Momentum, Impulse and Conservation of Momentum: PECTAA Class 11 Physics, Punjab Board

A cricket ball and a loaded truck both move at 10 m s⁻¹. Which is harder to stop? The truck, because it has more momentum. This note turns that everyday idea into three tools the paper keeps asking for: momentum, impulse and the conservation law.

Chapter 2 of the PECTAA Physics 11 book is Force and Motion. Section 2.7 covers momentum, impulse and the law of conservation of momentum. 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

Linear momentum. A moving body is hard to stop. The faster it moves, the harder; and of two bodies at the same speed, the heavier is harder to stop. Newton called this property momentum. The book defines linear momentum as

p = mv (equation 2.25)

where v is the velocity of mass m. Momentum is a vector, in the direction of v. Its unit is kg m s⁻¹, which can also be written N s.

Momentum and Newton's second law. Let a force F act on mass m for time t, changing its velocity from vᵢ to v_f. The acceleration is a = (v_f − vᵢ)/t, and Newton's second law gives a = F/m. Equating:

F × t = mv_f − mvᵢ (equation 2.26)

So force equals the time rate of change of momentum. The book notes this form is more general than F = ma, because it also handles cases where the mass changes, such as a rocket burning and ejecting fuel.

Impulse. When a bat hits a ball, the force changes from instant to instant and acts for a very short time. Then we use the average force. Impulse is the average force times the time it acts:

Impulse = F × t = mv_f − mvᵢ (equation 2.27)

Isolated system. A system on which no external agency exerts a force. The book's example is gas molecules in a closed glass vessel at constant temperature: they collide with each other, but nothing outside pushes them.

Conservation of momentum. Two balls m₁ and m₂ collide. During contact, ball 1 feels force F and ball 2 feels the reaction F′ = −F, for the same time t. Adding the impulse equations, the left side is zero, so

m₁v₁ + m₂v₂ = m₁v₁′ + m₂v₂′ (equation 2.28)

Total momentum before equals total momentum after. In words: the total linear momentum of an isolated system remains constant. Because momentum is a vector, signs for direction matter.

Key points

• Same change, longer time, smaller force. For a fixed change in momentum, stretching the contact time lowers the force. The book's side notes use this for seat belts, padded helmets and even hair cushioning the skull.
• Choose a positive direction first and give opposite velocities a minus sign.
• Momentum is conserved in every collision of an isolated system, whether or not kinetic energy is.

Worked examples

Example 1 (book Example 2.7). A 1500 kg car slows from 20 m s⁻¹ to 15 m s⁻¹ in 3.0 s. Find the average retarding force.

1. F × t = mv_f − mvᵢ.
2. F × 3.0 = 1500 × 15 − 1500 × 20 = −7500 kg m s⁻¹.
3. F = −2500 N = −2.5 kN. The minus sign shows the force opposes the motion.

Example 2 (book Example 2.8, with directions corrected). Balls of 2.0 kg and 3.0 kg move towards each other at 6.0 m s⁻¹ and 4.0 m s⁻¹. After the collision the bigger ball moves at 3.0 m s⁻¹. Find the velocity of the smaller ball.

1. Take the 2 kg ball's first direction as positive. Momentum before = 2 × 6 + 3 × (−4) = 12 − 12 = 0.
2. The bigger ball bounces back, so its velocity after is +3.0 m s⁻¹.
3. 0 = 2v + 3 × 3, so v = −4.5 m s⁻¹.
4. The smaller ball moves back at 4.5 m s⁻¹, opposite to its first direction.

Example 3. A 0.15 kg ball hits a bat at 20 m s⁻¹ and leaves at 30 m s⁻¹ in the opposite direction. Find the impulse.

1. Take the outgoing direction as positive: vᵢ = −20, v_f = +30.
2. Impulse = 0.15 × 30 − 0.15 × (−20) = 4.5 + 3.0 = 7.5 N s.

Common mistakes in the exam

1. Ignoring direction. Momentum is a vector. Opposite motions need opposite signs.
2. Writing the unit as kg m s⁻². Momentum is kg m s⁻¹ (or N s).
3. Using conservation when an outside force acts. The law holds only for an isolated system.
4. Mixing impulse and force. Impulse is force times time, and equals the change in momentum.
5. Dropping the minus sign in a retarding force. Keep it, or state the direction in words.

Exam technique

• State the law in the book's words before using it: "The total linear momentum of an isolated system remains constant."
• Write "take ... as positive" at the start of every collision problem.
• Show momentum before and after on separate lines, then equate.
• Check your board's notice for the topics in your paper.

Practice questions

1. Find the momentum of a 0.5 kg ball moving at 8 m s⁻¹. Answer: p = 0.5 × 8 = 4 kg m s⁻¹.
2. A force of 10 N acts for 0.2 s. Find the impulse. Answer: 10 × 0.2 = 2 N s.
3. A 2 kg trolley at 3 m s⁻¹ hits and sticks to a resting 1 kg trolley. Find their common velocity. Answer: 2 × 3 = 3v, so v = 2 m s⁻¹.
4. Why does a padded helmet reduce injury? Answer: It stretches the time of the collision, so the same change in momentum needs a smaller force.
5. State Newton's second law in terms of momentum. Answer: The time rate of change of momentum of a body equals the applied force.

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