🚀 Momentum — The Quantity of Motion

Part 1 of 7 — Momentum and Impulse

In everyday language, we say a speeding truck has "more momentum" than a bicycle. In physics, momentum has a precise mathematical definition that makes it one of the most powerful concepts in mechanics.

Momentum connects an object's mass and velocity into a single vector quantity that is central to understanding collisions, explosions, and all interactions between objects.

Defining Momentum

The linear momentum of an object is defined as:

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

where:

• $\vec{p}$ = momentum (kg·m/s)
• $m$ = mass (kg)
• $\vec{v}$ = velocity (m/s)

Key Properties

Example

A 0.145 kg baseball moving at 40 m/s:

$p = (0.145)(40) = 5.8 \text{ kg·m/s}$

A 1200 kg car moving at 0.005 m/s:

$p = (1200)(0.005) = 6.0 \text{ kg·m/s}$

The slow-moving car has more momentum than the fast baseball!

Momentum as a Vector

Because momentum is a vector, direction matters:

• In 1D: positive momentum means motion in the positive direction; negative means the opposite direction
• In 2D: momentum has components $p_x = mv_x$ and $p_y = mv_y$

Total Momentum of a System

For a system of multiple objects, the total momentum is the vector sum:

$\vec{p}_{\text{total}} = \vec{p}_1 + \vec{p}_2 + \cdots = m_1\vec{v}_1 + m_2\vec{v}_2 + \cdots$

Example

Object A (2 kg) moves right at +3 m/s; Object B (4 kg) moves left at −1 m/s:

$p_{\text{total}} = (2)(+3) + (4)(-1) = 6 - 4 = +2 \text{ kg·m/s (to the right)}$

Momentum vs. Kinetic Energy

Students often confuse momentum and kinetic energy. They are related but distinct:

Important Relationship

$KE = \frac{p^2}{2m}$

This means doubling velocity:

• Doubles momentum
• Quadruples kinetic energy

Concept Check — Momentum Basics 🎯

Momentum Calculations 🧮

1. A 0.50 kg ball moves at 12 m/s. What is its momentum? (in kg·m/s)

2. A 2000 kg car has a momentum of 30000 kg·m/s. What is its speed? (in m/s)

3. Object A (5 kg, +6 m/s) and Object B (3 kg, −4 m/s). What is the total momentum? (in kg·m/s, include sign)

Classify Momentum Properties 🔍

Exit Quiz — Momentum ✅

💥 Impulse — Changing Momentum

Part 2 of 7 — Momentum and Impulse

We know that momentum is $\vec{p} = m\vec{v}$. But what actually changes an object's momentum? The answer is impulse — a quantity that connects force and the time over which it acts to the resulting change in momentum.

Understanding impulse is key to explaining everything from catching a ball to designing safer cars.

Defining Impulse

Impulse ($\vec{J}$) is defined as:

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

where:

• $\vec{J}$ = impulse (N·s or kg·m/s)
• $\vec{F}$ = average net force (N)
• $\Delta t$ = time interval over which the force acts (s)

Key Properties

Units Check

$\text{N·s} = \text{(kg·m/s²)·s} = \text{kg·m/s}$

The units of impulse are identical to the units of momentum — this is not a coincidence!

Impulse Equals Change in Momentum

The fundamental relationship connecting impulse and momentum:

$\vec{J} = \Delta\vec{p} = \vec{p}_f - \vec{p}_i$

$\vec{F}\Delta t = m\vec{v}_f - m\vec{v}_i$

For constant mass:

$\vec{F}\Delta t = m(\vec{v}_f - \vec{v}_i) = m\Delta\vec{v}$

Why This Works (Newton's Second Law)

Starting from $\vec{F} = m\vec{a} = m\frac{\Delta\vec{v}}{\Delta t}$:

$\vec{F}\Delta t = m\Delta\vec{v} = \Delta\vec{p}$

Example

A 0.40 kg ball moving at +30 m/s is hit by a bat and reverses to −40 m/s:

$\Delta p = m(v_f - v_i) = (0.40)(-40 - 30) = (0.40)(-70) = -28 \text{ kg·m/s}$

The impulse delivered by the bat is $-28$ kg·m/s (in the negative direction).

