and \(60\,000\ \text{N}\) without it, a force 10 times larger.
  9. challenge

    A \(60\ \text{g}\) egg falls from \(1.8\ \text{m}\) and reaches the ground at \(6\ \text{m/s}\). It stops in \(0.1\ \text{s}\) on a mattress or in \(0.004\ \text{s}\) on concrete. What mean force does the surface exert on the egg in each case, taking its weight into account? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    The impulse of the net force is \(\Delta p = 0.06 \cdot 6 = 0.36\ \text{kg·m/s}\), and the net force is the force from the surface minus the weight, \(F - m\,g\).
    So \(F = \dfrac{\Delta p}{\Delta t} + m\,g\), with \(m\,g = 0.6\ \text{N}\).
    Mattress: \(F = \dfrac{0.36}{0.1} + 0.6 = 3.6 + 0.6 = 4.2\ \text{N}\)
    Concrete: \(F = \dfrac{0.36}{0.004} + 0.6 = 90 + 0.6 = 90.6\ \text{N}\)
    On concrete the weight accounts for less than 1% of the result, which is why we usually neglect it in short impacts.
    \(4.2\ \text{N}\) on the mattress and \(90.6\ \text{N}\) on concrete.
  10. challenge

    A hose sends \(2\ \text{kg}\) of water per second at \(10\ \text{m/s}\) against a wall. The water hits the wall and runs down it, losing all its horizontal velocity. What force does the water exert on the wall?

    Show solution
    Every second, \(2\ \text{kg}\) of water lose \(10\ \text{m/s}\) horizontally.
    \(\Delta p\) per second: \(2 \cdot 10 = 20\ \text{kg·m/s}\)
    The wall exerts \(F = \dfrac{\Delta p}{\Delta t} = \dfrac{20}{1} = 20\ \text{N}\) on the water and, by the 3rd law, the water pushes the wall with the same force.
    \(F = 20\ \text{N}\), pushing on the wall.
STEP 11

Collisions: what is conserved when two bodies collide

In an isolated system, where the net external force is zero (or negligible during the impact), the total momentum is conserved. During the collision each trolley pushes the other with equal and opposite forces (3rd law) for the same length of time, so the impulses cancel and whatever momentum one trolley gains, the other one loses.

The coefficient of restitution \(e\) compares the speed at which the bodies separate after the impact with the speed at which they approached before it. With \(e = 1\) the collision is elastic; with \(0 < e < 1\), partially elastic; and with \(e = 0\), perfectly inelastic, so the bodies move off stuck together.

Kinetic energy is conserved only in an elastic collision. In the others, part of it becomes heat, sound and deformation, even though the total momentum stays the same. If we throw two identical lumps of modelling clay at each other with opposite velocities, they stop together, with \(p = 0\) before and after and no \(E_k\) left.

\(p_{\text{before}} = p_{\text{after}}\)\(m_1 v_1 + m_2 v_2 = m_1 v_1^{\prime} + m_2 v_2^{\prime}\)\(e = \dfrac{v_2^{\prime} - v_1^{\prime}}{v_1 - v_2}\)\(e = 0:\ v^{\prime} = \dfrac{m_1 v_1 + m_2 v_2}{m_1 + m_2}\)Velocities carry a sign (positive to the right). We assume a frictionless track and a head-on impact along a single line. With \(e = 1\) and equal masses, the bodies swap velocities.

The same conservation law explains the recoil of a gun or a cannon. Before firing, \(p = 0\); afterwards, the shell moves forward with momentum \(m\,v\) and the cannon moves back with the same momentum, only at a much lower speed, because its mass is much larger. The skaters of step 4, pushing off from rest, are the same case.

