Interactive physics simulator
Inelastic Collision
In inelastic collisions, momentum is preserved while kinetic energy is lost. Explore air track gliders that stick together, vehicle crumple zones that absorb impact forces, and ballistic pendulums that measure bullet velocities.
Inelastic Collision Simulator
Configure velocities, mass values, and slider settings, then run the simulation to monitor conservation plots.
Live Telemetry
- Bullet Velocity
- 250 m/s
- Bullet Mass
- 0.02 kg
- Block Mass
- 2.5 kg
- Post-Collision Speed
- 0.0 m/s
- Swing Height (h)
- 0.00 m
- Swing Angle (θ)
- 0.0°
- Bullet Kinetic Energy
- 625.0 J
- Kinetic Energy Loss
- 0.00 J
What is an Inelastic Collision?
An inelastic collision is a collision in which total linear momentum is conserved, but the total kinetic energy of the system decreases. In any closed system free of external forces, the total momentum before collision equals the total momentum after collision:
However, the kinetic energy is not conserved because some portion of it is converted into internal work (plastically deforming materials, folding panels, creating sparks), thermal energy (heat generated by friction and shear), and acoustic vibrations (sound of impact). For a general inelastic collision, the post-collision velocities in one dimension are calculated using the coefficient of restitution (e), which satisfies 0 ≤ e < 1.0:
Perfectly Inelastic Limit
When the coefficient of restitution is zero, the maximum kinetic energy is converted:
- Gliders stick together: Upon contact, the gliders couple (e.g. using velcro or clay) and move with a single, joint final velocity: vf = (m1u1 + m2u2) / (m1 + m2).
- Maximum energy loss: While momentum conservation is strictly obeyed, the loss of kinetic energy is mathematically maximized.
- Zero rebound: The relative velocity of separation is exactly zero (v2 - v1 = 0).
Crumple Zones & Impulse
Safety structures exploit momentum changes to protect vehicle occupants:
- In a crash, the passenger compartment must be brought to rest, requiring a fixed change in momentum: J = Δp = F · Δt.
- A rigid steel bumper stops the car almost instantly (Δt is extremely small), yielding a massive peak impact force.
- Controlled crumpling deforms materials plastically, lengthening the deceleration window (Δt is 5 to 10 times longer). This dampens the peak impact force, protecting passengers.
The Ballistic Pendulum
A classic physics lab setup used to measure high-velocity bullets:
- A light bullet strikes a heavy, hanging wooden block and embeds inside (completely inelastic, e = 0.0).
- The collision is so fast that momentum is conserved before the block swings: m · u = (m + M) · vjoint.
- The combined block-bullet then swings upwards, converting all post-collision kinetic energy into gravitational potential energy: 1/2 · (m+M)vjoint2 = (m+M)g · h.
Coefficients of Restitution
Everyday objects display varying levels of collision elasticity:
- Superball: e ≈ 0.90 (highly elastic, bounces back near its initial height).
- Billiard Ball: e ≈ 0.95 (nearly elastic, loses very little energy to click sounds).
- Tennis Ball: e ≈ 0.75 (standard bounce on concrete).
- Lead Sphere or Clay: e ≈ 0.0 (deforms plastically, no rebound).
Solved Examples
A 1.5 kg air track glider (Glider A) moving East at +6.0 m/s makes a head-on completely inelastic collision with a stationary 2.5 kg glider (Glider B). Find their final velocity and kinetic energy loss.
- Identify initial values: mA = 1.5 kg, uA = +6.0 m/s, mB = 2.5 kg, uB = 0 m/s.
- Since the collision is completely inelastic, they stick together, meaning the coefficient of restitution e = 0.0.
- Calculate combined mass: mtotal = mA + mB = 1.5 + 2.5 = 4.0 kg.
- Apply momentum conservation: Pi = mA * uA + mB * uB = (1.5 * 6.0) + (2.5 * 0) = 9.0 kg·m/s.
- Solve for joint final velocity: vf = Pi / mtotal = 9.0 / 4.0 = +2.25 m/s.
- Calculate initial kinetic energy: KEi = 1/2 * mA * uA2 = 0.5 * 1.5 * 36.0 = 27.0 J.