Direction Matters — Sign Conventions

When working in 1D, assign a positive direction. Then:

Common Mistake Alert ⚠️

When an object reverses direction, the change in velocity is NOT just the difference in speeds:

$\Delta v = v_f - v_i = (-15) - (+20) = -35 \text{ m/s}$

NOT $|20 - 15| = 5$ m/s! Always use signed velocities.

Concept Check — Impulse 🎯

Impulse Calculations 🧮

1. A force of 200 N acts for 0.05 s on a tennis ball. What is the impulse? (in N·s)

2. A 0.15 kg ball moving at +20 m/s is caught (brought to rest). What is the magnitude of the impulse? (in N·s)

3. A 60 kg person jumps and is in the air for 0.5 s. If the average force from the ground during the jump was 1800 N, what impulse did the ground deliver? (in N·s)

Impulse Properties 🔍

Exit Quiz — Impulse ✅

⚡ The Impulse-Momentum Theorem

Part 3 of 7 — Momentum and Impulse

The Impulse-Momentum Theorem is one of the most important results in mechanics. It formally connects Newton's Second Law to the concepts of impulse and momentum, giving us a powerful tool for analyzing interactions where forces act over time.

The Theorem

The Impulse-Momentum Theorem states:

$\vec{J}_{\text{net}} = \Delta\vec{p}$

$\vec{F}_{\text{net}} \Delta t = m\vec{v}_f - m\vec{v}_i$

Derivation from Newton's Second Law

$\vec{F}_{\text{net}} = m\vec{a} = m\frac{\Delta\vec{v}}{\Delta t}$

Multiply both sides by $\Delta t$:

$\vec{F}_{\text{net}} \Delta t = m\Delta\vec{v} = \Delta\vec{p}$

What It Means

The net impulse on an object equals the change in its momentum.

This is actually a restatement of Newton's Second Law — just in a form that's especially useful when:

• Forces act for known time intervals
• Objects change velocity (especially direction)
• Collisions occur with brief, intense forces

Problem-Solving Strategy

Step-by-Step Approach

1. Choose a system — identify the object whose momentum changes
2. Define positive direction — stick with it throughout
3. Identify initial and final velocities — use signed values
4. Calculate $\Delta p$ — this equals the net impulse
5. If time is known — find average force: $F_{\text{avg}} = \frac{\Delta p}{\Delta t}$
6. If force is known — find time or velocity change

Example: Finding Average Force

A 0.145 kg baseball arrives at $+40$ m/s and is hit back at $-50$ m/s. Contact time = 0.001 s.

Step 1: System = baseball

Step 2: Positive = toward pitcher

Step 3: $v_i = +40$ m/s, $v_f = -50$ m/s

Step 4: $\Delta p = (0.145)(-50 - 40) = (0.145)(-90) = -13.05$ kg·m/s

Step 5: $F_{\text{avg}} = \frac{-13.05}{0.001} = -13{,}050$ N

The bat exerts about 13,000 N on the ball — roughly the weight of a car!

Variable Forces and the Theorem

When force is not constant, the impulse is the integral (area under the force-time curve):

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

But even with variable forces, the theorem still holds:

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

In AP Physics 1, we typically use the average force approximation:

$F_{\text{avg}} \Delta t = \Delta p$

This is exact for constant forces and a useful approximation for variable forces.

Concept Check — Impulse-Momentum Theorem 🎯

Impulse-Momentum Theorem Calculations 🧮

1. A 0.060 kg tennis ball is served at 50 m/s. If the racket contact time is 0.005 s, what is the average force exerted? (in N)

2. A 75 kg skydiver in free fall at 55 m/s opens a parachute. If the average drag force is 1200 N and the skydiver weighs 735 N, how long until the skydiver slows to 5 m/s? (in seconds, to 3 significant figures)

3. A 2000 kg car traveling at 25 m/s brakes to a stop in 5 s. What is the average braking force? (magnitude, in N)

Impulse-Momentum Theorem Relationships 🔍

Exit Quiz — Impulse-Momentum Theorem ✅

📊 Force-Time Graphs and Impulse

Part 4 of 7 — Momentum and Impulse

In real collisions, the force is rarely constant — it rises quickly, peaks, and then drops back to zero. Force-time graphs let us visualize these interactions and calculate impulse graphically.