In Newton's cradle, when we release one ball on one side, a single ball leaves the other side at the same speed. Two balls at half the speed would carry the same \(p\) but only half the kinetic energy, and the collisions between steel balls, which are nearly elastic, do not allow that loss.

trolley 1   trolley 2   total

Let's discuss

  • In ‘Elastic, equal masses’, look at the velocities before and after the impact. What have the trolleys exchanged?
  • In ‘Stick together’, compare the \(p\) and \(E_k\) bars before and after. Which total stayed the same, and where did the missing energy go?
  • In ‘Head-on’, both trolleys stop. Did the total momentum vanish, or was it already zero before the impact?
  • With the light trolley against the very heavy one, the light one bounces back at almost the speed it had. Why does this case resemble a ball hitting a wall?
  • Take \(e\) from 1 down to 0 without changing anything else and watch the \(E_k\) bar after the impact. Does the total \(p\) bar change at any point?
Total p before
Total p after
Total Ek before
Total Ek after
v₁ after
v₂ after
Exercises for step 11 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A \(2\ \text{kg}\) trolley at \(3\ \text{m/s}\) hits a \(1\ \text{kg}\) trolley at rest, and the two move off stuck together. At what speed do they move? Neglect friction.

    Show solution
    Without friction, the total momentum is conserved.
    \(2 \cdot 3 + 1 \cdot 0 = (2 + 1)\,v\)
    \(v = \dfrac{6}{3}\)
    \(v = 2\ \text{m/s}\), in the direction the first trolley was moving.
  2. basic

    What is conserved in an elastic collision (\(e = 1\)) and in a perfectly inelastic collision (\(e = 0\)) between two bodies in an isolated system?

    Show solution
    In an isolated system, the total momentum is conserved in any collision.
    Total kinetic energy is conserved only in the elastic case; in the perfectly inelastic one the bodies stick together and part of the \(E_k\) becomes heat, sound and deformation.
    Elastic: \(p\) and \(E_k\). Perfectly inelastic: only \(p\).
  3. basic

    A \(500\ \text{kg}\) cannon at rest fires a \(5\ \text{kg}\) shell at \(200\ \text{m/s}\). Neglecting friction with the ground, at what speed does the cannon recoil?

    Show solution
    Before firing, \(p = 0\), and the total is still zero afterwards.
    \(0 = 5 \cdot 200 + 500 \cdot v_c\)
    \(v_c = -\dfrac{1000}{500} = -2\ \text{m/s}\)
    The cannon recoils at \(2\ \text{m/s}\), opposite to the shell.
  4. basic

    A snooker ball at \(2\ \text{m/s}\) hits an identical ball at rest head-on, in an elastic collision. What happens to each ball?

    Show solution
    In an elastic collision between equal masses, the bodies swap velocities.
    The first ball stops and the second moves off at \(2\ \text{m/s}\).
    Checking: \(p\) before \(= m \cdot 2\) and after \(= m \cdot 2\); \(E_k\) before \(= \dfrac{m \cdot 4}{2}\) and the same after.
    The incoming ball stops, and the other moves off at \(2\ \text{m/s}\).
  5. intermediate

    Two \(1\ \text{kg}\) trolleys collide head-on. Before, trolley 1 moves at \(4\ \text{m/s}\) and trolley 2 is at rest; afterwards, trolley 1 moves at \(1\ \text{m/s}\) and trolley 2 at \(3\ \text{m/s}\), in the same direction. What is the coefficient of restitution, and how much kinetic energy was lost?

    Show solution
    We first check conservation: \(1 \cdot 4 = 1 \cdot 1 + 1 \cdot 3\) ✓
    \(e = \dfrac{v_2^{\prime} - v_1^{\prime}}{v_1 - v_2}\) \(= \dfrac{3 - 1}{4 - 0}\) \(= 0.5\)
    \(E_k\) before \(= \dfrac{1 \cdot 16}{2} = 8\ \text{J}\); after \(= \dfrac{1 \cdot 1}{2} + \dfrac{1 \cdot 9}{2} = 5\ \text{J}\)
    \(e = 0.5\); \(3\ \text{J}\) was lost (37.5% of the kinetic energy).
  6. intermediate

    In the first exercise of this step (\(2\ \text{kg}\) at \(3\ \text{m/s}\) against \(1\ \text{kg}\) at rest, sticking together), what is the kinetic energy before and after the impact?

