Momentum and Impulse

AP Physics 1· AP Physics 1 CED — Momentum· 14 min read

1. Linear Momentum★★☆☆☆⏱ 3 min

📘 Definition

Linear Momentum

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

Vector quantity describing the amount of motion of an object, equal to the product of the object's mass and velocity. Direction matches velocity, so use positive/negative signs for direction in 1D problems. Units are $kg \cdot m/s$ .

Example:

A 1000 kg car moving at 20 m/s right has momentum $+ 20000 kg \cdot m/s$ .

Momentum measures how hard it is to stop a moving object: a slow-moving semi-truck has more momentum than a fast-moving baseball, because mass has a larger effect than velocity here. For systems of multiple objects, total momentum is the vector sum of individual momentum values: always add signed values, not just magnitudes, to get the correct total.

📐 Worked Example

A 2.0 kg cart moving to the right at 3.0 m/s collides head-on with a 1.0 kg cart moving to the left at 4.0 m/s. What is the total momentum of the two-cart system before the collision, taking right as the positive direction?

1
Define the coordinate system (right = positive), so velocities are:
$v_1 = +3.0 \, \text{m/s}, \quad v_2 = -4.0 \, \text{m/s}$
2
Calculate momentum of the first cart:
$p_1 = m_1 v_1 = (2.0)(+3.0) = +6.0 \, \text{kg} \cdot \text{m/s}$
3
Calculate momentum of the second cart:
$p_2 = m_2 v_2 = (1.0)(-4.0) = -4.0 \, \text{kg} \cdot \text{m/s}$
4
Sum the signed momentum values to get total system momentum:
$p_{\text{total}} = p_1 + p_2 = 6.0 - 4.0 = +2.0 \, \text{kg} \cdot \text{m/s}$

2. Impulse★★☆☆☆⏱ 3 min

📘 Definition

Impulse

J

Quantity describing the effect of a net force acting over a time interval. For constant net force, $J = F_{net} Δ t$ ; for variable force, $J$ equals the area under a net force vs. time graph. Units are $N \cdot s$ , which is equivalent to $kg \cdot m/s$ .

Impulse follows the key force-time relationship: to get the same total impulse (same change in momentum), you can apply a large force over a short time or a small force over a long time. This principle explains the function of airbags, padded dashboards, and crash-absorbing bumpers: increasing collision time reduces the peak force experienced during impact.

AP Physics 1 does not require calculus for impulse calculation: the area under a force-time graph will always be made of simple geometric shapes (triangles, rectangles, trapezoids) that can be calculated with basic geometry.

📐 Worked Example

A student hits a 0.05 kg golf ball with a club. The force exerted by the club on the ball as a function of time forms a triangle with a peak force of 2000 N and total contact time of 0.005 s. What is the magnitude of the impulse delivered to the golf ball?

1
Impulse equals the area under the force vs. time graph, which is triangular in this case.
2
Area of a triangle is given by:
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
3
Base = contact time $Δ t = 0.005$ s, height = peak force $F_{max} = 2000$ N. Substitute to find impulse:
$J = \frac{1}{2} \times 0.005 \times 2000 = 5.0 \, \text{N} \cdot \text{s}$

3. The Impulse-Momentum Theorem★★★☆☆⏱ 4 min

📘 Definition

Impulse-Momentum Theorem

J_{\text{net}} = \Delta p

The net impulse delivered to an object (or system) equals the total change in the object's momentum. This is derived directly from Newton's second law.

Example:

A 1 kg object that speeds up from 2 m/s to 5 m/s has $Δ p = 3 kg \cdot m/s$ , so net impulse $J = 3 N \cdot s$ .

The full form of the theorem is:

J_{\text{net}} = \Delta p = p_f - p_i = m(v_f - v_i)

For systems of multiple objects, internal impulses (forces that objects within the system exert on each other) cancel out per Newton's third law, so only external forces contribute to the net impulse of the whole system. This relationship works for both constant and variable forces, since variable force impulse is calculated as area under the F-t graph.

📐 Worked Example

A 60 kg skateboarder moving right at 5.0 m/s hits a rough patch of ground that exerts an average net force of 120 N to the left on the skateboard for 1.5 s. What is the skateboarder’s final velocity?

1
Define right as positive, so all values are:
$F_{\text{net}} = -120 \, \text{N}, \, v_i = +5.0 \, \text{m/s}, \, \Delta t = 1.5 \, \text{s}, \, m = 60 \, \text{kg}$
2
Calculate net impulse:
$J_{\text{net}} = F_{\text{net}} \Delta t = (-120)(1.5) = -180 \, \text{N} \cdot \text{s}$
3
Rearrange the impulse-momentum theorem to solve for final velocity:
$v_f = v_i + \frac{J_{\text{net}}}{m}$
4
Substitute values and solve:
$v_f = 5.0 + \frac{-180}{60} = 2.0 \, \text{m/s}$

4. AP-Style Worked Practice Problems★★★☆☆⏱ 4 min

✓ Quick check

Try this multiple-choice question to test your understanding before looking at the solution:

A 0.2 kg ball is thrown straight toward a wall at 15 m/s, and bounces straight back at 10 m/s. What is the magnitude of the impulse delivered to the ball by the wall?
- A) 1 N·s
- B) 3 N·s
- C) 5 N·s
- D) 25 N·s
Reveal answer
C) 5 N·s —
You must account for the change in direction of the ball, so the velocity flips sign. Correct calculation: $J = Δ p = 0.2 (- 10 - 15) = - 5$ N·s, magnitude 5 N·s.