The key insight: the area under a force-time graph equals the impulse.

Area Under the Curve = Impulse

Since impulse is $J = F\Delta t$ for constant force, and more generally:

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

This integral is the area under the $F$-vs-$t$ graph.

Common Graph Shapes

Reading F-t Graphs

For any $F$-vs-$t$ graph:

1. Area above the time axis = positive impulse (force in + direction)
2. Area below the time axis = negative impulse (force in − direction)
3. Total impulse = sum of all areas (with signs)

Example: Rectangular Force Pulse

A constant force of $F = 500$ N acts on a ball for $\Delta t = 0.01$ s:

$J = F \times \Delta t = (500)(0.01) = 5 \text{ N·s}$

Example: Triangular Force Pulse

A force increases linearly from 0 to $F_{\text{max}} = 1000$ N over 0.01 s:

$J = \frac{1}{2} F_{\text{max}} \times \Delta t = \frac{1}{2}(1000)(0.01) = 5 \text{ N·s}$

Key Insight

Both pulses deliver the same impulse (5 N·s), but the peak forces are very different! The average force for the triangle is $F_{\text{avg}} = J/\Delta t = 5/0.01 = 500$ N — half the peak value.

Example: Multi-Part Graph

A force acts as follows:

• $t = 0$ to $t = 2$ s: $F = +10$ N (rectangle, area = $+20$ N·s)
• $t = 2$ to $t = 5$ s: $F = -5$ N (rectangle, area = $-15$ N·s)

Total impulse = $+20 + (-15) = +5$ N·s

Average Force from a Graph

The average force is the constant force that would deliver the same impulse in the same time:

$F_{\text{avg}} = \frac{J}{\Delta t} = \frac{\text{Area under curve}}{\text{Total time}}$

Real Collision Force Profiles

In a real collision (like a ball hitting a wall):

1. Force starts at zero
2. Rises rapidly to a peak value
3. Falls back to zero

The peak force is typically much larger than the average force. For a symmetric collision pulse:

$F_{\text{avg}} \approx \frac{F_{\text{peak}}}{2}$

This is why peak forces in collisions can be enormous even though the average force seems manageable.

Concept Check — F-t Graphs 🎯

F-t Graph Calculations 🧮

1. A rectangular force pulse of 400 N lasts 0.05 s. What is the impulse? (in N·s)

2. A triangular force pulse has peak 800 N and duration 0.04 s. What is the average force during the pulse? (in N)

3. A 2 kg object initially at rest receives a triangular impulse with peak 100 N lasting 0.5 s. What is the final speed? (in m/s)

F-t Graph Analysis 🔍

Exit Quiz — Force-Time Graphs ✅

🛡️ Applications: Airbags, Crumple Zones & Safety

Part 5 of 7 — Momentum and Impulse

The impulse-momentum theorem has life-saving applications! When a car crashes, the driver's momentum must change from $mv$ to zero. The impulse ($\Delta p$) is fixed by physics. But we can control how that impulse is delivered by manipulating force and time.

$F_{\text{avg}} = \frac{\Delta p}{\Delta t}$

Since $\Delta p$ is fixed: increasing $\Delta t$ decreases $F_{\text{avg}}$ — and that saves lives.

The Key Principle: Increase Time to Reduce Force

For a given change in momentum, force and time are inversely related:

$F_{\text{avg}} \times \Delta t = \Delta p = \text{constant}$

Example: Car Crash

A 70 kg driver at 30 m/s (67 mph) must stop. $\Delta p = -2100$ kg·m/s.

Real-World Applications

Catching a Ball

When you catch a fast baseball, you instinctively pull your hand back — this increases the time of contact, reducing the force on your hand.

• Stiff catch: $\Delta t \approx 0.01$ s → $F$ is very large (ouch!)
• "Soft" catch: $\Delta t \approx 0.1$ s → $F$ is 10× smaller

Egg Drop Competitions

The goal: land an egg without breaking it. Strategy: increase the stopping time using padding, parachutes, or collapsible structures.

Boxing

A boxer "rolls with the punch" — moving the head backward upon impact extends the contact time and reduces the peak force. Standing rigid allows the full force to transfer in a shorter time.

Bungee Jumping

The elastic cord stretches over several seconds, converting the jumper's momentum to zero over a long time rather than instantly (which would be like hitting the ground).