    Show solution
    Before: \(E_k = \dfrac{2 \cdot 3^2}{2} = 9\ \text{J}\)
    Afterwards both move together at \(2\ \text{m/s}\), and \(E_k = \dfrac{3 \cdot 2^2}{2} = 6\ \text{J}\)
    \(3\ \text{J}\), a third of the initial energy, went into heat, sound and deformation.
    \(9\ \text{J}\) before and \(6\ \text{J}\) after.
  7. intermediate

    A \(3\ \text{kg}\) trolley moving at \(2\ \text{m/s}\) to the right hits head-on a \(2\ \text{kg}\) trolley moving at \(4\ \text{m/s}\) to the left, and they stick together. What is the final velocity of the pair, and how much kinetic energy is left?

    Show solution
    Taking right as positive, \(p = 3 \cdot 2 + 2 \cdot (-4) = 6 - 8 = -2\ \text{kg·m/s}\).
    \(v = \dfrac{-2}{5} = -0.4\ \text{m/s}\)
    \(E_k\) before \(= \dfrac{3 \cdot 4}{2} + \dfrac{2 \cdot 16}{2} = 6 + 16 = 22\ \text{J}\); after \(= \dfrac{5 \cdot 0.16}{2} = 0.4\ \text{J}\)
    \(v = 0.4\ \text{m/s}\) to the left; only \(0.4\ \text{J}\) of the \(22\ \text{J}\) is left.
  8. intermediate

    In Newton's cradle, we release one ball on one side and a single ball leaves the other side at the same speed. Why do two balls not leave at half the speed, if that would also conserve momentum?

    Show solution
    With one ball at \(v\): \(p = m\,v\) and \(E_k = \dfrac{m\,v^2}{2}\).
    With two balls at \(v/2\): \(p = 2m \cdot \dfrac{v}{2} = m\,v\) ✓, but \(E_k = \dfrac{2m}{2}\left(\dfrac{v}{2}\right)^2 = \dfrac{m\,v^2}{4}\), only half.
    Collisions between steel balls are nearly elastic and also conserve kinetic energy, and the only outcome that conserves both \(p\) and \(E_k\) is one ball at the same speed.
    Two balls at \(v/2\) would lose half the kinetic energy, which a nearly elastic collision does not allow.
  9. challenge

    A \(10\ \text{g}\) bullet at \(400\ \text{m/s}\) embeds itself in a \(1.99\ \text{kg}\) block hanging from a string (a ballistic pendulum). At what speed does the block start to move, how high does it rise, and what fraction of the bullet's kinetic energy is lost in the impact? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    In the impact, momentum is conserved: \(0.01 \cdot 400 = (0.01 + 1.99)\,v\) \(\Rightarrow v = \dfrac{4}{2} = 2\ \text{m/s}\)
    After the impact, neglecting friction, mechanical energy is conserved: \(h = \dfrac{v^2}{2g} = \dfrac{4}{20} = 0.2\ \text{m}\)
    The bullet's \(E_k = \dfrac{0.01 \cdot 400^2}{2} = 800\ \text{J}\); just after the impact, \(\dfrac{2 \cdot 2^2}{2} = 4\ \text{J}\)
    \(v = 2\ \text{m/s}\); it rises \(0.2\ \text{m}\); \(99.5\%\) of the kinetic energy is lost.
  10. challenge

    A \(1\ \text{kg}\) trolley at \(6\ \text{m/s}\) hits a \(2\ \text{kg}\) trolley at rest head-on, in an elastic collision. Find the velocities after the impact and check the kinetic energy.

