Momentum and kinetic energy in interactions

Section: Dynamics  |  Syllabus: Cambridge AS Level Physics (9702)

Elastic and Inelastic Interactions

While momentum is always conserved in interactions, the total kinetic energy (E_k) may not be. Kinetic energy is calculated as E_k = (1/2)mv^2.

Definitions Worked Example: Snooker Ball (Elastic)

A 0.16 kg white ball at 1.0 m s^-1 hits a stationary 0.16 kg red ball. If the white ball stops and the red ball moves off at 1.0 m s^-1:

KE Calculation

Total E_k, before = (1/2)(0.16)(1.0)^2 = 0.08 J

Total E_k, after = (1/2)(0.16)(1.0)^2 = 0.08 J

Since E_k is conserved, the collision is perfectly elastic.

Worked Example: Railway Wagons (Inelastic)

A 1900 kg wagon at 3 m s^-1 hits a stationary 2200 kg wagon. They join and move together. Is it elastic?

Solution
  1. Find Velocity (v): 1900 × 3 = (1900 + 2200)v v ≈ 1.4 m s^-1
  2. Initial KE: (1/2)(1900)(3)^2 = 8550 J
  3. Final KE: (1/2)(4100)(1.4)^2 ≈ 4018 J

Significant KE is lost (8550 → 4018), so the collision is inelastic.

Relative Speed of Approach and Separation

Relative speed is the speed of one object compared to another. It provides a quick way to test for elasticity.

Perfectly Elastic Collision:

Relative Speed of Approach = Relative Speed of Separation

Calculating Relative Speed:

Diagram: Relative Speed Visualization

Show two cars on a road: (1) approaching each other (large relative speed) and (2) one overtaking the other (small relative speed).

Worked Example: Repulsion without Contact

An alpha particle (m) at v_1 approaches a stationary helium atom (m). They repel via electrostatic forces. If they don't touch, is it elastic?

Analysis

Interactions that involve non-contact forces (like electric repulsion or magnets) are often perfectly elastic because no energy is lost to heat or permanent deformation.

Explosions and Energy

In physics, a separation or explosion is the reverse of a collision. Separations are always inelastic.

The kinetic energy gained by the moving parts comes from stored potential energy (e.g., chemical energy in gunpowder or elastic potential in a compressed spring).

Worked Example: Gun Recoil Energetics

A 3 kg gun fires a 5 g bullet at 400 m s^-1. Initial KE is zero.

Solution

Total E_k, after = E_k, bullet + E_k, gun

E_k, bullet = (1/2)(0.005)(400)^2 = 400 J

E_k, after ≈ 400 J (neglecting small gun recoil KE).

Since KE increased from 0 to 400 J, energy was transferred from chemical stores. This is inelastic.

Core Summary

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