Conservation Of Momentum In Collisions
This animation walks through momentum conservation using three scenarios: an object exploding into two pieces, a perfectly inelastic 1D collision where two masses stick together, and a 2D collision requiring vector decomposition. Each scene pairs motion with live math overlays showing momentum sums before and after, demonstrating that total momentum stays constant even when kinetic energy is lost. Useful for physics students learning to set up and solve momentum conservation equations in one and two dimensions.
Narrated · 16:9 · Preview before teaching · automatic layout checks do not establish subject accuracy
The prompt that made it
Scene 4: Quick Application - Explosion / Separation in 1DVisual Prompt: Animated 3D model of a stationary $5.0\text{ kg}$ object. It suddenly splits into two pieces:Piece 1 ($3.0\text{ kg}$) moves East at $+1.0\text{ m/s}$.Piece 2 ($2.0\text{ kg}$) launches West at $-1.5\text{ m/s}$.Math Overlay:$$\Sigma p_i = 0$$$$\Sigma p_f = (3.0 \times +1.0) + (2.0 \times -1.5) = +3.0 - 3.0 = 0\text{ kg}\cdot\text{m/s}$$Voiceover / Text Prompts: "Total momentum before splitting was zero. After splitting, the eastward momentum (+3.0) perfectly cancels the westward momentum (-3.0), proving total momentum is strictly conserved!"Scene 5: Solved Example 1(a) - 1D Inelastic CollisionVisual Prompt: 1D animation along an X-axis. Object A ($1.2\text{ kg}$) moves right at $+3.0\text{ m/s}$. Object B ($2.8\text{ kg}$) moves left at $-2.0\text{ m/s}$. They collide, lock together, and move left as a combined $4.0\text{ kg}$ mass.Math Overlay:$$1.2(3.0) + 2.8(-2.0) = (1.2 + 2.8) v'$$$$3.6 - 5.6 = 4.0 v' \implies -2.0 = 4.0 v' \implies v' = -0.50\text{ m/s}$$Voiceover / Text Prompts: "Assigning the positive direction to the right, the initial negative momentum dominates. After sticking together, both objects move left at $0.50\text{ m/s}$."Scene 6: Solved Example 1(b) - 2D Collision DecompositionVisual Prompt: Top-down 2D grid animation. Object A ($0.20\text{ kg}$) moves right at $2.0\text{ m/s}$ towards stationary Object B ($0.80\text{ kg}$). Upon impact, A bounces vertically along the +Y axis. Object B deflects at an angle of $30^\circ$ below the +X axis.Math Overlay:X-axis: $0.20(2.0) + 0 = 0 + 0.80 v_B' \cos(30^\circ) \implies 0.40 = 0.80 v_B' (0.866) \implies v_B' \approx 0.58\text{ m/s}$Y-axis: $0 = 0.20 v_A' - 0.80 v_B' \sin(30^\circ) \implies 0.20 v_A' = 0.80(0.577)(0.5) \implies v_A' \approx 1.2\text{ m/s}$Voiceover / Text Prompts: "In two-dimensional collisions, conserve momentum independently along orthogonal X and Y axes. Object A moves at $1.2\text{ m/s}$ north, while Object B moves at $0.58\text{ m/s}$ at $30^\circ$ south of east."
Make the next one in this series
- Show the same 2D collision solved by separately conserving x and y momentum components
- Animate an elastic collision comparing momentum and kinetic energy conservation side by side
- Create a scene showing momentum conservation with friction and external forces breaking the ideal case
