Interactive physics simulator
Elastic Collision
In perfectly elastic collisions, both momentum and kinetic energy are conserved without loss. Explore velocity exchanges on 1D tracks, energy chains in Newton's Cradle, and thermodynamic distributions in 2D gas scattering.
Elastic Collision Simulator
Configure velocities, mass values, and slider settings, then run the simulation to monitor conservation plots.
Live Telemetry
- Active Balls Pulled
- 1
- Individual Ball Mass
- 0.2 kg
- Pull Amplitude
- 45.0°
- Current Peak Speed
- 0.0 m/s
- Kinetic Energy (Peak)
- 0.00 J
- Total Momentum P (x-dir)
- 0.00 kg·m/s
- Total energy (KE + PE)
- 0.00 J
What is an Elastic Collision?
An elastic collision is a collision in which there is no net loss of total kinetic energy in the system. Both linear momentum and kinetic energy are conserved quantities. During the collision, kinetic energy is briefly converted into elastic potential energy as the bodies compress, and then fully converted back into kinetic energy as they rebound without permanent deformation or thermal dissipation.
For a one-dimensional elastic collision, the equations of motion can be derived from the conservation of momentum and energy, yielding the final velocities:
Velocity swaps & Limits
Mass ratios dictate the velocity profiles of post-impact bodies:
- Identical Masses (m1 = m2): The incoming body stops dead, transferring 100% of its speed to the stationary target (as in billiard ball hits or Newton's Cradle).
- Heavy Projectile (m1 >> m2): The projectile barely slows down (v1f ≈ u1), while launching the light target at double speed (v2f ≈ 2u1).
- Light Projectile (m1 << m2): The light cart bounces back with opposite velocity (v1f ≈ -u1), leaving the heavy target nearly at rest (v2f ≈ 0).
Newton's Cradle Chains
A classic demonstration of energy and momentum propagation:
- When n spheres are lifted and released, exactly n spheres emerge on the other side with the same velocity.
- Why doesn't one ball swing out at twice the speed? It would conserve momentum (m · v = 2m · v/2) but violate kinetic energy conservation (KE_initial ≠ KE_final).
- Perfect elasticity requires both energy and momentum conservation, restricting the outcome to a symmetric swap.
Molecular Gas Kinetics
Elastic collisions form the backbone of the kinetic theory of gases:
- Gas molecules are modeled as perfectly elastic hard spheres in constant random motion.
- Collisions distribute energy among gas particles, driving the distribution of speeds to a thermodynamic balance.
- The peak of the distribution shifts with temperature, but the area under the curve (total molecules) and total kinetic energy (temperature) remain constant.
2D Oblique Billiard Scattering
Off-center collisions divide velocity vectors in a two-dimensional plane:
- Momentum is a vector, conserving X and Y directions separately: P_ix = P_fx and P_iy = P_fy.
- For equal mass elastic collisions where the target is at rest, the angle between the scattering paths is always exactly 90.0°.
- This allows pool players to predict target ball paths regardless of the cue ball angle.
Solved Examples
A 2.0 kg air track glider (Glider A) moving East at +4.0 m/s makes a head-on elastic collision with a stationary 2.0 kg glider (Glider B). Calculate their final velocities.
- Identify initial values: mA = 2.0 kg, uA = +4.0 m/s, mB = 2.0 kg, uB = 0 m/s.
- Since the masses are equal (mA = mB) and the collision is perfectly elastic (e = 1.0), they will completely swap velocities.
- Using the elastic velocity equations:
- vA,f = [(mA - mB)*uA + 2*mB*uB] / (mA + mB) = [(2.0 - 2.0)*4.0 + 2*2.0*0] / 4.0 = 0.0 m/s.
- vB,f = [(mB - mA)*uB + 2*mA*uA] / (mA + mB) = [(2.0 - 2.0)*0 + 2*2.0*4.0] / 4.0 = +4.0 m/s.
- Therefore, Glider A halts completely, and Glider B launches East at +4.0 m/s.
