Elastic and Inelastic Collisions: PECTAA Class 11 Physics, Punjab Board

Drop a tennis ball and it bounces back, but never quite to the height you dropped it from. Some of its kinetic energy has gone into heat and sound. Whether kinetic energy survives a collision is exactly what separates elastic from inelastic collisions.

Chapter 2 of the PECTAA Physics 11 book is Force and Motion. After momentum, impulse and the law of conservation of momentum (section 2.7), section 2.8 deals with elastic and inelastic collisions and section 2.9 with inelastic collision in one dimension. This note is for FSc Part 1 (1st year, Class 11) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

Two kinds of collision (section 2.8). Momentum of an isolated system is conserved in every collision. Kinetic energy is a different matter.

• Inelastic collision: the kinetic energy of the system is not conserved. Part of it is lost, for example as heat and sound when the molecules of a ball rub against each other on impact.
• Elastic collision: the kinetic energy of the system is conserved. A ball that loses a negligible amount of energy against the floor and returns almost to its starting height is close to elastic.

Elastic collision in one dimension. Two smooth, non-rotating balls of masses m₁ and m₂ move along one straight line with velocities v₁ and v₂, collide, and move on along the same line with v′₁ and v′₂. Take the right as positive.

1. Momentum: m₁v₁ + m₂v₂ = m₁v′₁ + m₂v′₂ (Eq. 2.29 in rearranged form).
2. Kinetic energy: ½m₁v₁² + ½m₂v₂² = ½m₁v′₁² + ½m₂v′₂² (Eq. 2.30).
3. Dividing one by the other gives v₁ + v′₁ = v₂ + v′₂, that is v₁ − v₂ = −(v′₁ − v′₂).

So the relative velocity of approach equals the relative velocity of separation, reversed.

Final velocities when v₂ = 0 (Eqs. 2.32 and 2.33):

• v′₁ = [(m₁ − m₂)/(m₁ + m₂)] v₁
• v′₂ = [2m₁/(m₁ + m₂)] v₁

Special cases (from the book's figures 2.11 to 2.14).

• Equal masses: the bodies exchange velocities.
• Equal masses, m₂ at rest: m₁ stops and m₂ moves off with v₁. This is the billiard-ball case.
• Light body hits a massive body at rest (m₂ ≫ m₁): v′₁ ≈ −v₁ and v′₂ ≈ 0. The light body bounces straight back. The book links this to a squash player's ball and the wall.
• Massive body hits a light body at rest (m₁ ≫ m₂): v′₁ ≈ v₁ and v′₂ ≈ 2v₁. The heavy body hardly changes speed; the light one shoots forward at about twice the incident velocity.

Perfectly inelastic collision (section 2.9). The bodies stick together and move as one mass m₁ + m₂ with common velocity v. Momentum is still conserved:

m₁v₁ + m₂v₂ = (m₁ + m₂)v, so v = (m₁v₁ + m₂v₂)/(m₁ + m₂).

If m₂ is at rest, v = [m₁/(m₁ + m₂)] v₁: the velocity is reduced by the mass ratio.

Key points

• Momentum is conserved in all collisions of an isolated system. Kinetic energy is conserved only in elastic ones.
• Keep signs: a velocity to the left is negative.
• A negative v′₁ means the first body rebounds.
• In a perfectly inelastic collision there is only one final velocity.

Worked examples

Example 1 (book Example 2.9). A 70 g ball collides with a 140 g ball at rest. The first ball moves at 9 m s⁻¹ to the right. If the collision is perfectly elastic, find both velocities afterwards.

1. v′₁ = [(70 − 140)/(70 + 140)] × 9 = (−70/210) × 9 = −3 m s⁻¹. The 70 g ball rebounds to the left.
2. v′₂ = [2 × 70/(70 + 140)] × 9 = (140/210) × 9 = 6 m s⁻¹ to the right.
3. Check kinetic energy (masses in kg): before, ½ × 0.07 × 81 = 2.835 J. After, ½ × 0.07 × 9 + ½ × 0.14 × 36 = 0.315 + 2.52 = 2.835 J. Conserved.

Example 2 (perfectly inelastic). A 2 kg trolley moving at 6 m s⁻¹ hits and sticks to a 1 kg trolley at rest.

1. v = (2 × 6)/(2 + 1) = 4 m s⁻¹.
2. Kinetic energy before = ½ × 2 × 36 = 36 J; after = ½ × 3 × 16 = 24 J.
3. 12 J of kinetic energy has changed into other forms, so the collision is inelastic.

Common mistakes in the exam

1. Saying momentum is lost in an inelastic collision. Only kinetic energy is lost; momentum is conserved.
2. Dropping the minus sign in m₁ − m₂ when m₂ is larger.
3. Swapping the formulas. The factor 2m₁ belongs to v′₂, the body that was at rest.
4. Mixing grams and kilograms in energy checks. The velocity formulas use a mass ratio, so units cancel, but energies need kg.
5. Writing two final velocities for bodies that stick together.

Exam technique

• Draw "before" and "after" sketches with arrows and a stated positive direction.
• Write the conservation law first, then substitute.
• For elastic collisions, a quick kinetic energy check catches arithmetic slips.
• Learn the four special cases with one line of reasoning each.
• Check your board's notice for the sections in your paper.

Practice questions

1. A 2 kg ball at 4 m s⁻¹ hits an identical ball at rest elastically. Find both final velocities. Answer: The first stops (0 m s⁻¹); the second moves at 4 m s⁻¹.
2. In an elastic collision, two balls approach each other with relative speed 5 m s⁻¹. What is their speed of separation? Answer: 5 m s⁻¹.
3. A 1 kg body at 10 m s⁻¹ sticks to a 4 kg body at rest. Find the common velocity. Answer: 10/5 = 2 m s⁻¹.
4. A table tennis ball hits a heavy wall elastically at 8 m s⁻¹. What is its velocity after the hit? Answer: About −8 m s⁻¹, straight back.
5. Which quantity is conserved in every collision of an isolated system? Answer: Linear momentum.

Ready to practise this? Continue in 1st Year Physics.