Concept Check — Safety Applications 🎯

Safety Physics Calculations 🧮

1. A 60 kg driver traveling at 20 m/s crashes. The airbag increases the stopping time to 0.2 s. What is the average force on the driver? (in N)

2. Without the airbag, the driver hits the steering wheel and stops in 0.01 s. What is the average force now? (in N)

3. What is the ratio of force without airbag to force with airbag? (give as a whole number)

Safety Design Principles 🔍

Exit Quiz — Safety Applications ✅

🔧 Problem-Solving Workshop

Part 6 of 7 — Momentum and Impulse

Let's put together everything we've learned about momentum and impulse. In this workshop, we'll work through AP-style problems that integrate multiple concepts: momentum calculation, impulse, the impulse-momentum theorem, F-t graphs, and safety applications.

Problem-Solving Framework

For Impulse-Momentum Problems

1. Identify the system (which object's momentum changes?)
2. Define positive direction (and stick with it)
3. List knowns: $m$, $v_i$, $v_f$, $F$, $\Delta t$
4. Write the equation: $F_{\text{avg}} \Delta t = m(v_f - v_i)$
5. Solve for the unknown
6. Check: Does the sign/direction make sense?

Common Pitfalls ⚠️

• Forgetting that velocity is signed (especially when objects reverse direction)
• Confusing momentum (vector) with kinetic energy (scalar)
• Using the wrong $\Delta t$ (contact time vs. total time)
• Forgetting to include all forces when finding $F_{\text{net}}$

Problem 1: Baseball Bat ⚾

A 0.145 kg baseball traveling at $+38$ m/s toward a batter is hit and travels at $-52$ m/s away. The bat-ball contact time is 0.0012 s. What is the average force of the bat on the ball?

Problem 2: Stopping a Car 🚗

A 1500 kg car is traveling at 25 m/s.

1. What is the car's momentum? (in kg·m/s)

2. If the brakes apply a constant force of 7500 N, how long does it take to stop? (in seconds)

3. If the car then accelerates from rest with a 4500 N force for 10 s, what is the final speed? (in m/s)

Problem 3: F-t Graph Analysis 📊

A 4 kg block at rest has a force applied to it. The force is +20 N from $t = 0$ to $t = 3$ s, then −10 N from $t = 3$ s to $t = 9$ s.

Problem 4: Bouncing Ball 🏀

A 0.60 kg ball is dropped from a height and hits the floor at 8.0 m/s (downward). It bounces back up at 6.0 m/s (upward). The contact time with the floor is 0.015 s. Take upward as positive.

1. What is the impulse from the floor on the ball? (in N·s)

2. What is the average force from the floor on the ball? (in N, to nearest whole number)

3. How does this force compare to the ball's weight ($mg$)? Give the ratio $F_{\text{avg}}/mg$ to 3 significant figures. (use $g = 9.8$ m/s²)

Problem-Solving Strategies 🔍

Exit Quiz — Problem Solving ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Momentum and Impulse

Let's bring together all the key ideas from this topic and practice AP-style questions. This review covers: definition of momentum, impulse, the impulse-momentum theorem, F-t graphs, and real-world applications.

Key Equations Summary

Key Concepts

1. Momentum is a vector — direction matters; use signed velocities
2. Impulse = area under F-t curve — works for any force profile
3. Increasing $\Delta t$ decreases $F$ — basis of all safety devices
4. Bouncing delivers more impulse than stopping — because $|\Delta v|$ is larger
5. The impulse-momentum theorem is Newton's 2nd Law — just rearranged

AP Review — Set 1 🎯

AP Review — Set 2 📝

AP Calculation Practice 🧮

1. A 0.40 kg ball hits a bat at +30 m/s and leaves at −45 m/s. The bat contact time is 0.002 s. What is the average force magnitude? (in N)

2. A 50 kg skater at rest pushes off a wall with a force of 200 N for 0.8 s. What is the skater's final speed? (in m/s)

3. A car's crumple zone extends the collision time from 0.05 s to 0.50 s. By what factor does the average force decrease? (whole number)

Round all answers to 3 significant figures.

Comprehensive Concept Review 🔍

Final Exit Quiz — Momentum & Impulse ✅