FULL REVIEW

Full Review: Momentum and Collisions

Review the essential ideas, relationships, and problem-solving tools for momentum and collisions.

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TIME

60 minutes

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BEST FOR

A complete topic review

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FINISH WITH

A readiness check

After this full review, you'll be able to...

recall the essential ideas, apply them to representative problems, and determine what to study next.

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Course Alignment

This bundle is designed to complement the chapter listed below. Use it to reinforce key concepts, prepare for homework, or review before a quiz or exam.

TEXTBOOK: Physics Sensei Mechanics

CHAPTER: Volume 1, MEC-U07

TOPIC: Momentum and Collisions

TREATMENT: Algebra-based introductory college physics

BEST USED

✓ After reading the chapter

✓ Before starting homework

✓ Before a quiz or exam

Your Review Plan

Complete these six stages in order. Each stage builds on the previous one and prepares you for the final readiness check.

6 Stages • Approximately 60 minutes.

①

Warm-Up Check

Activate prior knowledge.

②

Core Concepts

Review the essential ideas.

③

Guided Practice

Apply what you learned.

④

Confidence Check

Confirm your understanding.

Warm-Up Check

Before you begin, take a moment to see what you already remember. Do not worry about getting everything right. This is only a starting point.

ACTIVITY 1

Recall Activity

Subtract the initial momentum vector from the final momentum vector, then find the displacement magnitude.

A particle moves from r₁ = <2, −1, 3> m to r₂ = <8, 4, −1> m. Find the displacement vector and its magnitude.

Reveal Answers

Δr = <6, 5, −4> m; |Δr| = √77 m = 8.77 m.

Why it works: Displacement is final position minus initial position, component by component: <8 − 2, 4 − (−1), −1 − 3>. Its magnitude is √(6² + 5² + (−4)²).

ACTIVITY 2

Recall Activity

Treat horizontal and vertical motion independently, but evaluate both at the same time.

A collision has velocity components v_x = 12.0 m/s and v_y = 18.0 m/s at launch. Neglect air resistance. Find its velocity components, speed, and direction 1.50 s later. Use g = 9.80 m/s².

Reveal Answers

v_x = 12.0 m/s; v_y = 3.30 m/s; speed = 12.4 m/s; direction = 15.4° above +x.

Why it works: Horizontal acceleration is zero, so v_x stays 12.0 m/s. Vertically, v_y = 18.0 − (9.80)(1.50) = 3.30 m/s. The speed is √(12.0² + 3.30²) = 12.4 m/s, and atan2(3.30, 12.0) = 15.4°.

ACTIVITY 3

Recall Activity

Identify the direction and magnitude of the acceleration for uniform system momentum.

A cart moves at a constant speed of 15.0 m/s around a momentum track of radius 25.0 m. What is the acceleration?

Reveal Answers

The acceleration has magnitude 9.00 m/s² and points radially inward, toward the center.

Why it works: Constant speed does not mean zero acceleration because the velocity direction changes. The centripetal magnitude is a_rad = v²/r = (15.0)²/25.0 = 9.00 m/s².

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Core Concepts

Let's rebuild the key ideas one step at a time. Focus on understanding the relationships before worrying about solving problems.

KEY CONCEPT 1

Momentum Vectors in Two and Three Dimensions

Position, displacement, velocity, and acceleration are vectors. Write each in components and apply the one-dimensional definitions independently along x, y, and z. Average velocity is displacement divided by elapsed time. Instantaneous velocity is tangent to the path and equals the time derivative of position; acceleration is the time derivative of velocity. Speed is the magnitude of velocity, not a signed component.

r(t) = <x(t), y(t), z(t)>; Δr = r₂ − r₁; v_avg = Δr/Δt; v(t) = dr/dt; a(t) = dv/dt; speed = √(v_x² + v_y² + v_z²).

Example: If r(t) = <3t², 4t, 2 − t²> m, then v(t) = <6t, 4, −2t> m/s and a(t) = <6, 0, −2> m/s². At t = 2.00 s, v = <12, 4, −4> m/s and the speed is √176 = 13.3 m/s.

Sensei Note: A vector derivative is taken component by component. Never differentiate the magnitude of r and call that the velocity vector; doing so discards direction.

KEY CONCEPT 2

Momentum Conservation as Independent Component Motions

With negligible air resistance and a constant downward gravitational acceleration, horizontal motion has constant velocity while vertical motion has constant acceleration. Resolve the launch velocity before using kinematics. Both components share the same elapsed time. At the highest point, only the vertical velocity is zero; the horizontal velocity generally remains nonzero. Equal launch and landing heights give a symmetric trajectory, but unequal heights do not.

v_0,x = v₀ cos θ; v_0,y = v₀ sin θ; x = x₀ + v_0,xt; y = y₀ + v_0,yt − ½gt²; v_y = v_0,y − gt. For equal heights: T = 2v_0,y/g, R = v₀² sin(2θ)/g, H = v_0,y²/(2g).

Example: A ball launched at 20.0 m/s and 35.0° from a height of 1.80 m has v_0,x = 16.4 m/s and v_0,y = 11.5 m/s. Solving 0 = 1.80 + 11.5t − 4.90t² gives t = 2.49 s. The horizontal range is 40.8 m. The maximum height above the ground is 8.51 m.

Sensei Note: Do not use equal-height range or flight-time formulas when launch and landing heights differ. Return to the component equations and solve for the shared time.

KEY CONCEPT 3

System Momentum and Collision Momentum

In system momentum, velocity is tangent to the path. The radial acceleration points toward the center and changes the velocity direction. If speed also changes, tangential acceleration changes the speed; radial and tangential components are perpendicular. Collision momentum relates measurements made in different frames through vector addition. Choose the frame labels first, then add components consistently.

a_rad = v²/r = ω²r (inward); a_tan = dv/dt; |a| = √(a_rad² + a_tan²); v_A/C = v_A/B + v_B/C.

