Lesson 1: What is Center of Mass in a System of Particles?

1. What is the actual difference between Center of Mass and Center of Gravity, or are they the same thing?

They are almost always the same in standard physics problems, but technically different.

• Center of Mass (CoM): The average position of the mass of an object. It depends only on density and shape.
• Center of Gravity (CoG): The point where the total force of gravity effectively acts on the object.
• The Difference: If an object is so huge (like a skyscraper or a moon) that gravity is stronger at the bottom than the top, the CoG will be lower than the CoM. For everyday lab objects, assume they are the same.

2. When a question asks for the 'velocity of the Center of Mass,' is that just the average velocity of everything?

Yes, but it is a weighted average. You cannot just add up velocities and divide by the number of particles. You must multiply each object's mass by its velocity, sum them up, and divide by the total mass of the system.

Formula: v_cm = (m1v1 + m2v2 + ...) / M_total

3. Can the Center of Mass of an object be located in empty space where there is no actual material (like a donut)

Yes, absolutely. The Center of Mass is a geometric point, not necessarily a physical part of the object. For a donut, a ring, or a boomerang, the CoM is in the empty space in the middle.

4. When calculating Center of Mass for a continuous object (like a rod with changing density), how do I set up the integral for 'dm' in terms of 'dx'?

You need the Linear Mass Density, usually written as lambda (λ).

1. We know that dm = lambda * dx.
2. If the rod has uniform density, lambda is just Total Mass / Total Length.
3. If density changes (non-uniform), you are given a function, e.g., lambda = kx.
4. Substitute dm in the integral formula: X_cm = (1/M) * integral(x * lambda * dx)

Lesson 2: How to Apply Newton’s Second Law to a System of Particles?

5. In simple terms, what counts as a 'system' of particles, and how do I know if a system is isolated?

• System: Any collection of objects you choose to analyze inside a "box" (a mental boundary).
• Isolated: If no net external force acts on your box from the outside (no gravity, friction, or hands pushing), it is isolated. If the only forces are objects inside hitting each other, momentum is conserved.

6. During an explosion (like a firework), does the Center of Mass stop moving since the pieces fly everywhere?

No. The CoM continues strictly on its original path (a parabola). An explosion is an internal force. Internal forces cannot change the momentum of the system's Center of Mass. If the firework was a dud and fell to the ground, the CoM of the exploded pieces would land in that exact same spot at the exact same time.

7. If internal forces (like a spring between two carts) cancel out, why do the individual carts still accelerate?

Internal forces cancel out for the system as a whole, meaning the CoM doesn't accelerate due to the spring. However, Newton’s laws still apply to individual objects. Cart A feels a push left; Cart B feels a push right. They accelerate individually, but the system's average motion remains unchanged.

8. If a person walks from one end of a floating boat to the other, how do I calculate how far the boat moves relative to the water?

• Use the "Center of Mass stays put" trick. Since there are no external horizontal forces (assuming frictionless water), the CoM of the "Person + Boat" system cannot move.
• If the person moves distance d_p right, the boat must move distance d_b left so that m_person * d_p = m_boat * d_b.

9. How does Newton's Second Law apply to a system of particles (F_ext = M_total * a_cm) if the particles are all moving in different directions?

Newton's Second Law for systems ignores the crazy internal jiggling. It treats the entire system as a single "ghost particle" located at the Center of Mass with mass M_total. You only care about the External Forces.

Sum of External Forces = Total Mass * Acceleration of CoM.

Lesson 3: Impulse-Momentum Theorem: Force vs. Time in Collisions

10. I know Momentum is p=mv, but how is Impulse different from Momentum physically?

Think of Momentum as a status (how much "oomph" you have right now) and Impulse as a transfer (how much "oomph" you gave or received).

• Impulse (J) is the cause. Change in momentum (delta p) is the effect.
• Equation: Impulse = Force * Time = Change in Momentum.

11. If a collision happens instantaneously, is it physically accurate to say gravity doesn't act on the object during that split second (The Impulse Approximation)?

Yes. Gravity is a weak force compared to the massive forces of a collision (like a bat hitting a ball). Because the collision time dt is so tiny (milliseconds), the impulse from gravity (mg * dt) is practically zero compared to the collision impulse. We ignore gravity during the impact.