📐 Worked Example

A 2.0 kg block slides right along a frictionless horizontal surface at 6.0 m/s. A variable force (positive when pointing right) is applied to the block over 4.0 seconds, with the following force vs. time shape: from t=0 to t=1 s, force increases linearly from 0 to 4 N; from t=1 s to t=3 s, force is constant at 4 N; from t=3 s to t=4 s, force decreases linearly back to 0. (a) Calculate the total impulse delivered to the block over 4.0 seconds. (b) Calculate the final velocity of the block after 4.0 seconds. (c) Explain why increasing the time over which a fixed total impulse is applied reduces the maximum force on the block.

1
(a) Split the F-t graph into three regions and calculate area for each:
- Left triangle (0-1 s): $A_{1} = \frac{1}{2} (1) (4) = 2$ N·s
- Middle rectangle (1-3 s): $A_{2} = (2) (4) = 8$ N·s
- Right triangle (3-4 s): $A_{3} = \frac{1}{2} (1) (4) = 2$ N·s
2
Total impulse is the sum of all areas:
$J = 2 + 8 + 2 = 12 \, \text{N·s}$
3
(b) Use the impulse-momentum theorem to solve for final velocity:
$J = m(v_f - v_i) \implies v_f = v_i + \frac{J}{m} = 6.0 + \frac{12}{2.0} = 12 \, \text{m/s}$
4
(c) For a fixed total impulse, average force is inversely proportional to the time interval ( $F_{a v g} = J /Δ t$ ). Increasing $Δ t$ reduces the average force, and maximum force scales with average force for a similar force profile. This is the core principle behind automotive crash safety design.

📐 Worked Example

A car manufacturer tests crash safety by running a 1500 kg car into a barrier at 15 m/s. The car comes to a complete stop after impact. Two setups are tested: a rigid barrier stops the car in 0.08 s, and an energy-absorbing barrier stops the car in 0.30 s. Calculate the average force exerted on the car by the barrier for both setups.

1
Define initial direction of motion as positive, so $v_{i} = + 15 m/s$ , $v_{f} = 0$ , $m = 1500 kg$ .
2
Calculate total change in momentum:
$\Delta p = m(v_f - v_i) = 1500(0 - 15) = -22500 \, \text{kg·m/s}$
3
Rearrange the impulse-momentum theorem to solve for average force:
$F_{\text{avg}} = \frac{\Delta p}{\Delta t}$
4
Rigid barrier average force:
$F_{\text{avg, rigid}} = \frac{-22500}{0.08} = -2.8 \times 10^5 \, \text{N}$
5
Energy-absorbing barrier average force:
$F_{\text{avg, absorb}} = \frac{-22500}{0.30} = -7.5 \times 10^4 \, \text{N}$
6
The negative sign indicates force acts opposite the car's initial direction. The energy-absorbing barrier reduces average force by nearly 75%.

5. Common Pitfalls

Wrong move:

Calculating impulse for a triangular force vs. time graph as $F_{ma x} Δ t$ instead of $\frac{1}{2} F_{ma x} Δ t$

Why:

Students confuse the graph shape and use the rectangle area formula, resulting in twice the correct impulse, which is a common MCQ distractor.

Correct move:

Always explicitly identify the shape of the F-t region and write the correct area formula before plugging in numbers.

Wrong move:

Adding magnitudes of momentum for objects moving in opposite directions, instead of adding signed values

Why:

Students forget momentum is a vector and treat it like a scalar quantity, leading to wrong total momentum.

Correct move:

Write your coordinate system at the start of the problem, assign signs to all velocities before calculating momentum.

Wrong move:

Equating impulse to total momentum, instead of change in momentum

Why:

Students abbreviate the theorem to 'impulse equals momentum' when memorizing, leading to wrong answers for final velocity.

Correct move:

Always write the full theorem $J_{n e t} = Δ p = p_{f} - p_{i}$ at the start of your calculation to remind yourself it is a change.

Wrong move:

Calculating $Δ p = p_{i} - p_{f}$ instead of $Δ p = p_{f} - p_{i}$

Why:

Students mix up the 'final minus initial' rule for change, leading to a sign error that propagates through the whole problem.

Correct move:

Double-check the order of subtraction for any change in quantity, and always put the final value first.

Wrong move:

Using area under force vs. position to find impulse

Why:

Students confuse impulse (uses F-t graphs) and work (uses F-x graphs), because both are areas under force graphs.

Correct move:

Always check the x-axis label before calculating area: x = time → impulse, x = position → work.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Linear Momentum	$p = m v$	Vector, direction matches velocity; use signs for direction in 1D
Total System Momentum	$p_{total} = p_{1} + p_{2} + ... + p_{n}$	Add signed momentum values, not just magnitudes
Impulse (Constant Force)	$J = F_{net} Δ t$	Units: $N \cdotp s = kg \cdotp m/s$ , same as momentum
Impulse (Variable Force)	$J = Area under F vs. t graph$	Use triangle area $A = \frac{1}{2} bh$ , rectangle area $A = bh$
Impulse-Momentum Theorem	$J_{net} = Δ p = p_{f} - p_{i} = m (v_{f} - v_{i})$	Core relation, applies to all constant-mass problems
Average Force	$F_{avg} = \frac{J _{net}}{Δ t}$	Used to find average force for variable impulse over time

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · MCQ
Impulse from triangular F-t graph
2023 · FRQ
Impulse-momentum velocity calculation

What's Next

This sub-topic gives you the foundational tools for the rest of AP Physics 1 Unit 5: Momentum. The most immediate application is conservation of momentum for closed systems, where net external impulse is zero, so total momentum remains constant. Without mastering the impulse-momentum theorem and consistent sign conventions for vector momentum, you will not be able to correctly solve collision and explosion problems, which make up the majority of Unit 5 FRQ points on the AP exam. This topic also connects directly to energy conservation, where you will learn when to use momentum vs. energy to solve different types of collision problems, and core logic extends to angular momentum in rotational motion.