Momentum and Impulse

Definition of momentum, impulse, and impulse-momentum theorem

💨 Momentum and Impulse

What is Momentum?

Momentum is the product of mass and velocity:

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

where:

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

Key Properties

Vector quantity - has magnitude and direction (same direction as velocity)
Depends on both mass and velocity - heavy slow object can have same momentum as light fast object
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:

Identify initial and final velocities
Calculate change in momentum: $\Delta p = m(v_f - v_i)$ $Δ p = m (v_{f} - v_{i})$
- Remember: momentum is a vector! Watch signs!
Apply impulse-momentum theorem: $J = \Delta p$ or $F\Delta t = \Delta p$
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:

| Property | Momentum | Kinetic Energy | |----------|----------|----------------| | Formula | $p = mv$ | $KE = \frac{1}{2}mv^2$ | | Vector? | Yes | No (scalar) | | Velocity dependence | Linear ( $v$ ) | Quadratic ( $v^2$ ) | | Always conserved? | Yes (isolated system) | No (can convert to other forms) | | Can be negative? | Yes (direction) | No (always ≥ 0) |

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

| Concept | Formula | Units | |---------|---------|-------| | Momentum | $\vec{p} = m\vec{v}$ | kg·m/s | | Impulse | $\vec{J} = \vec{F}\Delta t$ | N·s | | Impulse-Momentum Theorem | $\vec{J} = \Delta \vec{p}$ | kg·m/s | | Newton's 2nd Law (momentum form) | $\vec{F} = \frac{d\vec{p}}{dt}$ | N |

📚 Practice Problems

1Problem 1medium

❓ Question:

A 0.15 kg baseball traveling at 40 m/s is hit by a bat and reverses direction, leaving at 50 m/s. The bat is in contact with the ball for 0.001 seconds. (a) What is the change in momentum? (b) What is the impulse on the ball? (c) What is the average force exerted by the bat?

💡 Show Solution

Solution:

Given: m = 0.15 kg, v_i = 40 m/s, v_f = -50 m/s (reversed), Δt = 0.001 s

(a) Change in momentum: Choose initial direction as positive. p_i = mv_i = 0.15 × 40 = 6.0 kg·m/s p_f = mv_f = 0.15 × (-50) = -7.5 kg·m/s Δp = p_f - p_i = -7.5 - 6.0 = -13.5 kg·m/s

Magnitude: |Δp| = 13.5 kg·m/s

(b) Impulse: J = Δp = -13.5 kg·m/s (or 13.5 kg·m/s in direction of bat)

The negative sign indicates force is in direction of final velocity (opposite to initial).

2Problem 2medium

❓ Question:

A 0.15 kg baseball traveling at 40 m/s is hit by a bat and reverses direction, leaving at 50 m/s. If the bat and ball are in contact for 0.002 s, what is the average force exerted by the bat on the ball?

💡 Show Solution

Given Information:

Mass: $m = 0.15$ kg
Initial velocity: $v_i = +40$ m/s (choose toward bat as positive)
Final velocity: $v_f = -50$ m/s (reverses direction)
Contact time: $\Delta t = 0.002$ s

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!

3Problem 3medium

❓ Question:

💡 Show Solution

Solution:

Given: m = 0.15 kg, v_i = 40 m/s, v_f = -50 m/s (reversed), Δt = 0.001 s

(a) Change in momentum: Choose initial direction as positive. p_i = mv_i = 0.15 × 40 = 6.0 kg·m/s p_f = mv_f = 0.15 × (-50) = -7.5 kg·m/s Δp = p_f - p_i = -7.5 - 6.0 = -13.5 kg·m/s

Magnitude: |Δp| = 13.5 kg·m/s

(b) Impulse: J = Δp = -13.5 kg·m/s (or 13.5 kg·m/s in direction of bat)

The negative sign indicates force is in direction of final velocity (opposite to initial).

4Problem 4medium

❓ Question:

A 1200 kg car traveling at 25 m/s crashes into a barrier and comes to rest in 0.10 seconds. (a) What is the impulse on the car? (b) What is the average force on the car? (c) How would increasing the crash time to 0.20 s affect the average force?

💡 Show Solution

Solution:

Given: m = 1200 kg, v_i = 25 m/s, v_f = 0, Δt = 0.10 s

(a) Impulse: J = Δp = m(v_f - v_i) = 1200(0 - 25) = -30,000 kg·m/s Magnitude: 3.0 × 10⁴ kg·m/s

(b) Average force: J = F_avg Δt -30,000 = F_avg (0.10) F_avg = -300,000 N or -3.0 × 10⁵ N Magnitude: 3.0 × 10⁵ N (opposing motion)

Conclusion: Doubling the crash time halves the average force. This is why cars have crumple zones - they increase collision time, reducing force on passengers.

