Visighta

Momentum, Impulse, and Conservation Laws

8.1 Linear Momentum and Impulse

The Quantity That Survives the Chaos

A billiard ball rolls toward a cluster and sends balls flying in every direction. The contact lasts about a millisecond. During that millisecond, the forces between the balls are enormous, constantly changing, and impossible to measure in any practical way. You cannot track what happens during the collision with a free-body diagram and Newton's second law --- the forces are too complex and too brief.

And yet, something remarkable is true. A quantity that you can compute before the collision --- without knowing anything about the contact forces --- turns out to be exactly the same after the collision. Balls scatter in every direction, speeds change, trajectories change, but this one quantity is perfectly predictable.

What is it? It is not speed. The total speed of all the balls is different before and after. It is not kinetic energy --- as you will see in Section 8.4, collisions can lose kinetic energy. It is something else. Something that combines how much mass is moving with how fast it is going and in what direction.

That quantity is momentum, and it is the subject of this section.

Before you read on: Consider two scenarios.

Scenario 1: A tennis ball (mass 0.06 kg) and a bowling ball (mass 7 kg) are both rolling at the same speed, say 5 m/s. Which one has more momentum?

Scenario 2: A tennis ball is moving at 100 m/s and a bowling ball is moving at 1 m/s. Now which one has more momentum?

Commit to your answers before continuing. Does your intuition shift between the two scenarios?

[Interactive: Predict-Then-Reveal. Student selects which object has more momentum in each scenario and types a brief justification. After submitting, the answers are revealed: In Scenario 1, the bowling ball has far more momentum (35 kg m/s vs. 0.3 kg m/s) --- mass dominates when speeds are equal. In Scenario 2, the bowling ball still has more momentum (7 kg m/s vs. 6 kg m/s), but the gap is surprisingly narrow. A fast, light object can rival a slow, heavy one. Momentum depends on both mass and velocity.]

The Guiding Question

Why is mass times velocity a useful quantity for describing interactions?

You already have powerful tools for analyzing motion. Newton's second law (Chapter 5) tells you what forces do to acceleration. The work-energy theorem (Chapter 7) tells you how forces transfer energy through displacement. But both of those tools require you to know the forces --- their magnitudes, directions, and how long they act.

In many of the most dramatic physical events --- car crashes, bat hitting ball, two hockey players colliding --- the forces are enormous, act for tiny fractions of a second, and are practically impossible to measure. You need a tool that lets you say something useful about these events without knowing the force details.

That tool is momentum. And the reason it works is deeply connected to Newton's second law, as you will see.

Exploration: Collisions and a Mysterious Constant

[Interactive: Collision Momentum Explorer. Two objects sit on a frictionless track. Students can adjust the mass of each object (sliders from 0.5 kg to 10 kg) and their initial velocities (sliders from $-10$ m/s to $+10$ m/s, where positive is to the right). A "Collide" button runs the collision animation. Before the collision, the system displays the total momentum $\sum m_i v_i$ with a bold number and colored bar (separate bars for each object's momentum, stacked to show the total). After the collision, it displays the total momentum again.

Key feature: the total momentum is always the same before and after.

Guided prompts:

  • "Set both masses equal and give the left object a velocity of 5 m/s while the right object is at rest. Run the collision. What is the total momentum before? After?"
  • "Now make the left object ten times heavier. Same initial velocities. Run it. Is the total momentum still the same before and after?"
  • "Try giving both objects velocities in opposite directions. Is the total momentum conserved? (Watch the signs carefully.)"
  • "Try every combination of masses and speeds you can think of. Can you find any setup where the total momentum changes?"]

Spend time with this interactive before reading on. Try at least five different combinations. You should find that no matter what you do --- no matter how different the masses, no matter how the objects bounce or stick --- the total momentum before the collision equals the total momentum after.

This is not a coincidence. It is a law of nature, and its formal statement will come in Section 8.3. For now, the observation raises a question: what is this quantity that refuses to change?

The Concept: Momentum

The momentum of an object is defined as:

$$\vec{p} = m\vec{v}$$

where $m$ is the object's mass and $\vec{v}$ is its velocity.

Let's unpack this.

Momentum is a vector. It has both magnitude and direction. A 2 kg ball moving to the right at 3 m/s has momentum $\vec{p} = (6, 0)$ kg m/s. A 2 kg ball moving to the left at 3 m/s has momentum $\vec{p} = (-6, 0)$ kg m/s. Same mass, same speed, but different momenta --- because direction matters.

Momentum scales with mass. At the same velocity, a more massive object has more momentum. This matches intuition: a freight train at 1 m/s is harder to stop than a tennis ball at 1 m/s.

