Collisions in Two Dimensions and Rocket Propulsion: PECTAA Class 11 Physics, Punjab Board

On a carrom board, the striker hits a piece and both fly off at angles. Nothing moves along a single line any more, yet momentum is still conserved, one direction at a time. The same idea, pushed backwards through an exhaust, lifts a rocket.

Chapter 2 of the PECTAA Physics 11 book is Force and Motion. After collisions in one dimension (sections 2.8 and 2.9), the book treats collisions in two dimensions (section 2.10 onward) and then rocket propulsion (section 2.12). 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

Elastic collision in two dimensions (section 2.10). Two balls of masses m₁ and m₂ move along the x-axis with v₁ and v₂ and collide elastically. Afterwards they move off with v′₁ and v′₂ at angles θ₁ and θ₂ to the x-axis, on opposite sides of it. Because the collision is elastic, both momentum and kinetic energy are conserved. Momentum is a vector, so it is resolved along the axes:

• Along x (Eq. 2.35): m₁v₁ + m₂v₂ = m₁v′₁ cos θ₁ + m₂v′₂ cos θ₂
• Along y (Eq. 2.36): 0 = m₁v′₁ sin θ₁ − m₂v′₂ sin θ₂
• Kinetic energy (Eq. 2.37): ½m₁v₁² + ½m₂v₂² = ½m₁v′₁² + ½m₂v′₂²

These three equations solve two-dimensional elastic problems.

Perfectly inelastic collision in two dimensions. Everyday (macroscopic) collisions are generally inelastic. In a perfectly inelastic one the bodies stick together. Let m₁ move along the x-axis with v₁, and m₂ move at angle θ to the x-axis with v₂. They meet at the origin and move on as one mass M = m₁ + m₂ with velocity v_f at angle φ.

• x-direction (Eq. 2.38): m₁v₁ + m₂v₂ cos θ = Mv_f cos φ
• y-direction (Eq. 2.39): 0 + m₂v₂ sin θ = Mv_f sin φ

These two equations give v_f and φ.

Kinetic energy loss. Kinetic energy is not conserved here. Since it is a scalar, speeds are used directly:

• initial: ½m₁v₁² + ½m₂v₂²
• final: ½Mv_f²
• loss: ΔK.E. = initial − final

The lost energy becomes heat, sound or deformation. The book's examples: a karate chop breaking bricks, a car crash where crumpling stops the cars bouncing back, and a bat striking a ball that deforms.

Rocket propulsion (section 2.12). A rocket expels burning gas through engines at its rear. The gas gains momentum backwards, and the rocket gains equal momentum forwards. As long as the engines keep firing, the rocket keeps gaining momentum, so it goes faster and faster rather than at steady speed.

• It carries its own fuel and oxygen, so it can work at heights with little or no air.
• The book states that a typical rocket burns about 10 000 kg of fuel per second, ejects gas at over 4000 m s⁻¹, and that more than 80% of its launch mass is fuel.
• To deal with the mass of fuel, several rockets are linked; each is discarded when its fuel is used.

Thrust and acceleration (Eq. 2.42). If mass m of gas is ejected per second with velocity v relative to the rocket, the momentum given to the gas each second is mv. This equals the thrust on the rocket, so its acceleration is a = mv/M, where M is the rocket's mass. As fuel burns, M falls, so the acceleration increases.

Key points

• In 2D, write one momentum equation per axis.
• Elastic: add the kinetic energy equation. Inelastic: compute the energy loss.
• Thrust = mv; acceleration a = mv/M rises as M falls.

Worked examples

Example 1 (perfectly inelastic, at right angles). A 2 kg ball moving at 3 m s⁻¹ along x sticks to a 1 kg ball moving at 6 m s⁻¹ along y (θ = 90°).

1. x: 2 × 3 = 6 = 3v_f cos φ.
2. y: 1 × 6 = 6 = 3v_f sin φ.
3. tan φ = 1, so φ = 45°; v_f = √(6² + 6²)/3 = √72/3 ≈ 2.83 m s⁻¹.
4. Initial K.E. = ½(2)(9) + ½(1)(36) = 9 + 18 = 27 J. Final K.E. = ½(3)(2.83)² ≈ 12 J. Loss ≈ 15 J.

Example 2 (rocket thrust). Using the book's figures, 10 000 kg s⁻¹ ejected at 4000 m s⁻¹:

1. Thrust = mv = 10 000 × 4000 = 4 × 10⁷ N.
2. For a rocket of mass 2 × 10⁶ kg, a = 4 × 10⁷ / 2 × 10⁶ = 20 m s⁻² (before allowing for gravity).

Common mistakes in the exam

1. Adding momenta as plain numbers in 2D. Resolve into x and y first.
2. Resolving kinetic energy. It is a scalar; use speeds.
3. Assuming kinetic energy is conserved when bodies stick together.
4. Saying the rocket pushes on the air. It pushes on its own exhaust gas.
5. Keeping M constant for a rocket over a long burn.

Exam technique

• Draw before-and-after sketches with angles marked from the x-axis.
• Label each equation "x-direction" or "y-direction".
• In rocket answers, state that thrust equals the rate of change of momentum of the gas.
• Check your board's notice for the sections in your paper.

Practice questions

1. Which quantities are conserved in a 2D elastic collision? Answer: Momentum (along each axis) and kinetic energy.
2. Why can a rocket work where there is no air? Answer: It carries its own oxygen and pushes against its own exhaust.
3. Gas leaves a rocket at 50 kg s⁻¹ and 2000 m s⁻¹. Find the thrust. Answer: 50 × 2000 = 1 × 10⁵ N.
4. What happens to a rocket's acceleration as fuel is used, if thrust stays the same? Answer: It increases, because M decreases.
5. Where does lost kinetic energy go in a car crash? Answer: Into heat, sound and damage to the vehicles.

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