Answer: Glider A: 0.0 m/s | Glider B: +4.0 m/s
A heavy subatomic particle of mass 4.0 u moving at +300 m/s collides head-on elastically with a stationary light target particle of mass 1.0 u. Find the final velocity of both particles.
- Initial conditions: m1 = 4.0 u, u1 = +300 m/s, m2 = 1.0 u, u2 = 0 m/s.
- Calculate Glider 1 final velocity: v1 = [(m1 - m2)/(m1 + m2)]*u1 + [2*m2/(m1 + m2)]*u2 = [(4.0 - 1.0) / (4.0 + 1.0)] * 300 = (3/5) * 300 = +180 m/s.
- Calculate Glider 2 final velocity: v2 = [2*m1/(m1 + m2)]*u1 + [(m2 - m1)/(m1 + m2)]*u2 = [2*4.0 / (4.0 + 1.0)] * 300 = (8/5) * 300 = +480 m/s.
- Notice that the target is propelled to a velocity greater than the initial speed of the projectile (+480 m/s), while the heavy projectile slows down but continues forward at +180 m/s.
Answer: Heavy Particle: +180 m/s | Target Particle: +480 m/s
A 0.15 kg billiard ball (Ball 1) traveling at 6.0 m/s strikes an identical stationary ball (Ball 2) off-center. After this elastic oblique collision, Ball 1 scatters at an angle of 30.0°. Calculate the final speeds of both balls and the scatter angle of Ball 2.
- In a 2D elastic collision between two identical masses (m1 = m2) where the target is initially at rest, the scatter angle vectors after impact are always perpendicular (θ1 + θ2 = 90.0°).
- Target ball scatter angle: θ2 = 90.0° - 30.0° = 60.0° (on the opposite side of the initial trajectory, representing -60.0°).
- Resolve using trigonometric relations derived from momentum and energy conservation:
- v1f = u1 * cos(θ1) = 6.0 * cos(30.0°) = 6.0 * 0.8660 = 5.20 m/s.
- v2f = u1 * sin(θ1) = 6.0 * sin(30.0°) = 6.0 * 0.5000 = 3.00 m/s.
- Double check kinetic energy: KE_initial = 1/2 * m * u1^2 = 0.5 * 0.15 * 36 = 2.70 J. KE_final = 1/2 * m * (v1f^2 + v2f^2) = 0.5 * 0.15 * (27.04 + 9.00) = 2.70 J. Both momentum and energy are conserved.
Answer: Ball 1: 5.20 m/s (at 30°) | Ball 2: 3.00 m/s (at -60°)
Common Misconceptions
- "Elastic collisions occur in car accidents": False. Vehicles deform, generate crumple zones, and emit loud crash sounds. These are inelastic. Perfectly elastic collisions are micro-scale (atoms, subatomic particles).
- "Velocity is always swapped": False. Velocity swaps occur only when the colliding masses are identical. If one mass is larger, both velocities change non-symmetrically.
- "Momentum is only conserved if it is elastic": False. Total momentum is conserved in all collisions (elastic, inelastic, completely inelastic) as long as there are no external forces.
Quick Summary
- Both momentum (P) and kinetic energy (KE) are conserved: ΔP = 0, ΔKE = 0.
- The coefficient of restitution is exactly e = 1.0.
- Equal mass elastic collisions result in complete velocity swaps.
- In Newton's Cradle, the number of balls launched matches the number released.
- Random elastic molecular collisions maintain a constant total energy while generating the Maxwell-Boltzmann speed curve.
Practice Questions
1. What mathematical conditions define a perfectly elastic collision?
A perfectly elastic collision must satisfy two conditions simultaneously: (1) Total linear momentum is conserved: Σp_i = Σp_f, or m1*u1 + m2*u2 = m1*v1 + m2*v2. (2) Total kinetic energy is conserved: ΣKE_i = ΣKE_f, or 1/2*m1*u1² + 1/2*m2*u2² = 1/2*m1*v1² + 1/2*m2*v2². This corresponds to a coefficient of restitution (e) of exactly 1.0.