12. For a Force vs. Time graph, I know the area is Impulse, but does that area tell me the final velocity or just the change in velocity?

1. The area tells you the Change in Momentum (m * delta_v). To find the final velocity, you must:
2. Calculate the Area.
3. Set Area = m(v_final - v_initial).
4. Solve for v_final. (Don't forget to include the initial velocity!)

13. When using the Impulse-Momentum theorem with a changing force, how do I integrate F(t) if the function isn't a simple straight line?

• You must solve the definite integral of the function given.
• Impulse = Integral from t1 to t2 of F(t) dt.
• Example: If F(t) = 3t^2, you integrate to get t^3 and plug in your time limits.

Lesson 3: Impulse-Momentum Theorem: Force vs. Time in Collisions

10. I know Momentum is p=mv, but how is Impulse different from Momentum physically?

Think of Momentum as a status (how much "oomph" you have right now) and Impulse as a transfer (how much "oomph" you gave or received).

• Impulse (J) is the cause. Change in momentum (delta p) is the effect.
• Equation: Impulse = Force * Time = Change in Momentum.

11. If a collision happens instantaneously, is it physically accurate to say gravity doesn't act on the object during that split second (The Impulse Approximation)?

Yes. Gravity is a weak force compared to the massive forces of a collision (like a bat hitting a ball). Because the collision time dt is so tiny (milliseconds), the impulse from gravity (mg * dt) is practically zero compared to the collision impulse. We ignore gravity during the impact.

12. For a Force vs. Time graph, I know the area is Impulse, but does that area tell me the final velocity or just the change in velocity?

1. The area tells you the Change in Momentum (m * delta_v). To find the final velocity, you must:
2. Calculate the Area.
3. Set Area = m(v_final - v_initial).
4. Solve for v_final. (Don't forget to include the initial velocity!)

13. When using the Impulse-Momentum theorem with a changing force, how do I integrate F(t) if the function isn't a simple straight line?

• You must solve the definite integral of the function given.
• Impulse = Integral from t1 to t2 of F(t) dt.
• Example: If F(t) = 3t^2, you integrate to get t^3 and plug in your time limits

Lesson 5: 1D Elastic Collisions: Deriving Final Velocity Formulas (v1 & v2)

21. "How do I actually derive the long final velocity formulas for 1D Elastic Collisions without getting stuck in a page of algebra?"

• The trick is to use the Relative Velocity Equation instead of the Kinetic Energy equation (because squaring velocities is messy).
• Start with Momentum Conservation: m1v1 + m2v2 = m1v1' + m2v2'.
• Use the "Elasticity Condition": v1 - v2 = -(v1' - v2') (Speed of approach = Speed of separation).
• Solve the second (simpler) equation for v2' and substitute it into the momentum equation. This avoids all quadratic equations

Lesson 6: Variable Mass Systems: The Rocket Equation Explained

22. In the Rocket Equation derivation, why is the mass change 'dm' considered negative, and how does that fix the signs?

The mass of the rocket M is decreasing as it burns fuel. So, the change in mass dM is technically a negative number (e.g., -5 kg/s). In the equation, we often write -dm/dt to represent the positive rate of fuel burn (burn rate R).

23. How do I calculate the thrust force of a rocket if I'm only given the burn rate (kg/s) and the exhaust velocity?

• Thrust is simply the force generated by shooting mass out the back.
• Thrust = |v_exhaust * (dm/dt)|
• Multiply the speed of the gas (v_exhaust) by the rate at which you burn it (dm/dt).

24. Why can't I just use F = ma for a rocket? Why do I need a special 'Variable Mass' equation?

1. F = ma assumes 'm' is constant. For a rocket, 'm' changes every second. You must use the original form of Newton's Second Law:
2. F = dp/dt (Force is the rate of change of momentum).
3. Since p = mv, and both change, the derivative involves the product rule, leading to the rocket equation.

25. In the Rocket Equation, is the 'exhaust velocity' measured relative to the ground or relative to the rocket, and why does that matter?

• It is measured relative to the rocket. The engine shoots gas out at a fixed speed compared to the nozzle.
• If the rocket is moving forward at 1000 m/s, and shoots gas backward at 500 m/s (relative to the rocket), an observer on the ground sees the gas moving forward at 500 m/s! The equation accounts for this.