Center of Mass, Momentum, Impulse, and Collisions — Comprehensive Notes (Grade 12 General Physics 1)

Center of Mass

• Center of Mass (CoM) is the point where the mass of an object is considered to be concentrated for purposes of balance and motion analysis.
• In solids, the CoM is related to stability: weight acts downward and the upward support from the ground balances the forces in equilibrium.
• The CoM can be used as the balancing point of an object (the point at which it can be balanced).

Center of Mass for Solids

• Definition (from slide):
  • "The point of an object at which all the mass of an object is thought to be concentrated is called center of mass."
• Centre of Mass and Stability:
  • Weight acts downward; upward force from ground.
  • A box in equilibrium has balanced forces and balanced turning effects (torques).

Center of Mass as Balancing Point

• The center of mass is the point at which the object can be balanced.
• Relationship between center of mass and weight distribution.

Center of Mass in Motion: Parabolic Paths

• For extended bodies like a baseball and a spinning baseball bat, the centers of mass follow parabolic paths (projectile-like motion for the CoM under gravity when the object is in flight or in rotation).

Center of Mass: Simple Shapes vs. Complicated Shapes

• For simple rigid objects with uniform density, the CoM is located at the centroid of the shape.
• For complicated shapes, the CoM can be found using a plumb line: the intersection point where vertical lines (plumb line) through all parts cross marks the CoM.
• With a thin, irregular sheet, the CoM is the crossing point of the lines drawn by balancing methods (e.g., plumb line).

Center of Mass vs Geometric Center

• Geometric center depends only on the shape.
• Center of mass depends on the density distribution.
• Example: A circular disk with a hole
  • The geometric center lies at the center of the disk.
  • The hole creates non-uniform density, so the CoM shifts away from the geometric center.
  • Diagrammatic idea: geometric center vs. CoM shift due to density distribution.

Mathematical Definition of Center of Mass (General Form)

• For a system with N particles of masses mi located at position vectors ri, the center of mass XCM is: $X</em>{CM} = \frac{\sum<em>{i=1}^{N} m</em>i \mathbf{r}<em>i}{M}, \quad M = \sum</em>{i=1}^{N} m_i$
• 2-mass special case (1D):
  $X<em>{CM} = \frac{m</em>1 x<em>1 + m</em>2 x<em>2}{m</em>1 + m_2}$
• 2-mass 3D generalization (X, Y, Z components):
  $X<em>{CM} = \frac{\sum m</em>i x<em>i}{M}, \quad Y</em>{CM} = \frac{\sum m<em>i y</em>i}{M}, \quad Z<em>{CM} = \frac{\sum m</em>i z_i}{M}$

Worked Examples: Center of Mass (1D and 2D)

• Example 1 (3 point masses on a line):
  • masses and positions: m1 = 1 kg at x1 = 0 m; m2 = 4 kg at x2 = 2 m; m3 = 2 kg at x3 = 5 m
  • Total mass: M = 1 + 4 + 2 = 7 kg
  • Center of mass:
    $X_{CM} = \frac{(1)(0) + (4)(2) + (2)(5)}{7} = \frac{0 + 8 + 10}{7} = \frac{18}{7} \approx 2.57\text{ m}$
• Example 2 (3 masses on a line with x-coordinates):
  • masses: m1 = 1.0 kg at x1 = 0 m; m2 = 4.0 kg at x2 = 2 m; m3 = 2.0 kg at x3 = 5 m; (Note: values reflect the slide’s setup with total mass 7.0 kg and x_cm ≈ 2.57 m as above.)
• Example 3 (2D coordinates):
  • masses and coordinates: m1 = 1.2 kg at (0, 0); m2 = 2.5 kg at (140, 0); m3 = 3.4 kg at (0, 121)
  • Total mass: M = 1.2 + 2.5 + 3.4 = 7.1 kg
  • x-coordinate of CoM:
    $X_{CM} = \frac{(1.2)(0) + (2.5)(140) + (3.4)(0)}{7.1} = \frac{350}{7.1} \approx 49.29\text{ m}$
  • y-coordinate of CoM:
    $Y_{CM} = \frac{(1.2)(0) + (2.5)(0) + (3.4)(121)}{7.1} = \frac{0 + 0 + 412.\,?}{7.1} \approx 58\text{ m}$
  • (Note: The slide lists Y_CM = 58 m for the given setup.)

