Solving Momentum Conservation Practice Problems
A worked-problem walkthrough covering conservation of momentum in three scenarios: a 1D collision with rebound, an explosion of a stationary block into two fragments, and a symmetric 2D collision at 45 degrees. Vector arrows and algebraic overlays show how momentum equations are set up and solved for unknown velocities. Useful for physics students practicing quantitative problem-solving and for teachers demonstrating step-by-step application of conservation laws in both one and two dimensions.
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The prompt that made it
Scene 7: Practice Problem 1 - 1D Collision & ReboundVisual Prompt: Object A ($2.0\text{ kg}$) moving right at $+4.0\text{ m/s}$ collides with Object B ($3.0\text{ kg}$) moving left at $-4.0\text{ m/s}$. After collision, Object B rebounds right at $+2.0\text{ m/s}$. Animated vector arrows illustrate the momentum shift for Object A.Math Overlay:$$2.0(4.0) + 3.0(-4.0) = 2.0 v_A' + 3.0(2.0)$$$$8.0 - 12.0 = 2.0 v_A' + 6.0 \implies -4.0 - 6.0 = 2.0 v_A' \implies v_A' = -5.0\text{ m/s}$$Voiceover / Text Prompts: "Solving the 1D momentum equation shows Object A rebounds to the left at a speed of $5.0\text{ m/s}$."Scene 8: Practice Problem 2 - Explosion of Stationary ObjectVisual Prompt: A stationary $8.0\text{ kg}$ block splits into two pieces: Object A ($3.0\text{ kg}$) and Object B ($5.0\text{ kg}$). Object A shoots left at $-10\text{ m/s}$. Object B recoils to the right.Math Overlay:$$0 = 3.0(-10) + 5.0 v_B' \implies 0 = -30 + 5.0 v_B' \implies v_B' = +6.0\text{ m/s}$$Voiceover / Text Prompts: "To maintain zero net momentum, the heavier $5.0\text{ kg}$ Object B moves to the right at $+6.0\text{ m/s}$."Scene 9: Practice Problem 3 - Symmetric 2D Collision ($45^\circ$)Visual Prompt: Top-down view on a smooth surface. Sphere A ($1.0\text{ kg}$) moving at $4.0\text{ m/s}$ strikes stationary Sphere B ($2.0\text{ kg}$). Sphere A deflects upwards at $45^\circ$, and Sphere B deflects downwards at $45^\circ$.Math Overlay:Y-axis: $0 = 1.0 v_A' \sin(45^\circ) - 2.0 v_B' \sin(45^\circ) \implies v_A' = 2 v_B'$X-axis: $1.0(4.0) = 1.0(2 v_B') \cos(45^\circ) + 2.0 v_B' \cos(45^\circ)$$4.0 = 4.0 v_B' (0.707) \implies v_B' \approx 1.41\text{ m/s}$, $v_A' \approx 2.83\text{ m/s}$.Voiceover / Text Prompts: "Resolving components gives Sphere A a speed of $2.83\text{ m/s}$ at $45^\circ$ above horizontal, and Sphere B a speed of $1.41\text{ m/s}$ at $45^\circ$ below horizontal."
Make the next one in this series
- Show a fully inelastic 2D collision where both objects stick together after impact
- Animate a perfectly elastic collision and compare kinetic energy before and after
- Derive the general equations for elastic collisions between two objects of different masses