- Calculate final kinetic energy: KEf = 1/2 * mtotal * vf2 = 0.5 * 4.0 * (2.25)2 = 2.0 * 5.0625 = 10.125 J.
- Determine kinetic energy loss: ΔKE = KEi - KEf = 27.0 - 10.125 = 16.875 J (62.5% of initial energy lost to deformation and heat).
Answer: Final Velocity: +2.25 m/s | Kinetic Energy Loss: 16.875 J (62.5%)
A 1200 kg passenger vehicle traveling at 15 m/s crashes into a solid concrete barrier. If the bumper has a rigid steel frame, it stops in 0.05 seconds. If it features a controlled crumple zone, it stops in 0.25 seconds. Compare the peak forces experienced in both crashes.
- Calculate initial momentum: pi = m * u = 1200 * 15 = 18,000 kg·m/s. Final momentum: pf = 0.
- Calculate change in momentum (impulse): Δp = pf - pi = -18,000 kg·m/s.
- Case 1: Rigid Steel Bumper (Δt = 0.05 s). Average force: Favg = Δp / Δt = -18,000 / 0.05 = -360,000 N.
- Case 2: Crumple Zone Bumper (Δt = 0.25 s). Average force: Favg = Δp / Δt = -18,000 / 0.25 = -72,000 N.
- Adding a deforming crumple zone increases contact time 5-fold, which yields a 5x reduction in peak impact force (from 360 kN to 72 kN), preserving passenger cabin integrity.
Answer: Rigid Steel Bumper: 360,000 N | Controlled Crumple Zone: 72,000 N
A 10 g (0.01 kg) bullet is fired horizontally into a 2.49 kg wooden block suspended by a light cord 2.0 m long. The bullet embeds itself in the block, and the joint system swings up to a peak height of 0.20 m. Find the muzzle velocity of the bullet.
- Let bullet mass m = 0.01 kg and block mass M = 2.49 kg. Combined mass: m + M = 2.50 kg.
- The swing phase conserves mechanical energy: 1/2 * (m+M) * vjoint2 = (m+M) * g * h.
- Solve for joint velocity immediately after collision: vjoint = √(2 * g * h) = √(2 * 9.8 * 0.20) = √(3.92) = 1.98 m/s.
- The collision phase conserves linear momentum: m * ubullet = (m + M) * vjoint.
- Solve for initial bullet speed: ubullet = [(m + M) * vjoint] / m = [2.50 * 1.98] / 0.01 = 495 m/s.
Answer: Initial Bullet Speed: 495 m/s
Common Misconceptions
- "Momentum is only conserved in elastic collisions": False. Momentum is conserved in all isolated collisions, regardless of whether they are elastic, inelastic, or completely inelastic.
- "Lost kinetic energy is destroyed": False. Energy is never destroyed. The lost mechanical kinetic energy is converted into thermal energy, acoustic wave energy, and chemical/structural bonds within the deformed materials.
- Inelastic collisions must stick together: False. Sticking together is only the extreme limit of inelastic collisions (perfectly inelastic, e = 0). Most everyday collisions are partially inelastic (0 < e < 1), where objects bounce but lose energy.
Practice Questions
1. Explain why kinetic energy is not conserved in an inelastic collision, even though total energy is conserved.
According to the Law of Conservation of Energy, total energy in a closed system is always constant. In an inelastic collision, while total linear momentum is preserved, mechanical kinetic energy is transformed into non-mechanical forms of energy. These include thermal energy (frictional heat inside the deforming bodies), sound waves (the acoustic emission of the impact), and internal work (elastic/plastic stress deforming structural materials). thus, kinetic energy decreases, but the energy itself is not destroyed, keeping total energy conserved.
2. Show that a completely inelastic collision between a moving mass and an identical stationary mass loses exactly 50% of the initial kinetic energy.
Let the moving mass be m1 = m with initial velocity u, and the target mass be m2 = m with u2 = 0. From momentum conservation: m * u = (m + m) * vf, which gives a final joint velocity vf = u / 2. The initial kinetic energy is KEi = 1/2 * m * u2. The final kinetic energy is KEf = 1/2 * (2m) * vf2 = m * (u/2)2 = m * u2 / 4 = 1/2 * KEi. Therefore, exactly 50% of the system kinetic energy is converted into other forms.
