Momentum and Collisions Simulation

About this experiment

Send two carts toward one another and compare their velocities before and after contact. Choose an elastic collision or reduce the coefficient of restitution to remove kinetic energy while conserving total momentum. The carts start 4 m apart with 0.5 m-wide bodies; the camera follows them so both remain visible.

Learning goals

Apply conservation of momentum to an isolated two-body collision
Use restitution to distinguish elastic and inelastic collisions
Separate kinetic energy conversion from momentum conservation
Compare the velocity-time graphs of elastic and inelastic collisions

How to use it

1Play the default collision and inspect the velocity jumps at the same contact time.
2Set both masses equal with restitution 1. Check that the carts exchange velocities.
3Set restitution to zero and compare the common final velocity with total momentum divided by total mass.
4Choose initial velocities that separate the carts. Check that no collision occurs and no kinetic energy is lost.
5Make cart B four times heavier than cart A and send A into B at rest with restitution 1. Describe what happens to A.
6Use the time slider to stop just before and just after the collision and check that total momentum stays the same.

The model

m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}

Momentum conservation during the short collision

v_{2} - v_{1} = e (u_{1} - u_{2})

Relative separation speed from restitution e

Δ E = \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} (1 - e^{2}) (u_{1} - u_{2})^{2}

Kinetic energy converted during an approaching collision

t_{c} = \frac{4 - 0.5}{u _{1} - u _{2}} (u_{1} > u_{2})

Contact time from the starting center separation and cart widths

Assumptions

Motion is one-dimensional on a frictionless horizontal track, with no external horizontal impulse.
The collision is instantaneous. Cart deformation, sound production and the finite contact-force profile are not simulated.
Both cart widths are fixed at 0.5 m independently of mass. The moving camera changes the picture scale.
For e = 0, the carts share a common final velocity and remain in contact. Converted kinetic energy is not assigned to a specific internal process.

Try this in class

Q1Can total momentum be zero while total kinetic energy is positive?
Q2For an elastic collision between equal masses, why are the velocities exchanged?
Q3Why does a zero-restitution collision conserve momentum even though kinetic energy decreases?
Q4A heavy truck and a small car collide head on and stick together. Which one has the larger change in velocity? Which has the larger change in momentum?
Q5Why do crash safety engineers want cars to crumple rather than bounce?

Frequently asked questions

Why is there no collision for some velocity settings?

The left cart must move toward the right cart faster than the right cart moves away. If u1 is less than or equal to u2, their separation never decreases.

Why does the picture zoom as the carts separate?

The camera follows both carts. Numerical velocities, collision time and conserved momentum remain in physical units and do not depend on the camera scale.

Where does the lost kinetic energy go?

In a real collision it becomes heat, sound and the energy needed to permanently deform the objects. Momentum cannot be lost in the same way because the two carts push on each other with equal and opposite forces for the same time.