1. Elastic Collisions in One Dimension
For an elastic collision, both momentum and mechanical energy are conserved.
In one dimension,
𝑚 𝐴 𝑣 𝐴1𝑥 + 𝑚 𝐵 𝑣 𝐵1𝑥 = 𝑚 𝐴 𝑣 𝐴2𝑥 + 𝑚 𝐵 𝑣 𝐵2𝑥
1 2 𝑚 𝐴 𝑣 𝐴1𝑥 𝑚 𝐵 𝑣 𝐵1𝑥 2 = 𝑚 𝐴 𝑣 𝐴2𝑥 𝑚 𝐵 𝑣 𝐵2𝑥 2
Given the masses and initial velocities, we can solve for the final velocities.
We find that the relative velocities before and after the collision are equal magnitude but opposite sign:
𝑣 𝐴1𝑥 − 𝑣 𝐵1𝑥 =− 𝑣 𝐴2𝑥 − 𝑣 𝐵2𝑥

2. Elastic Collisions in One Dimension
For the special case where one body is initially at rest, the velocities after the collision reduce to:
𝑣 𝐴2𝑥 = 𝑚 𝐴 − 𝑚 𝐵 𝑚 𝐴 + 𝑚 𝐵 𝑣 𝐴1𝑥
𝑣 𝐵2𝑥 = 2 𝑚 𝐴 𝑚 𝐴 + 𝑚 𝐵 𝑣 𝐴1𝑥

3. Collision Example #2

4. Center of Mass
For a collection of particles, we can define the center of mass as the mass-weighted average position of the particles.
𝑥 cm = 𝑚 1 𝑥 1 + 𝑚 2 𝑥 2 + 𝑚 3 𝑥 3 +… 𝑚 1 + 𝑚 2 + 𝑚 3 +… = 𝑖 𝑚 𝑖 𝑥 𝑖 𝑖 𝑚 𝑖
𝑦 cm = 𝑚 1 𝑦 1 + 𝑚 2 𝑦 2 + 𝑚 3 𝑦 3 +… 𝑚 1 + 𝑚 2 + 𝑚 3 +… = 𝑖 𝑚 𝑖 𝑦 𝑖 𝑖 𝑚 𝑖
In vector form,
𝒓 cm = 𝑚 1 𝒓 1 + 𝑚 2 𝒓 𝟐 + 𝑚 3 𝒓 3 +… 𝑚 1 + 𝑚 2 + 𝑚 3 +… = 𝑖 𝑚 𝑖 𝒓 𝑖 𝑖 𝑚 𝑖
The sum of the masses is the total mass 𝑀.
𝑀= 𝑖 𝑚 𝑖

5. Center of Mass Example

6. Center of Mass of Solid Bodies
To find the center of mass of a solid body, we replace the sums with integrals.
For uniform bodies, the center of mass is the geometric center.
The center of mass will lie on an axis of symmetry, if there is one.
The center of mass does not have to be within the body.

7. Motion of the Center of Mass
Taking the time derivative of the center of mass position gives the center of mass velocity.
𝒗 cm = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 𝟐 + 𝑚 3 𝒗 3 +… 𝑚 1 + 𝑚 2 + 𝑚 3 +…
𝒗 cm = 𝑖 𝑚 𝑖 𝒗 𝑖 𝑀
We then see that the total momentum is the total mass times the velocity of the center of mass.
𝑀 𝒗 cm = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 𝟐 + 𝑚 3 𝒗 3 +…
𝑷 =𝑀 𝒗 cm

8. External Forces and Motion of the Center of Mass
Similarly, taking the time derivative of the center of mass velocity gives the center of mass acceleration.
𝒂 cm = 𝑚 1 𝒂 1 + 𝑚 2 𝒂 𝟐 + 𝑚 3 𝒂 3 +… 𝑚 1 + 𝑚 2 + 𝑚 3 +… = 𝑖 𝑚 𝑖 𝒂 𝑖 𝑖 𝑚 𝑖
The sum of all of the forces, internal and external is equal to the mass times acceleration.
By Newton’s third law, for each internal force, there is an equal but opposite reaction force, so the internal forces cancel out.
Σ 𝑭 ext =𝑀 𝒂 cm
So, the collection of particles moves as if all of the mass were located at the center of mass, and were acted on by only the external forces.

9. External Forces and Motion of the Center of Mass
We can also express this in terms of momentum.
Σ 𝑭 ext = 𝑑 𝑷 𝑑𝑡
Only external forces can change the total momentum.
This is a restatement of conservation of momentum.

10. Rocket Propulsion
In outer space, there is no atmosphere, so how do space ships maneuver?
Rocket propulsion
Conservation of momentum
Burned fuel is ejected at a high velocity from the rear of the rocket, propelling it forward.
The mass of the rocket is therefore decreasing.

11. Rocket Propulsion
Consider a rocket in outer space (no gravity or air resistance) with a mass 𝑚 and velocity of magnitude 𝑣.
In time 𝑑𝑡, the rocket ejects burned fuel and its mass changes by 𝑑𝑚, a negative quantity.
A positive mass of −𝑑𝑚 is ejected at a relative velocity of − 𝑣 ex
From conservation of momentum, we can find the thrust, or force exerted on the rocket, and its acceleration.
The mass of the rocket is therefore decreasing.

12. Rocket Propulsion The thrust is given by: 𝐹=− 𝑣 ex 𝑑𝑚 𝑑𝑡
From Newton’s second law, the acceleration in the absence of any other forces, such as gravity, is:
𝑎=− 𝑣 ex 𝑚 𝑑𝑚 𝑑𝑡
Integrating, we find that the change in velocity is given by:
𝑣− 𝑣 0 = 𝑣 ex ln 𝑚 0 𝑚

13. Rocket Propulsion Example

14. Chapter 8 Summary Momentum, Impulse, and Collisions
Newton’s second law: 𝑭 = 𝑑 𝒑 𝑑𝑡
Impulse: 𝑱 = 𝒑 2 − 𝒑 1 = 𝑡 1 𝑡 2 Σ 𝑭 𝑑𝑡 = 𝑭 ave Δ𝑡
Conservation of momentum
No external forces: 𝑷 = 𝑚 𝐴 𝒗 𝐴 + 𝑚 𝐵 𝒗 𝐵 +…=constant
Collisions – Momentum is conserved
Elastic – Kinetic energy is conserved
Inelastic – Loss of kinetic energy

15. Chapter 8 Summary Momentum, Impulse, and Collisions
Center of mass
𝒓 cm = 𝑚 1 𝒓 1 + 𝑚 2 𝒓 𝟐 + 𝑚 3 𝒓 3 +… 𝑚 1 + 𝑚 2 + 𝑚 3 +… = 𝑖 𝑚 𝑖 𝒓 𝑖 𝑖 𝑚 𝑖
Σ 𝑭 ext =𝑀 𝒂 cm = 𝑑 𝑷 𝑑𝑡
Rocket propulsion
Thrust: 𝐹=− 𝑣 ex 𝑑𝑚 𝑑𝑡
𝑣− 𝑣 0 = 𝑣 ex ln 𝑚 0 𝑚