Momentum scales with velocity. At the same mass, a faster object has more momentum. Also intuitive: catching a baseball pitched at 40 m/s demands more of you than catching one tossed at 2 m/s.

Units. Momentum has units of $\text{kg} \cdot \text{m/s}$. There is no special named unit for momentum in SI, unlike energy (joule) or force (newton). You just write kg m/s.

Pause and think: An object's kinetic energy is $K = \frac{1}{2}mv^2$. Its momentum is $p = mv$. Both depend on mass and speed. But one is a scalar and the other is a vector. Can two objects have the same kinetic energy but different momenta? Can two objects have the same momentum but different kinetic energies?

Check your answer Yes to both. **Same kinetic energy, different momenta:** A 1 kg object moving right at 4 m/s has $K = 8$ J and $\vec{p} = 4$ kg m/s to the right. A 1 kg object moving *left* at 4 m/s has $K = 8$ J and $\vec{p} = 4$ kg m/s to the *left*. Same energy, opposite momenta. Kinetic energy is direction-blind; momentum is direction-aware. **Same momentum magnitude, different kinetic energies:** A 2 kg object at 3 m/s has $p = 6$ kg m/s and $K = 9$ J. A 6 kg object at 1 m/s has $p = 6$ kg m/s and $K = 3$ J. Same momentum magnitude, very different energies. This is because $K = p^2/(2m)$ --- for the same momentum, a heavier object has less kinetic energy.

The Concept: Impulse

How do you change an object's momentum? You apply a force.

Start with Newton's second law:

$$\vec{F} = m\vec{a} = m\frac{d\vec{v}}{dt}$$

For an object with constant mass, $m\,d\vec{v}/dt = d(m\vec{v})/dt = d\vec{p}/dt$. So Newton's second law can be written:

$$\vec{F} = \frac{d\vec{p}}{dt}$$

This is actually the form Newton himself used. It says: force is the rate of change of momentum. A force acting on an object changes the object's momentum, and the rate of that change equals the force.

Now integrate both sides over a time interval from $t_i$ to $t_f$:

$$\int_{t_i}^{t_f} \vec{F}\,dt = \int_{t_i}^{t_f} \frac{d\vec{p}}{dt}\,dt = \vec{p}_f - \vec{p}_i = \Delta\vec{p}$$

The left side of this equation has a name. The impulse delivered by a force is:

$$\vec{J} = \int_{t_i}^{t_f} \vec{F}\,dt$$

Impulse is the total "push" accumulated over time. A large force acting briefly can deliver the same impulse as a small force acting for a long time. Impulse is a vector --- it has the direction of the net force.

And the equation that connects impulse to momentum change is the impulse-momentum theorem:

$$\boxed{\vec{J} = \Delta\vec{p}}$$

The impulse delivered to an object equals the change in its momentum. This is not a new law --- it is Newton's second law, integrated over time. But the integrated form is enormously useful because it lets you connect the "before" and "after" of an interaction without tracking the force at every instant in between.

When the force is constant

If the force $\vec{F}$ is constant during the interval $\Delta t = t_f - t_i$, the integral simplifies:

$$\vec{J} = \vec{F}\,\Delta t$$

This is the simplest version: impulse equals force times time. It is the starting point for many problems, though real collisions rarely involve constant forces.

Connection: Impulse and Work --- Two Faces of Force

You have seen this mathematical structure before.

In Chapter 7, you met work: the integral of force over displacement.

$$W = \int \vec{F} \cdot d\vec{r}$$

Work accumulates force along a spatial path. It changes an object's kinetic energy.

Now you meet impulse: the integral of force over time.

$$\vec{J} = \int \vec{F}\,dt$$

Impulse accumulates force along a temporal path. It changes an object's momentum.

Work (Chapter 7) Impulse (Chapter 8)
Accumulation variable Displacement $d\vec{r}$ Time $dt$
Result Scalar (energy transfer) Vector (momentum change)
What changes Kinetic energy $K$ Momentum $\vec{p}$
Key equation $W = \Delta K$ $\vec{J} = \Delta\vec{p}$
Simple case $W = Fd\cos\theta$ $\vec{J} = \vec{F}\,\Delta t$

The calculus structure is identical. The same force, integrated over different variables, produces different physical quantities. This is a pattern worth noticing: force is a single physical entity, but it has multiple effects that are captured by different integrals.

Returning to the Prediction

Let's revisit the opening scenarios with the formal definition.

Scenario 1: Both objects at 5 m/s.

