Momentum and Impulse — AP Physics 1
AP Physics 1 · AP Physics 1 CED Unit 5 · 14 min read
Covers: This guide covers core definitions of linear momentum and impulse, impulse calculation for constant/variable forces, the impulse-momentum theorem, and 1D vector sign conventions, aligned to AP Physics 1 CED for Unit 5.
You should already know:
Newton's laws of motion
1D vector sign conventions
Area calculation for basic geometric shapes
Define linear momentum and impulse as vector quantities
Calculate impulse for constant and variable forces from force-time graphs
Apply the impulse-momentum theorem to solve 1D motion problems
Correctly use vector sign conventions for 1D momentum problems
1. Linear Momentum ★★☆☆☆ ⏱ 3 min
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 $\text{kg} \cdot \text{m/s}$.
Example: A 1000 kg car moving at 20 m/s right has momentum $+20000 \, \text{kg} \cdot \text{m/s}$.
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.
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?
Define the coordinate system (right = positive), so velocities are:v_1 = +3.0 \, \text{m/s}, \quad v_2 = -4.0 \, \text{m/s}
Calculate momentum of the first cart:p_1 = m_1 v_1 = (2.0)(+3.0) = +6.0 \, \text{kg} \cdot \text{m/s}
Calculate momentum of the second cart:p_2 = m_2 v_2 = (1.0)(-4.0) = -4.0 \, \text{kg} \cdot \text{m/s}
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}
Always explicitly define your positive direction at the start of every momentum problem. AP graders require this for FRQ credit, and it eliminates 90% of common sign errors.
2. Impulse ★★☆☆☆ ⏱ 3 min
Impulse
J
Quantity describing the effect of a net force acting over a time interval. For constant net force, $J = F_{\text{net}} \Delta t$; for variable force, $J$ equals the area under a net force vs. time graph. Units are $\text{N} \cdot \text{s}$, which is equivalent to $\text{kg} \cdot \text{m/s}$.
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.
The area under a force vs. position graph equals work, not impulse. This is a very common multiple-choice distractor trap.
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?
Impulse equals the area under the force vs. time graph, which is triangular in this case.
Area of a triangle is given by:\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
Base = contact time $\Delta t = 0.005$ s, height = peak force $F_{\text{max}} = 2000$ 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
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 $\Delta p = 3 \, \text{kg} \cdot \text{m/s}$, so net impulse $J = 3 \, \text{N} \cdot \text{s}$.
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.
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?
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}
Calculate net impulse:J_{\text{net}} = F_{\text{net}} \Delta t = (-120)(1.5) = -180 \, \text{N} \cdot \text{s}
Rearrange the impulse-momentum theorem to solve for final velocity:v_f = v_i + \frac{J_{\text{net}}}{m}
Substitute values and solve:v_f = 5.0 + \frac{-180}{60} = 2.0 \, \text{m/s}
When asked for average force, always use net impulse (sum of impulse from all forces acting over the interval), not just the impulse from the applied force. Friction or gravity often contribute a non-negligible impulse that changes the result.
4. AP-Style Worked Practice Problems ★★★☆☆ ⏱ 4 min
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.
(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
Total impulse is the sum of all areas:J = 2 + 8 + 2 = 12 \, \text{N·s}
(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}
(c) For a fixed total impulse, average force is inversely proportional to the time interval ($F_{avg} = J/\Delta t$). Increasing $\Delta t$ 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.
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.
Define initial direction of motion as positive, so $v_i = +15 \, \text{m/s}$, $v_f = 0$, $m = 1500 \, \text{kg}$.
Calculate total change in momentum:\Delta p = m(v_f - v_i) = 1500(0 - 15) = -22500 \, \text{kg·m/s}
Rearrange the impulse-momentum theorem to solve for average force:F_{\text{avg}} = \frac{\Delta p}{\Delta t}
Rigid barrier average force:F_{\text{avg, rigid}} = \frac{-22500}{0.08} = -2.8 \times 10^5 \, \text{N}
Energy-absorbing barrier average force:F_{\text{avg, absorb}} = \frac{-22500}{0.30} = -7.5 \times 10^4 \, \text{N}
The negative sign indicates force acts opposite the car's initial direction. The energy-absorbing barrier reduces average force by nearly 75%.
Common Pitfalls
Why: Students confuse the graph shape and use the rectangle area formula, resulting in twice the correct impulse, which is a common MCQ distractor.
Why: Students forget momentum is a vector and treat it like a scalar quantity, leading to wrong total momentum.
Why: Students abbreviate the theorem to 'impulse equals momentum' when memorizing, leading to wrong answers for final velocity.
Why: Students mix up the 'final minus initial' rule for change, leading to a sign error that propagates through the whole problem.
Why: Students confuse impulse (uses F-t graphs) and work (uses F-x graphs), because both are areas under force graphs.
Quick Reference Cheatsheet