Linear Momentum

Task 1 - Impulse from a force-time graph

Hook

A baseball bat is in contact with the ball for about one thousandth of a second. During that time the force rises from nothing to a peak and falls away again. With a force that changes so quickly, how can you work out what it does to the ball?

Intuition

Split the contact time into many tiny slices. In each slice the force is nearly constant, so the impulse in that slice is force times a tiny time, a thin rectangle of area. Adding all the slices gives the total area under the curve. Whatever the shape, the area is the impulse, and the impulse is the change in momentum.

Reshape the force curve below. A tall narrow pulse and a low wide pulse can have exactly the same area and give the ball exactly the same change in velocity.

Formalise

The integral sign means "add up the area under". The average force is the impulse spread evenly over the contact time.

Worked example

A 0.145 kg baseball arrives at 40 m/s and leaves at 45 m/s in the opposite direction. Contact lasts 1 ms.

1. Change in momentum. Taking the outgoing direction as positive, $\Delta p = 0.145 \times 45 - 0.145 \times (-40) = 0.145 \times 85 = 12.3\text{ N s}$.
2. Average force. $F_{avg} = 12.3 / 0.001 = 12\,300\text{ N}$.
3. Peak force for a triangular pulse. A triangle's area is half its base times its height, so the same area needs a peak of twice the average, about 24 600 N.

Your turn

Task 2 - Centre of mass

Hook

A firework rocket explodes at the top of its flight, sending fragments in every direction. Is there anything about the fragments as a group that still follows the rocket's original curved path?

Intuition

The centre of mass is the mass-weighted average position of a system. For two equal masses it is halfway between them. If one is heavier, it is closer to the heavier one. Its velocity is the total momentum divided by the total mass. Because internal forces cannot change total momentum, they cannot change the centre of mass's velocity either. The fragments of the firework fly apart, but their centre of mass carries on along the parabola the rocket was following, until the first fragment hits the ground.

Formalise

Worked example

1. Two masses. 2 kg at $x = 0$ and 3 kg at $x = 5\text{ m}$. $x_{cm} = (0 + 15) / 5 = 3\text{ m}$, closer to the heavier mass.
2. Skaters pushing apart. 40 kg and 80 kg skaters start at rest. Total momentum is zero, so $v_{cm} = 0$. After pushing apart they move in opposite directions, but the point between them, two thirds of the way toward the heavier skater, stays exactly where it was.

Your turn

Task 3 - Elastic collisions in one dimension

Hook

In a Newton's cradle, one steel ball swings in, stops dead, and the ball at the far end swings out at the same speed. Why does the incoming ball stop, rather than bouncing back or carrying on?

Intuition

An elastic collision conserves both momentum and kinetic energy. Writing out both conservation laws for two objects gives two equations in the two unknown final velocities. Solving them leads to a useful fact. In an elastic collision the relative velocity reverses. The speed at which the objects approach equals the speed at which they separate. Combining that with conservation of momentum gives the final velocities directly.

Formalise

From $u_1 - u_2 = -(v_1 - v_2)$ and $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$.

Worked example

A 1 kg cart at 6 m/s hits a stationary 2 kg cart elastically.

1. First cart. $v_1 = \dfrac{(1 - 2) \times 6 + 0}{3} = -2\text{ m/s}$. It bounces back.
2. Second cart. $v_2 = \dfrac{0 + 2 \times 1 \times 6}{3} = 4\text{ m/s}$.
3. Check momentum. Before 6, after $-2 + 8 = 6$. Check kinetic energy. Before 18 J, after $2 + 16 = 18\text{ J}$.

Your turn

Task 4 - Coefficient of restitution

Hook

Drop a tennis ball and a lump of clay from the same height. The tennis ball bounces back up to about half its height. The clay does not bounce at all. How can a game describe both with one number?

Intuition

The coefficient of restitution $e$ is the ratio of separation speed to approach speed. For a perfectly elastic collision $e = 1$. For a perfectly inelastic, sticky collision $e = 0$. Real materials sit in between. For a ball bouncing on a fixed floor, the ball leaves at $e$ times the speed it arrived at. Since height depends on speed squared, each bounce reaches $e^2$ of the previous height.

Change $e$ below and drop the ball.

Formalise

With $u$ for velocities before and $v$ for after, and for a ball on a fixed floor.

Worked example

A ball dropped from 2 m bounces to 1.28 m.

1. Height ratio. $1.28 / 2 = 0.64 = e^2$.
2. Coefficient. $e = \sqrt{0.64} = 0.8$.
3. Next bounce. $1.28 \times 0.64 = 0.82\text{ m}$.

In a game engine, $e$ is usually a property set on each material, often called bounciness or restitution. When two materials meet, the engine combines their values, commonly by taking the larger, the smaller or the average.

Your turn

Task 5 - Collisions in two dimensions

Hook

In pool, the cue ball strikes another ball off centre and the two balls fly apart at an angle. How do you predict both directions?

Intuition

At the instant of contact, the only force between two smooth balls acts along the line joining their centres, the contact normal. So split each velocity into a part along the normal and a part along the tangent. The tangent parts are untouched. The normal parts behave exactly like a one-dimensional collision. Recombine the parts and you have the final velocities.

Game engines go one step further and fold all of that into a single impulse $j$ along the normal, calculated from the closing speed, the restitution and the two masses. Move the offset slider below to go from a head-on hit to a glancing blow. With equal masses and $e = 1$, notice the angle between the two outgoing velocities.

Formalise

With $\hat{n}$ the unit normal pointing from A to B. If the closing speed $(\vec{v}_B - \vec{v}_A)\cdot\hat{n}$ is positive the bodies are already separating and no impulse is applied.