  • Tennis ball: $p = 0.06 \times 5 = 0.30$ kg m/s
  • Bowling ball: $p = 7 \times 5 = 35$ kg m/s

The bowling ball has over 100 times more momentum. When speeds are equal, mass dominates completely.

Scenario 2: Tennis ball at 100 m/s, bowling ball at 1 m/s.

  • Tennis ball: $p = 0.06 \times 100 = 6.0$ kg m/s
  • Bowling ball: $p = 7 \times 1 = 7.0$ kg m/s

Now they are nearly equal. A very fast, light object can have momentum comparable to a slow, heavy one. The bowling ball still wins, but barely.

This illustrates why momentum is more nuanced than "heavier means harder to stop." Both mass and velocity matter, and they trade off against each other.

Practice

Layer 1: Concrete

Problem 1. A 0.145 kg baseball is pitched at 40 m/s (about 90 mph). A bat hits the ball, and it leaves at 50 m/s in the opposite direction. Assume the positive direction is toward the pitcher.

(a) What is the baseball's momentum before the hit?

(b) What is the baseball's momentum after the hit?

(c) What is the impulse delivered by the bat?

(d) If the bat is in contact with the ball for 1.0 ms (0.001 s), what is the average force on the ball?

Check your answer **(a)** Before the hit, the ball moves toward the batter, which is the negative direction: $$p_i = 0.145 \times (-40) = -5.80 \text{ kg m/s}$$ **(b)** After the hit, the ball moves toward the pitcher (positive direction): $$p_f = 0.145 \times (+50) = +7.25 \text{ kg m/s}$$ **(c)** The impulse is the change in momentum: $$J = p_f - p_i = 7.25 - (-5.80) = 13.05 \text{ kg m/s}$$ Notice that the impulse is *not* just $0.145 \times 50$ or $0.145 \times 40$. The ball reversed direction, so the full change includes both the stopping and the re-accelerating. **(d)** Using the constant-force approximation: $$F_{\text{avg}} = \frac{J}{\Delta t} = \frac{13.05}{0.001} = 13{,}050 \text{ N}$$ That is roughly 1.3 tons of force --- enormous but brief. This is why tracking the force during a collision is impractical. The impulse-momentum theorem lets you bypass the force details entirely.

Problem 2. A 1200 kg car is traveling east at 25 m/s. Find its momentum. If a braking force brings the car to rest in 8 s, what impulse was delivered, and what was the average braking force?

[Interactive: Momentum Calculator. Students input mass and velocity. The system displays a momentum vector arrow scaled to the value. Then students input the final velocity (zero) and the time interval. The system shows the impulse as a vector in the opposite direction and computes the average force.]

Check your answer Momentum: $\vec{p} = 1200 \times 25 = 30{,}000$ kg m/s, directed east. Impulse to stop: $\vec{J} = \Delta\vec{p} = 0 - 30{,}000 = -30{,}000$ kg m/s (directed west, opposing motion). Average braking force: $$F_{\text{avg}} = \frac{J}{\Delta t} = \frac{-30{,}000}{8} = -3{,}750 \text{ N}$$ The negative sign confirms the force acts westward (opposing the eastward motion), as expected for braking.

Layer 2: Pattern

Problem 3. Three objects need to be brought to rest. Rank them by the impulse required, from greatest to least.

  • Object A: mass 2 kg, speed 10 m/s
  • Object B: mass 5 kg, speed 3 m/s
  • Object C: mass 0.5 kg, speed 40 m/s

Then rank them by the average force required if each is stopped in the same time interval $\Delta t = 0.5$ s.

Check your answer The impulse required to stop each object equals its initial momentum (in magnitude): - **A:** $J = mv = 2 \times 10 = 20$ kg m/s - **B:** $J = mv = 5 \times 3 = 15$ kg m/s - **C:** $J = mv = 0.5 \times 40 = 20$ kg m/s Ranking by impulse: **A = C > B**. Objects A and C require the same impulse despite having very different masses and speeds. This is because momentum depends on the *product* $mv$, and $2 \times 10 = 0.5 \times 40$. Since each is stopped in the same time interval, the average force ranking is the same: - **A:** $F = 20/0.5 = 40$ N - **B:** $F = 15/0.5 = 30$ N - **C:** $F = 20/0.5 = 40$ N Ranking by force: **A = C > B**.

Layer 3: Structure

Problem 4. Momentum is a vector. Kinetic energy is a scalar. Both depend on mass and velocity. Why the difference?