2. Prove that the relative speed of separation equals the relative speed of approach in a 1D elastic collision.
From KE conservation: m1(u1² - v1²) = m2(v2² - u2²). Factor both sides: m1(u1 - v1)(u1 + v1) = m2(v2 - u2)(v2 + u2). From momentum conservation: m1(u1 - v1) = m2(v2 - u2). Divide the KE equation by the momentum equation: u1 + v1 = v2 + u2. Rearranging gives: u1 - u2 = v2 - v1. The term (u1 - u2) is the relative speed of approach, and (v2 - v1) is the relative speed of separation.
3. Explain why identical steel spheres in a Newton's Cradle transfer momentum without stopping in the middle.
In a Newton's Cradle, steel balls undergo successive elastic collisions. Because the steel balls are identical in mass and elastic properties (e ≈ 0.99), each collision completely exchanges velocity. When ball 1 hits ball 2, ball 1 stops and transfers all its velocity to ball 2. This sequence repeats instantly through balls 3 and 4, culminating in ball 5 swinging out with the original velocity.
4. How does the Maxwell-Boltzmann distribution emerge from elastic collisions in a gas?
In an ideal gas, molecules behave as elastic spheres colliding randomly. While each individual collision is perfectly elastic (conserving total system KE), momentum transfers constantly distribute velocities in all directions. Over billions of events, this statistical scattering drives the speeds of the particles to populate a characteristic probability curve called the Maxwell-Boltzmann distribution, maintaining a constant temperature.
FAQ
Frequently Asked Questions
What is an elastic collision in physics?
An elastic collision is an interaction between two bodies in which both total linear momentum and total kinetic energy are conserved, meaning no kinetic energy is converted into other forms like heat, sound, or permanent deformation.
How does an elastic collision differ from an inelastic collision?
In both elastic and inelastic collisions, momentum is conserved. However, kinetic energy is conserved ONLY in an elastic collision. In an inelastic collision, some kinetic energy is converted into other forms (heat, sound, deformation).
What is the coefficient of restitution (e) for an elastic collision?
For a perfectly elastic collision, the coefficient of restitution is exactly e = 1.0. This means the relative speed of separation after collision equals the relative speed of approach before collision.
Are there any perfectly elastic collisions in real life?
Macroscopic collisions (like billiard balls or steel spheres) are never perfectly elastic because a tiny fraction of energy is always lost to sound and heat. However, collisions between subatomic particles (like electrons or gas molecules) can be perfectly elastic.
What happens when two identical masses collide elastically in 1D?
If one identical mass is stationary and another strikes it elastically in 1D, they completely exchange velocities. The moving mass stops completely, and the stationary mass moves off with the initial velocity of the first mass.
How does Newton's Cradle demonstrate elastic collisions?
Newton's Cradle consists of suspended identical steel spheres. When one sphere is pulled back and released, it strikes the others, passing momentum and energy through elastic collisions so that exactly one sphere swings out on the opposite end.
What is the formula for final velocities in a 1D elastic collision?
The final velocities v1f and v2f are calculated using: v1f = [(m1 - m2)*u1 + 2*m2*u2] / (m1 + m2) and v2f = [(m2 - m1)*u2 + 2*m1*u1] / (m1 + m2).
What does a 90-degree scattering angle mean in a 2D elastic collision?
When an object strikes an identical stationary object off-center in a perfectly elastic 2D collision, the two objects will always travel along directions that are perpendicular (90° apart) after the collision.
Is total momentum conserved in an elastic collision?
Yes. Total linear momentum is always conserved in all closed systems, including both elastic and inelastic collisions.
How does the gas molecule simulator demonstrate elastic collisions?
The simulator models gas particles as elastic hard discs. They bounce off each other and the walls with no energy loss, demonstrating how chaotic collisions maintain a constant total kinetic energy while redistributing individual speeds into a Maxwell-Boltzmann distribution.
Why does kinetic energy remain constant in the simulator?
Because the coefficient of restitution is set to e = 1.0, which mathematically enforces that the relative speed of separation equals the relative speed of approach, conserving KE_total = 1/2 * m1 * v1^2 + 1/2 * m2 * v2^2.