Example: A car moving at 12.0 m/s on a 30.0 m radius curve has a_rad = 4.80 m/s² inward. If its speed increases at 1.60 m/s², its total acceleration is √(4.80² + 1.60²) = 5.06 m/s². Separately, a cart moving 4.00 m/s north relative to water in a 2.50 m/s eastward current moves at <2.50 east, 4.00 north> m/s relative to shore.

Sensei Note: “Centripetal” describes the inward component of net acceleration; it is not an additional force. In relative-motion equations, every velocity must state “of what relative to what.”

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You've reinforced the essential concepts. Now it's time to put them into practice by working through guided examples and building your problem-solving confidence. Need a quick reminder?

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Guided Practice

Now it's time to apply what you've reviewed.

Work through each activity in order. The examples become gradually more challenging, and each one prepares you for the final readiness check.

PRACTICE 1

Worked Example

Differentiate each position component and evaluate the resulting vectors.

A particle has r(t) = <3t² − 2, 4t + 1, −t² + 6t> m. Find r, v, a, and speed at t = 2.00 s.

Reveal Answers

r = <10, 9, 8> m; v = <12, 4, 2> m/s; a = <6, 0, −2> m/s²; speed = √164 = 12.8 m/s.

Why it works: Differentiate component by component: v(t) = <6t, 4, −2t + 6> and a(t) = <6, 0, −2>. Substitute t = 2.00 s. The speed is the magnitude of v, √(12² + 4² + 2²).

PRACTICE 2

Guided Problem

Resolve the launch velocity, solve the vertical equation for the positive flight time, and use that time horizontally.

A ball is launched from a 1.80 m platform at 20.0 m/s and 35.0° above horizontal. Neglect air resistance. Find the flight time, horizontal range, maximum height above the ground, and impact velocity.

Reveal Answers

Flight time = 2.49 s; range = 40.8 m; maximum height = 8.51 m; impact velocity = <16.4, −12.9> m/s, with speed 20.9 m/s at 38.3° below +x.

Why it works: Use v_0,x = 16.4 m/s and v_0,y = 11.5 m/s. The positive root of 0 = 1.80 + 11.5t − 4.90t² is 2.49 s. Then x = v_0,xt. The rise above launch is v_0,y²/(2g). At impact, v_y = v_0,y − gt = −12.9 m/s.

PRACTICE 3

Independent Problem

Add relative-velocity vectors and use the crossing component to determine time.

A collision track is 180 m wide. A cart points due north and moves at 4.00 m/s relative to the water; the current is 2.50 m/s east relative to shore. Find the cart velocity relative to shore, crossing time, and downstream drift.

Reveal Answers

v_cart/shore = <2.50 east, 4.00 north> m/s; speed = 4.72 m/s at 32.0° east of north; crossing time = 45.0 s; drift = 112.5 m east.

Why it works: Add v_cart/water and v_water/shore. Only the north component crosses the 180 m width, so t = 180/4.00 = 45.0 s. The current produces an eastward displacement (2.50)(45.0) = 112.5 m.

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You've practiced the essential skills with guidance. Now it's time to solve a few short problems on your own and confirm you're ready to move forward. Need a quick reminder?

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Confidence Check

You've rebuilt the key ideas and practiced them with guidance. Now try these short questions on your own to check your understanding before moving on.

QUICK CHECK 1

Read a Vector Motion Function

Answer without expanding the vector magnitude until the end.

For r(t) = <2t³, 5t², 7> m, what are v(1.00 s) and a(1.00 s)?

Reveal Answers

v(1.00 s) = <6, 10, 0> m/s; a(1.00 s) = <12, 10, 0> m/s².

Why it works: v(t) = <6t², 10t, 0>; a(t) = <12t, 10, 0>. Evaluate each component at 1.00 s.

QUICK CHECK 2

Check Collision Independence

Choose the correct statement and justify it physically.

Two balls leave the same height at the same instant. One is dropped and the other is launched horizontally. With no air resistance, which reaches the ground first?

Reveal Answers

They reach the ground at the same time.

Why it works: They have the same initial vertical velocity, the same vertical displacement, and the same vertical acceleration. Horizontal velocity does not change the time required for the vertical fall.

QUICK CHECK 3

Separate Radial, Tangential, and Relative Effects

State the relevant vector direction in each case.

A car moves clockwise around a circle while slowing down. Where do its radial and tangential accelerations point? If an observer moves with the car, how is another object’s velocity found in that observer’s frame?

Reveal Answers

Radial acceleration points inward. Tangential acceleration points opposite the car’s instantaneous velocity. The other object’s velocity relative to the car is its ground velocity minus the car’s ground velocity.

Why it works: The radial component changes direction; the tangential component changes speed. Collision momentum is a vector difference when both velocities are expressed relative to the same frame.

How did it go?

You've checked your understanding. Take one final look at the essential ideas before deciding what to do next. Need a quick reminder?

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Summary

Before moving on, take one final look at the most important ideas from this review.

KEY TAKEAWAY 1

Differentiate Components, Then Recombine

Represent motion with vectors, apply derivatives or finite changes to each component, and compute magnitudes only after the vector is known.

KEY TAKEAWAY 2

One Time Connects Independent Collision Components

Horizontal and vertical collision equations are independent in dynamics but linked by the same elapsed time.

KEY TAKEAWAY 3

Direction Changes Require Acceleration and Frame Labels

Momentum motion requires inward acceleration even at constant speed; relative-motion calculations require explicit reference frames and vector addition.