Check your answer The difference traces back to how each quantity is defined and what physical process it describes. **Momentum** $\vec{p} = m\vec{v}$ is the product of mass and velocity. Velocity is a vector, and multiplying a vector by a scalar (mass) gives a vector. Momentum inherits its vector nature directly from velocity. Direction matters because momentum describes the "directional content" of motion --- how much motion is happening, and which way. **Kinetic energy** $K = \frac{1}{2}mv^2$ involves the *square* of the speed, $v^2 = \vec{v} \cdot \vec{v}$. The dot product of a vector with itself is a scalar --- it loses all directional information. Kinetic energy does not care which way the object moves, only how fast. The physical reason for the difference is connected to how each quantity changes: - Momentum changes by impulse: $\vec{J} = \int \vec{F}\,dt$. The direction of the force determines the direction of the momentum change. Direction is essential. - Kinetic energy changes by work: $W = \int \vec{F} \cdot d\vec{r}$. The dot product projects force onto displacement, collapsing the vector information into a scalar. The result tells you "how much energy," not "which direction." A concrete illustration: two objects of equal mass moving in opposite directions at the same speed have the same kinetic energy but opposite momenta. If they collide and stick, the total kinetic energy drops (it becomes zero), but the total momentum was already zero and stays zero. The scalar quantity (energy) changed; the vector quantity (momentum) was conserved. The two quantities track different aspects of the motion.

Layer 4: Debug

Problem 5. Two objects move along a line. Object A (3 kg) moves to the right at 4 m/s. Object B (2 kg) moves to the left at 6 m/s. A student computes the total momentum as: $p_{\text{total}} = 3 \times 4 + 2 \times 6 = 12 + 12 = 24$ kg m/s. What is wrong?

Check your answer The student added the *magnitudes* of the momenta instead of treating momentum as a vector. Momentum has direction. If we take rightward as positive, then: - Object A: $p_A = 3 \times (+4) = +12$ kg m/s - Object B: $p_B = 2 \times (-6) = -12$ kg m/s The total momentum is: $$p_{\text{total}} = +12 + (-12) = 0 \text{ kg m/s}$$ The total momentum is *zero*, not 24 kg m/s. The two momenta point in opposite directions and happen to cancel exactly. This is a critical error. Adding magnitudes is equivalent to ignoring direction --- it treats momentum as a scalar. But momentum is a vector, and opposite momenta subtract, not add. In a collision between these two objects, the total momentum of zero would be conserved: whatever happens after the collision, the total momentum would remain zero. If the student used 24 kg m/s, every subsequent calculation would be wrong. **The rule:** when combining momenta, always assign signs (or use vector components) to account for direction. Never add magnitudes blindly.

Spaced Retrieval

Before you move on, test your recall of earlier material.

Recall prompt 1: What is Newton's third law, and what does it say about the forces that two interacting objects exert on each other? (Section 5.3)

Recall prompt 2: When computing work, what role does the angle between force and displacement play? (Section 7.1)

Recall prompt 3: What is the difference between a definite integral and the antiderivative? What does a definite integral compute geometrically? (Section 2.3)

Reflection

Think back over what you explored and read in this section.

What makes momentum a more useful quantity than just velocity alone?

Velocity tells you how fast something is moving and in what direction. But it says nothing about how hard it is to change that motion. A mosquito at 10 m/s and a truck at 10 m/s have the same velocity, but they are not equivalent in any meaningful physical sense. The truck is vastly harder to stop, to redirect, to speed up. Momentum captures this by folding mass into the picture: $\vec{p} = m\vec{v}$ encodes both how much stuff is moving and how fast it is going.

This is why momentum is what survives collisions, not velocity. When two objects interact, the forces between them are equal and opposite (Newton's third law). Those forces change each object's momentum by equal and opposite amounts. The individual momenta change, but the total does not. Velocity alone does not have this conservation property. Momentum does --- and that is what makes it the right quantity for analyzing interactions.

Looking Ahead

You now know what momentum is, what impulse is, and how they are connected through the impulse-momentum theorem. The key equation $\vec{J} = \Delta\vec{p}$ tells you that the impulse delivered to an object --- the accumulated force over time --- equals the change in its momentum.

But that raises a question. Two very different forces can deliver the same impulse: a gentle push over a long time, or a violent blow in a split second. The impulse is the same, but the experience is completely different. One is comfortable; the other shatters bones.

In the next section, you will explore how the shape of the force-time profile determines what happens during an interaction. The area under the $F(t)$ curve is the impulse --- that fixes the momentum change. But the height of the curve is the peak force --- and that is what breaks eggs, deploys airbags, and decides whether a landing is survivable. Same impulse, different consequences. Section 8.2 will show you why.