    Show solution
    Conservation: \(1 \cdot 6 = 1 \cdot v_1^{\prime} + 2\,v_2^{\prime}\)
    With \(e = 1\): \(v_2^{\prime} - v_1^{\prime} = 6 - 0 = 6\)
    Substituting \(v_1^{\prime} = v_2^{\prime} - 6\): \(6 = v_2^{\prime} - 6 + 2\,v_2^{\prime}\) \(\Rightarrow v_2^{\prime} = 4\ \text{m/s}\) and \(v_1^{\prime} = -2\ \text{m/s}\)
    \(E_k\): before \(\dfrac{1 \cdot 36}{2} = 18\ \text{J}\); after \(\dfrac{1 \cdot 4}{2} + \dfrac{2 \cdot 16}{2} = 2 + 16 = 18\ \text{J}\) ✓
    The \(1\ \text{kg}\) trolley bounces back at \(2\ \text{m/s}\) and the \(2\ \text{kg}\) trolley moves off at \(4\ \text{m/s}\).
WRAP-UP

Challenges

Incline + energy

A block slides from rest down a frictionless ramp \(1.8\ \text{m}\) high. With what speed does it reach the bottom, and does anything change if the ramp is steeper? (\(g = 10\ \text{m/s}^2\))

Show solution
\(v = \sqrt{2\,g\,h} = \sqrt{36} = 6\ \text{m/s}\)
The final speed does not depend on the angle; a steeper ramp only makes the block get there sooner (larger \(a = g\sin\theta\)).
Action and reaction + inertia

Skaters of \(45\ \text{kg}\) and \(90\ \text{kg}\) push each other with \(180\ \text{N}\) for \(0.5\ \text{s}\). With what speed does each move off, and what happens next on the frictionless ice?

Show solution
\(v_1 = \dfrac{180}{45} \cdot 0.5 = 2\ \text{m/s}\); \(v_2 = \dfrac{180}{90} \cdot 0.5 = 1\ \text{m/s}\)
After the push the net force is zero, so each keeps moving at constant velocity (1st law).
Weight and normal force

A \(50\ \text{kg}\) person on a lift scale reads \(40\ \text{kg}\). What is the lift's acceleration? (\(g = 10\ \text{m/s}^2\))

Show solution
\(N = 400\ \text{N}\); \(N = m\,(g + a)\) \(\Rightarrow 400 = 50\,(10 + a)\)
\(a = -2\ \text{m/s}^2\), that is, \(2\ \text{m/s}^2\) downward (going down and speeding up, or going up and braking).
Pulleys + energy

In an Atwood machine with \(3\ \text{kg}\) and \(1\ \text{kg}\), released from rest, what is the speed after the heavy block has moved down \(1\ \text{m}\)? Solve with the 2nd law and check with energy. (\(g = 10\ \text{m/s}^2\))

Show solution
\(a = \dfrac{(3 - 1) \cdot 10}{4} = 5\ \text{m/s}^2\); \(v = \sqrt{2 \cdot 5 \cdot 1} = \sqrt{10} \approx 3.16\ \text{m/s}\)
In terms of energy, \(30 - 10 = 20\ \text{J}\) of \(E_p\) is lost, and \(\dfrac{4\,v^2}{2} = 20\) \(\Rightarrow v^2 = 10\) ✓
Work done by friction

A \(1000\ \text{kg}\) car at \(30\ \text{m/s}\) brakes to a stop with friction \(\mu = 0.6\). Use energy to find the braking distance. (\(g = 10\ \text{m/s}^2\))

Show solution
\(E_k = \dfrac{1000 \cdot 900}{2} = 450\,000\ \text{J}\); \(f = 0.6 \cdot 10\,000 = 6000\ \text{N}\)
\(f\,d = E_k\) \(\Rightarrow d = \dfrac{450\,000}{6000} = 75\ \text{m}\)
Impulse + weight

A \(60\ \text{kg}\) person jumps from a height of \(1.25\ \text{m}\). What is the mean force from the ground if they stop in \(0.02\ \text{s}\) with straight legs, or in \(0.25\ \text{s}\) by bending their knees? Compare it with their weight. (\(g = 10\ \text{m/s}^2\))