Worked example

Body A, 2 kg at 3 m/s, meets body B, 1 kg at -1 m/s, head on along x, so $\hat{n} = (1, 0)$. Take $e = 0.5$.

1. Closing speed. $(-1 - 3) \times 1 = -4\text{ m/s}$. Negative, so they are approaching.
2. Impulse. $j = -(1.5)(-4) / (0.5 + 1) = 6 / 1.5 = 4\text{ N s}$.
3. New velocities. $v_A' = 3 - 4/2 = 1\text{ m/s}$, $v_B' = -1 + 4/1 = 3\text{ m/s}$.
4. Checks. Momentum before $6 - 1 = 5$, after $2 + 3 = 5$. Separation speed $3 - 1 = 2 = 0.5 \times 4$.

Your turn

Two 1 kg discs, A moving at (3, 0) m/s and B at rest, touch with contact normal $\hat{n} = (0.8, 0.6)$. Take $e = 1$.

Task 6 - Impulse-based collision response in code

Collision response in a physics engine happens in two stages. Collision detection finds which bodies are touching and the contact normal between them. Collision response then changes their velocities. This task builds the response stage for circular bodies using the impulse formula from Task 5.

Create a new Console App (C++) project named CollisionResponse and replace its source file with the skeleton below, saved as CollisionResponse.cpp. Every scenario in main places two discs of radius 0.5 m exactly touching, so detection is already done.

1. Calculate the contact normal n, the vector from a.position to b.position divided by its length.
2. Calculate the closing speed as the dot product of b.velocity - a.velocity with n.
3. If the closing speed is positive, return straight away.
4. Calculate the impulse magnitude j.
5. Subtract n * (j / a.mass) from a.velocity and add n * (j / b.mass) to b.velocity.

Success condition: in every scenario the momentum before matches the momentum after. Equal masses head on swap velocities. The glancing blow gives (1.08, -1.44) and (1.92, 1.44), matching your answers to Task 5. The worked example gives 1 and 3 m/s. The already separating pair is left unchanged.

#include <cmath>
#include <iomanip>
#include <iostream>
#include <string>

struct Vec2
{
    double x;
    double y;
};

Vec2 operator+(Vec2 a, Vec2 b) { return { a.x + b.x, a.y + b.y }; }
Vec2 operator-(Vec2 a, Vec2 b) { return { a.x - b.x, a.y - b.y }; }
Vec2 operator*(Vec2 a, double s) { return { a.x * s, a.y * s }; }
double dot(Vec2 a, Vec2 b) { return a.x * b.x + a.y * b.y; }
double length(Vec2 a) { return std::sqrt(dot(a, a)); }

struct Body
{
    double mass;
    Vec2 position;
    Vec2 velocity;
};

Vec2 momentum(const Body& a, const Body& b)
{
    return a.velocity * a.mass + b.velocity * b.mass;
}

double kineticEnergy(const Body& a, const Body& b)
{
    return 0.5 * a.mass * dot(a.velocity, a.velocity)
         + 0.5 * b.mass * dot(b.velocity, b.velocity);
}

void resolveCollision(Body& a, Body& b, double e)
{
    // TODO: the contact normal n is the unit vector from a's centre to b's

    // TODO: the closing speed is the relative velocity (b minus a) along n

    // TODO: if that value is positive the bodies are already moving apart,
    // so return without changing anything

    // TODO: impulse magnitude j = -(1 + e) * closing / (1 / ma + 1 / mb)

    // TODO: push a back along n by j / ma and b forward along n by j / mb
}

void runScenario(const std::string& name, Body a, Body b, double e)
{
    Vec2 pBefore = momentum(a, b);
    double keBefore = kineticEnergy(a, b);

    resolveCollision(a, b, e);

    Vec2 pAfter = momentum(a, b);
    std::cout << name << "  (e = " << e << ")\n";
    std::cout << "  A after  (" << a.velocity.x << ", " << a.velocity.y << ")\n";
    std::cout << "  B after  (" << b.velocity.x << ", " << b.velocity.y << ")\n";
    std::cout << "  momentum before (" << pBefore.x << ", " << pBefore.y << ")"
              << "  after (" << pAfter.x << ", " << pAfter.y << ")\n";
    std::cout << "  kinetic energy before " << keBefore << " J"
              << "  after " << kineticEnergy(a, b) << " J\n\n";
}

int main()
{
    std::cout << std::fixed << std::setprecision(3);

    // Each pair of discs has radius 0.5 m and is placed exactly touching
    runScenario("Head-on, equal masses",
        { 1.0, { 0.0, 0.0 }, { 3.0, 0.0 } }, { 1.0, { 1.0, 0.0 }, { 0.0, 0.0 } }, 1.0);
    runScenario("Head-on, heavy hits light",
        { 4.0, { 0.0, 0.0 }, { 3.0, 0.0 } }, { 1.0, { 1.0, 0.0 }, { 0.0, 0.0 } }, 1.0);
    runScenario("Glancing blow, equal masses",
        { 1.0, { 0.0, 0.0 }, { 3.0, 0.0 } }, { 1.0, { 0.8, 0.6 }, { 0.0, 0.0 } }, 1.0);
    runScenario("Worked example from Task 5",
        { 2.0, { 0.0, 0.0 }, { 3.0, 0.0 } }, { 1.0, { 1.0, 0.0 }, { -1.0, 0.0 } }, 0.5);
    runScenario("Already separating",
        { 1.0, { 0.0, 0.0 }, { -1.0, 0.0 } }, { 1.0, { 1.0, 0.0 }, { 1.0, 0.0 } }, 1.0);
    return 0;
}