3. Define the coefficient of restitution (e) and explain its limits.
The coefficient of restitution (e) is a dimensionless ratio describing the elasticity of a collision: e = (v2 - v1) / (u1 - u2), representing the relative separation speed divided by the relative approach speed. The limits are: (1) e = 1.0 represents a perfectly elastic collision where kinetic energy is fully conserved. (2) 0 < e < 1.0 represents inelastic collisions where objects bounce apart but lose kinetic energy. (3) e = 0.0 represents a completely inelastic collision where objects stick together.
4. Why is a ballistic pendulum a highly reliable device for measuring high-speed projectiles?
High-speed projectiles are difficult to measure directly without complex electronic chronographs. A ballistic pendulum solves this by converting a high-velocity, microsecond impact into a slow, easily measurable gravitational swing height (h) of a heavy block. Because the collision is completely inelastic (e = 0.0), momentum conservation applies during the impact, and conservation of mechanical energy applies during the swing. Measuring the height displacement lets us calculate bullet velocity with high accuracy.
FAQ
Frequently Asked Questions
What is an inelastic collision in physics?
An inelastic collision is a collision in which total linear momentum is conserved, but total kinetic energy decreases. The lost kinetic energy is converted into heat, sound, and work done to permanently deform the colliding bodies.
How does an inelastic collision differ from an elastic collision?
In both elastic and inelastic collisions, momentum is conserved. However, kinetic energy is conserved ONLY in an elastic collision (coefficient of restitution e = 1.0). In an inelastic collision, kinetic energy is lost (e < 1.0).
What is a completely (or perfectly) inelastic collision?
A completely inelastic collision occurs when the colliding objects stick together after the impact and move with a single final velocity. This collision mode yields the maximum possible kinetic energy loss consistent with momentum conservation, and corresponds to a coefficient of restitution of e = 0.0.
What is the coefficient of restitution (e) for an inelastic collision?
For inelastic collisions, the coefficient of restitution is in the range 0 ≤ e < 1.0. Completely inelastic collisions have e = 0.0, while partially inelastic collisions fall between 0.0 and 1.0.
Where does the lost kinetic energy go in an inelastic collision?
The lost kinetic energy is transformed into non-mechanical energy: heat (raising the temperature of the materials), acoustic energy (sound of impact), and plastic work (bending, denting, or crumpling structures).
How does a ballistic pendulum demonstrate inelastic collision?
A ballistic pendulum fires a bullet into a suspended block, where it embeds itself (completely inelastic, e = 0.0). By transferring momentum to the block, it swings upwards. Measuring the height of the swing allows us to calculate the initial bullet velocity.
What is a crumple zone and how does it protect passengers?
A crumple zone is a structural safety feature in cars designed to deform plastically during a crash. By crumpling, it increases the collision duration (Δt), which dramatically lowers the peak impact force (F = Δp / Δt) transferred to the occupants.
Is total energy conserved in an inelastic collision?
Yes. Total energy is always conserved in the universe according to the First Law of Thermodynamics. While kinetic energy is not conserved (it decreases), that energy is converted into heat, sound, and structural deformation, never destroyed.
Can a collision be inelastic if no objects stick together?
Yes. Any collision with 0.0 < e < 1.0 is inelastic. For example, two billiard balls or rubber balls bouncing off each other lose a tiny fraction of kinetic energy to clicking sounds and friction, which makes them inelastic.
What is the formula for a completely inelastic collision in one dimension?
Since the two objects stick together, they share a final velocity vf. The formula derived from momentum conservation is: m1 * u1 + m2 * u2 = (m1 + m2) * vf, solving to vf = (m1 * u1 + m2 * u2) / (m1 + m2).
Why does a rigid bumper lead to higher impact forces than a crumple zone?
A rigid bumper does not deform, causing the vehicle to halt almost instantaneously. This extremely brief contact duration (Δt) means momentum changes rapidly, creating a massive, hazardous peak impact force (F = dp/dt) on the cabin.