Show solution
\(v = \sqrt{2 \cdot 10 \cdot 1.25} = 5\ \text{m/s}\); \(\Delta p = 60 \cdot 5 = 300\ \text{kg·m/s}\)
The ground has to stop the person and also support their weight, so \(F = \dfrac{\Delta p}{\Delta t} + m\,g\).
Straight legs: \(F = 15\,000 + 600 = 15\,600\ \text{N}\), about 26 times the weight.
Bent knees: \(F = 1200 + 600 = 1800\ \text{N}\), 3 times the weight.
Collision + friction

A \(1000\ \text{kg}\) car at \(20\ \text{m/s}\) runs into the back of a stationary \(1500\ \text{kg}\) car, and the two move on locked together, slowed by friction with \(\mu = 0.5\). How far do they slide, and how much kinetic energy was lost in the crash? (\(g = 10\ \text{m/s}^2\))

Show solution
In the impact, \(1000 \cdot 20 = 2500\,v\) \(\Rightarrow v = 8\ \text{m/s}\)
Afterwards friction decelerates them at \(a = \mu\,g = 5\ \text{m/s}^2\), so \(d = \dfrac{8^2}{2 \cdot 5} = 6.4\ \text{m}\)
\(E_k\): \(200\ \text{kJ}\) before and \(\dfrac{2500 \cdot 64}{2} = 80\ \text{kJ}\) after, so \(120\ \text{kJ}\) (60%) was lost in the crash.
Think, no calculation

In a tug of war, by the 3rd law, the rope pulls both teams with the same force, so how can one team win?

Show solution
The rope pulls both equally, but each team also receives the friction force from the ground on their feet.
The team that pushes the ground harder (and has more friction) wins, because for that team the net force points towards its side.

ENEM-style questions

The ENEM is Brazil’s national secondary-school exam, the one most students sit to get into university, and we wrote these five questions in its format, each with a short everyday text, a question and five options of which only one is right. The wrong options repeat mistakes we see in class, so it is worth reading the solution even when you get the answer right. Further down there are real questions from the exam to practise with.

  1. Weight, normal force and the 2nd law · Steps 3 and 5

    A video that does the rounds on social media shows someone standing on bathroom scales inside a lift, with the number on the display changing during the ride. Lucas, whose mass is 60 kg, repeated the experiment in the block of flats where he lives. He rode up from the ground floor to the 10th floor and wrote down what the scales showed on each stretch.

    Scale reading on the way up
    StretchReading (kg)
    At rest on the ground floor60.0
    Starting upwards67.2
    Rising at constant speed60.0
    Braking near the 10th floor51.0

    We know that the scales measure the normal force and display that value divided by 10. With g = 10 m/s², what is the lift’s acceleration on the last stretch?

    1. 1.5 m/s², downwards.
    2. 1.5 m/s², upwards.
    3. 8.5 m/s², downwards.
    4. 9.0 m/s², downwards.
    5. Zero, because Lucas’s mass does not change during the ride.
    Show solution
    Answer: A.
    A reading of 51.0 "kg" corresponds to a normal force of \(510\ \text{N}\), less than the weight of \(600\ \text{N}\).
    \(N - W = m\,a\)
    \(510 - 600 = 60\,a\)
    \(a = -1.5\ \text{m/s}^2\)
    The minus sign means the acceleration points downwards, as we would expect of a lift that is going up and braking.
    B gets the size right and loses the direction, C mistakes the acceleration for \(g + a\), D takes the 9 "kg" difference as if it were the acceleration, and E assumes the scales measure mass. It may come as a surprise that, while rising at constant speed, the reading goes back to 60.0.
  2. Inclined plane with friction · Steps 6 and 7

    To unload boxes from a lorry without carrying them in his arms, a delivery driver leans a 3.0 m wooden plank against the back of the lorry, 1.8 m above the ground, and lets the boxes slide down it. Let’s assume that, between cardboard and wood, the coefficients of friction are \(\mu_s = 0.5\) and \(\mu_k = 0.4\), and use g = 10 m/s².