5Problem 5medium

❓ Question:

💡 Show Solution

Solution:

Given: m = 1200 kg, v_i = 25 m/s, v_f = 0, Δt = 0.10 s

(a) Impulse: J = Δp = m(v_f - v_i) = 1200(0 - 25) = -30,000 kg·m/s Magnitude: 3.0 × 10⁴ kg·m/s

(b) Average force: J = F_avg Δt -30,000 = F_avg (0.10) F_avg = -300,000 N or -3.0 × 10⁵ N Magnitude: 3.0 × 10⁵ N (opposing motion)

Conclusion: Doubling the crash time halves the average force. This is why cars have crumple zones - they increase collision time, reducing force on passengers.

6Problem 6medium

❓ Question:

A 60 kg person jumps from a height and lands on the ground, coming to rest in 0.1 s. If the person was traveling at 5 m/s just before landing, what average force does the ground exert on the person?

💡 Show Solution

Given Information:

Mass: $m = 60$ kg
Initial velocity (downward): $v_i = -5$ m/s (taking down as negative)
Final velocity: $v_f = 0$ m/s (comes to rest)
Time to stop: $\Delta t = 0.1$ s

Find: Average force from ground

Step 1: Calculate change in momentum

$\Delta p = m(v_f - v_i) = 60(0 - (-5))$

$\Delta p = 60(5) = 300 \text{ kg·m/s}$

(Positive means upward direction)

Step 2: Apply impulse-momentum theorem

$F_{net}\Delta t = \Delta p$

$F_{net}(0.1) = 300$

$F_{net} = 3,000 \text{ N (upward)}$

Step 3: Identify forces

Two forces act on the person:

Weight: $W = mg = 60(9.8) = 588$ N (downward, so negative)
Normal force from ground: $N$ (upward, positive)

Net force: $F_{net} = N - W = N - 588$

Step 4: Solve for normal force

$N - 588 = 3,000$

$N = 3,588 \text{ N}$

Answer: The ground exerts an average force of 3,588 N upward (about 6 times the person's weight).

Note: If the person bends their knees and takes 0.2 s to stop instead, the force would be cut in half to about 1,794 N. This is why we bend our knees when landing!

7Problem 7hard

❓ Question:

A 1200 kg car traveling at 25 m/s collides with a wall and comes to rest in 0.15 s. (a) What is the impulse on the car? (b) What is the average force on the car? (c) If the car had an airbag that increased the stopping time to 0.3 s, how would the force change?

💡 Show Solution

Given Information:

Mass: $m = 1200$ kg
Initial velocity: $v_i = 25$ m/s
Final velocity: $v_f = 0$ m/s
Stopping time: $\Delta t = 0.15$ s

(a) Find impulse on the car

Step 1: Calculate change in momentum

$\Delta p = m(v_f - v_i) = 1200(0 - 25)$

$\Delta p = -30,000 \text{ kg·m/s}$

Step 2: Impulse equals change in momentum

$J = \Delta p = -30,000 \text{ kg·m/s (or N·s)}$

The negative sign indicates the impulse is opposite to the initial motion (backward).

Magnitude: $|J| = 30,000$ N·s

(b) Find average force

Step 3: Apply impulse-momentum theorem

$F\Delta t = J$

$F(0.15) = -30,000$

$F = \frac{-30,000}{0.15} = -200,000 \text{ N}$

Answer (a): Impulse = -30,000 N·s (or 30,000 N·s backward)

Answer (b): Average force = 200,000 N (about 45,000 pounds!)

(c) With airbag ( $\Delta t = 0.3$ s)

Step 4: Calculate new force

The impulse remains the same ( $J = -30,000$ N·s), but time doubles:

$F'\Delta t' = J$

$F'(0.3) = -30,000$

$F' = \frac{-30,000}{0.3} = -100,000 \text{ N}$

Step 5: Compare forces

$\frac{F'}{F} = \frac{100,000}{200,000} = \frac{1}{2}$

The force is reduced by half when the time is doubled!

Answer (c): With airbag, force = 100,000 N (half the original force)

Key Insight: Since $F\Delta t =$ constant, increasing stopping time decreases the force. This is why airbags, crumple zones, and padded dashboards save lives - they increase $\Delta t$ to decrease $F$ .

$\text{If } \Delta t \text{ doubles} \Rightarrow F \text{ is cut in half}$

🎴

Practice with Flashcards

Review key concepts with our flashcard system

📖

Browse All Topics

Explore other calculus topics