where:

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

Key Properties

1. Vector quantity - has magnitude and direction (same direction as velocity)
2. Depends on both mass and velocity - heavy slow object can have same momentum as light fast object
3. Units: kg·m/s (or N·s)

💡 Physical Meaning: Momentum measures "quantity of motion" - how hard it is to stop a moving object.

Understanding Momentum

Large Momentum

• Heavy object moving fast: Truck at highway speed
• Hard to stop: Requires large force or long time

Small Momentum

• Light object moving slowly: Tennis ball rolling
• Easy to stop: Small force does the job

Zero Momentum

• Object at rest ($v = 0$)
• Velocities cancel (two objects moving opposite directions with same $|mv|$)

Impulse

Impulse is the product of force and time:

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

where:

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

For Variable Forces

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

Graphically: Area under force vs. time graph

Impulse-Momentum Theorem

Impulse equals change in momentum:

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

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

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

Derivation

From Newton's 2nd Law: $\vec{F} = m\vec{a} = m\frac{\Delta \vec{v}}{\Delta t}$

Multiply both sides by $\Delta t$:

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

Applications of Impulse-Momentum Theorem

Same Change in Momentum, Different Ways

To stop a moving object ($\Delta p$ is fixed):

Option 1: Large force, short time

• Example: Hitting a wall
• $F$ is large, $\Delta t$ is small
• Result: Large force can cause injury/damage

Option 2: Small force, long time

• Example: Catching with hands that "give"
• $F$ is small, $\Delta t$ is large
• Result: Safer, less damage

Since $F\Delta t = \Delta p$ (constant), increasing $\Delta t$ decreases $F$!

Real-World Examples

Air Bags

• Increase collision time $\Delta t$
• Same momentum change, but force reduced
• Prevents injury

Padded Dashboards

• Increase stopping time
• Reduce peak force on passengers

Following Through (Sports)

• Golf, tennis, baseball
• Increase contact time
• Greater impulse → greater momentum change

Karate: Breaking Boards

• Short contact time
• Large force generated
• Follow through maximizes impulse

Catching a Ball

• Pull hands back (increase $\Delta t$)
• Reduce force on hands
• More comfortable catch

Problem-Solving Strategy

For Impulse Problems:

1. Identify initial and final velocities
2. Calculate change in momentum: $\Delta p = m(v_f - v_i)$
   • Remember: momentum is a vector! Watch signs!
3. Apply impulse-momentum theorem: $J = \Delta p$ or $F\Delta t = \Delta p$
4. Solve for unknown (usually $F$ or $\Delta t$)

Important Notes:

• Direction matters! Use + and - for 1D problems
• If $F$ varies, use average force: $F_{avg}\Delta t = \Delta p$
• Impulse has same units as momentum: N·s = kg·m/s

Momentum vs. Kinetic Energy

Both depend on mass and velocity, but differently:

Example: If you double velocity:

• Momentum doubles ($p \to 2p$)
• Kinetic energy quadruples ($KE \to 4KE$)

Force-Time Graphs

The area under a Force vs. Time graph equals the impulse.

For a constant force: $J = F\Delta t$ (rectangle area)

For a variable force: $J = \int F\,dt$ (total area)

⚠️ Common Mistakes

Mistake 1: Forgetting Vector Nature

❌ Wrong: Using speeds instead of velocities (losing direction info) ✅ Right: Use velocity with proper signs: $\Delta p = m(v_f - v_i)$

Mistake 2: Impulse vs. Force

Impulse is NOT force! Impulse = force × time

Mistake 3: Units Confusion

Remember: 1 N·s = 1 kg·m/s (same units as momentum)

Mistake 4: Sign Errors

If object reverses direction:

• Initial: $v_i = +10$ m/s
• Final: $v_f = -5$ m/s
• $\Delta v = -5 - 10 = -15$ m/s (NOT -5 m/s!)

Special Cases

Object Bouncing Off Wall

If object hits wall and bounces back elastically:

• Before: $v_i = +v$
• After: $v_f = -v$
• $\Delta v = -v - v = -2v$
• $|\Delta p| = 2mv$ (twice what you might expect!)

Object Sticking to Wall

• Before: $v_i = +v$
• After: $v_f = 0$
• $\Delta v = -v$
• $|\Delta p| = mv$

Bouncing off creates twice the momentum change of sticking!

Key Formulas Summary

Find: Average force $F$

Step 1: Calculate initial momentum

$p_i = mv_i = (0.15)(40) = 6 \text{ kg·m/s}$

Step 2: Calculate final momentum

$p_f = mv_f = (0.15)(-50) = -7.5 \text{ kg·m/s}$

Step 3: Calculate change in momentum

$\Delta p = p_f - p_i = -7.5 - 6 = -13.5 \text{ kg·m/s}$

The negative sign indicates the direction (away from bat).

Magnitude: $|\Delta p| = 13.5$ kg·m/s

Step 4: Apply impulse-momentum theorem

$F\Delta t = \Delta p$

$F(0.002) = -13.5$

$F = \frac{-13.5}{0.002} = -6,750 \text{ N}$

Answer: The average force is 6,750 N in the direction opposite to the initial motion (the bat pushes the ball backward).

Magnitude: 6,750 N (about 1,500 pounds of force!)

Note: The change in momentum is $13.5$ kg·m/s, not just $7.5$ kg·m/s, because the ball reversed direction. The velocity change is $50 - (-40) = 90$ m/s!