    Does a box placed at rest at the top of the plank start to slide? If it does, with what acceleration?

    1. It slides, at 2.0 m/s².
    2. It slides, at 2.8 m/s².
    3. It slides, at 5.6 m/s².
    4. It slides, at 6.0 m/s².
    5. It does not slide, because static friction holds the box.
    Show solution
    Answer: B.
    The plank makes an angle \(\theta\) with the ground such that \(\sin\theta = \dfrac{1.8}{3.0} = 0.6\), so \(\cos\theta = 0.8\) and \(\tan\theta = 0.75\).
    Since \(\tan\theta = 0.75\) is greater than \(\mu_s = 0.5\), static friction cannot hold the box, and it slides.
    \(a = g\,(\sin\theta - \mu_k\cos\theta)\)
    \(a = 10\,(0.6 - 0.4 \cdot 0.8)\)
    \(a = 2.8\ \text{m/s}^2\)
    A subtracts the friction without the \(\cos\theta\) \((0.6 - 0.4)\), C swaps sine and cosine \((0.8 - 0.4 \cdot 0.6)\), D ignores friction, and E seems to come from the intuition that a box at rest only moves if someone pushes it. The mass of the box never entered the calculation, something many students find surprising.
  3. Mechanical energy and dissipation · Step 9

    On a U-shaped skate ramp, the half-pipe, the skater trades height for speed on every descent. Pedro, whose mass is 50 kg, starts from rest at the edge of a 3.2 m high ramp, and a friend with a speed gun measures 7.0 m/s at the lowest point. Without friction or air resistance, Pedro would reach the bottom at 8.0 m/s.

    What fraction of Pedro’s initial mechanical energy was dissipated on this descent?

    1. 0%
    2. 12.5%
    3. About 23%
    4. About 77%
    5. 87.5%
    Show solution
    Answer: C.
    If we take the lowest point as the reference level, all of Pedro’s initial energy is potential.
    \(E_p = 50 \cdot 10 \cdot 3.2 = 1600\ \text{J}\)
    \(E_k = \dfrac{50 \cdot 7^2}{2} = 1225\ \text{J}\)
    \(\dfrac{1600 - 1225}{1600} = \dfrac{375}{1600} \approx 0.23\)
    B compares the speeds \((1 - 7/8)\) and forgets that the energy depends on \(v^2\), D is the fraction left over as kinetic energy, E is the ratio \(7/8\), and A assumes mechanical energy is conserved, which only holds if we neglect friction and the air. Losing 1 m/s may seem small, and yet it takes almost a quarter of the energy.
  4. Impulse and momentum · Step 10

    Silicone cases tend to save phones from falls that would otherwise crack the screen. A 200 g phone slips out of a pocket, falls 1.25 m and stops on the floor without bouncing. Let’s assume that, without a case, the impact with the floor lasts 4 ms and that, with the case, it lasts 20 ms, and neglect air resistance (g = 10 m/s²).

    What average force does the floor exert on the phone with the case during the impact?

    1. 0.05 N
    2. 2 N
    3. 50 N
    4. 100 N
    5. 250 N
    Show solution
    Answer: C.
    The phone reaches the floor at the speed of a 1.25 m free fall.
    \(v = \sqrt{2\,g\,h} = \sqrt{25} = 5\ \text{m/s}\)
    To stop, it has to lose all its momentum, \(\Delta Q = 0.2 \cdot 5 = 1\ \text{kg·m/s}\), with or without the case.
    \(F_{\text{mean}} = \dfrac{\Delta Q}{\Delta t} = \dfrac{1}{0.020} = 50\ \text{N}\)
    The phone’s weight, 2 N, is small next to this, so we treat the force from the floor as the net force, as we did with the egg in the simulation.
    A leaves the time in milliseconds, B keeps only the weight, D assumes the phone bounces back at the same speed and E is the force without the case \((1/0.004)\). With the case, \(\Delta Q\) stays the same, and an impact five times longer divides the force by five.
  5. Collisions · Step 11

    In railway marshalling yards, wagons are coupled by running them into each other, and the automatic coupler keeps them joined from then on. A 30 t wagon moving at 2.0 m/s hits a 20 t wagon standing on the track, and the two move off together. Friction with the rails is taken to be negligible during the collision.

    Just after coupling, what is the speed of the two wagons, and what fraction of the initial kinetic energy was lost in the collision?

    1. 2.0 m/s, with no loss of kinetic energy.
    2. 1.2 m/s, with no loss of kinetic energy.
    3. 3.0 m/s, with a loss of 40%.
    4. 1.2 m/s, with a loss of 40%.
    5. 1.2 m/s, with a loss of 60%.
    Show solution
    Answer: D.
    During the collision the two wagons form an isolated system, and the total momentum is conserved.
    \(30 \cdot 2.0 = (30 + 20)\,v\)
    \(v = \dfrac{60}{50} = 1.2\ \text{m/s}\)
    \(E_{k,\text{before}} = \dfrac{30\,000 \cdot 2^2}{2} = 60\ \text{kJ}\)
    \(E_{k,\text{after}} = \dfrac{50\,000 \cdot 1.2^2}{2} = 36\ \text{kJ}\)
    \(\dfrac{60 - 36}{60} = 0.40\)
    A imagines the stationary wagon moves off with all of the other’s speed, which only happens in an elastic collision between equal masses, B assumes kinetic energy is conserved along with momentum, C swaps the masses in the calculation and E is the fraction left over. The missing 24 kJ becomes heat, sound and deformation of the coupler.

Real ENEM questions on this topic

The official papers and answer keys are on the INEP website, where each question can be found by year, day, booklet colour and number. Until 2016, the Natural Sciences paper was sat on Day 1.

  • ENEM 2011, Day 1, blue booklet, question 86. The stages of a pole vault are used to discuss which energy conversion lets the athlete rise as high as possible, if we neglect dissipative forces.
  • ENEM 2013, Day 1, blue booklet, question 76. A person walks up a ramp by pushing the ground with their feet, and the question asks for the direction of the friction force the ground exerts on them.
  • ENEM 2016, Day 1, blue booklet, question 77. On an air track, a trolley of known mass hits a stationary trolley and the two move off together, and the times from the sensors are used to find the mass of the second trolley.
  • ENEM 2017, Day 2, blue booklet, question 99. Force graphs for five seat-belt models in a head-on collision are used to choose the one with the lowest risk for the driver, and the choice depends on the contact time.
  • ENEM 2023, Day 2, blue booklet, question 114. A load tied with two ropes on the back of a lorry, with friction on the floor, has the tensions in the ropes worked out as the lorry pulls away and as it brakes.

ENEM papers and answer keys on the INEP website (in Portuguese) →

Cheat sheet

Net force (perpendicular)\(F_R = \sqrt{F_1^2 + F_2^2}\)
1st and 2nd laws\(F_R = m\,a\) (with \(F_R = 0\), \(v\) constant)
3rd law\(\vec F_{A \to B} = -\vec F_{B \to A}\)
Weight and normal force in a lift\(W = m\,g \quad N = m\,(g + a)\)
Friction\(f_s \le \mu_s N \quad f_k = \mu_k N\)
Inclined plane\(W_x = W\sin\theta \quad N = W\cos\theta\)
Starts to slip\(\tan\theta = \mu_s\)
Atwood\(a = \dfrac{(m_1 - m_2)\,g}{m_1 + m_2}\)
Work\(\tau = F\,d\cos\theta\)
Energies\(E_k = \tfrac{m v^2}{2} \quad E_p = m g h\)
Momentum and impulse\(\vec p = m\,\vec v \quad I = F\,\Delta t = \Delta p\)
Collisions (isolated system)\(p_{\text{before}} = p_{\text{after}} \quad e = \dfrac{v_{\text{sep}}}{v_{\text{app}}}\)