Practice Center-of-Mass Problems (Selected)

• Problem A: Three point masses on a number line
  • m1 = 2 kg at x = 0 m; m2 = 3 kg at x = 2 m; m3 = 5 kg at x = 5 m
  • Find X_CM using 1D formula above.
• Problem B: Three objects on a flat table
  • m1 = 2 kg at (0,0); m2 = 4 kg at (2,0); m3 = 2 kg at (0,2)
  • Find (XCM, YCM).
• Problem C: Three masses on the x-axis
  • m1 = 4 kg at x1 = 0 m; m2 = 3 kg at x2 = 2 m; m3 = 5 kg at x3 = 4 m
  • Find X_CM.
• Problem D: Four masses at corners of a rectangle
  • m1 = 2 kg at (0,0); m2 = 3 kg at (3,0); m3 = 4 kg at (0,4); m4 = 1 kg at (3,4)
  • Find (XCM, YCM).
• Problem E: Placement for a desired CoM
  • Two masses: 5 kg at x = 0 and 3 kg at x = 6. Find where to place a third mass m3 = 2 kg so that x_COM = 2.
  • Solve: $\frac{5\cdot 0 + 3\cdot 6 + 2\cdot x}{5+3+2} = 2 \Rightarrow x = 1\text{ m}.$
• Problem F: Rectangle corners with weights
  • 1 kg at (0,0), 2 kg at (4,0), 3 kg at (4,2), 4 kg at (0,2)
  • Find COM:$X_{CM} = \frac{1(0) + 2(4) + 3(4) + 4(0)}{1+2+3+4} = \frac{0 + 8 + 12 + 0}{10} = 2\text{ m}$
  • $Y_{CM} = \frac{1(0) + 2(0) + 3(2) + 4(2)}{10} = \frac{0 + 0 + 6 + 8}{10} = 1.4\text{ m}$

Momentum (Linear Momentum)

• Momentum (p) is the quantity of motion of an object and is related to how hard it is to stop the object.
• In physics, momentum is defined as:
  $\mathbf{p} = m\mathbf{v}$
• For a system of multiple particles:
  • Total momentum: $\mathbf{P}<em>{\text{tot}} = \sum</em>i \mathbf{p}<em>i = \sum</em>i m<em>i \mathbf{v}</em>i$
  • In 2D: $p<em>x = m v</em>x, \quad p<em>y = m v</em>y$
• Units: $\text{kg} \cdot \text{m}/\text{s}$

Momentum and Newton's Second Law (in Momentum Form)

• Newton’s second law can be written as:
  $\mathbf{F}_{\text{net}} = \frac{d\mathbf{p}}{dt}$
• For a particle with constant mass: $\mathbf{F}_{\text{net}} = m\frac{d\mathbf{v}}{dt} = \frac{d}{dt}(m\mathbf{v}) = \frac{d\mathbf{p}}{dt}$
• In 1D, the relation reduces to: $F_{\text{net}} = \frac{d}{dt}(mv)$
• Isolated or closed systems: there is no external interaction with the environment.

Conservation of Linear Momentum

• In an isolated system, momentum is conserved over time:
  $P<em>{\text{initial}} = P</em>{\text{final}}\,.$
• Total momentum remains constant even when non-conservative forces act on the system internally.

Examples: Momentum and Collisions (Selected)

• Rifle-bullet recoil (conservation of momentum)
  • Rifle mass: mR = 3 kg; bullet mass: mB = 0.005 kg; bullet speed: v_B = +300 m/s; initial momentum is zero.
  • Final momentum: mR vR + mB vB = 0 ⇒ vR = -\frac{mB vB}{mR} = -\frac{0.005\cdot 300}{3} = -0.5\text{ m/s}.
• Elastic collision (two identical masses, one initially at rest)
  • If m1 = m2 and u1 = u, u2 = 0, then after an elastic collision: v1' = 0, v2' = u (velocities swap).
• Perfectly inelastic collision (example with 85 kg and 105 kg masses)
  • Initial: mA = 85 kg with uA = 7 m/s; mB = 105 kg at rest (uB = 0).
  • After collision, common velocity:
    $v<em>f = \frac{m</em>A u<em>A + m</em>B u<em>B}{m</em>A + m_B} = \frac{85\cdot 7 + 105\cdot 0}{85+105} = \frac{595}{190} \approx 3.13\text{ m/s}.$
  • Initial Kinetic Energy: $KE<em>i = \tfrac{1}{2} m</em>A u_A^2 = \tfrac{1}{2} (85)(7^2) = 2082.5\text{ J}.$
  • Final Kinetic Energy: $KE<em>f = \tfrac{1}{2} (m</em>A + m<em>B) v</em>f^2 = \tfrac{1}{2} (190) (3.13)^2 \approx 930\text{ J}.$
  • KE lost: $KE<em>i - KE</em>f \approx 2082.5 - 930 \approx 1150\text{ J}.$
• Elastic collision (two billiard balls, equal masses, one moving):
  • Initial: u1 = 6 m/s, u2 = 0; After elastic collision: v1' = 0, v2' = 6 m/s.

Conservation of Momentum: 2-Body Collisions

• For a two-particle system with initial momenta pA,i = mA uA and pB,i = mB uB:
  $m<em>A u</em>A + m<em>B u</em>B = m<em>A v</em>A + m<em>B v</em>B.$
• This can be used to solve for the final velocities given the other values.

Types of Collisions (Kinetic Energy Perspective)

• Elastic collision: kinetic energy is conserved (KEi = KEf).
• Inelastic collision: kinetic energy is not conserved.
• Perfectly inelastic collision: the colliding bodies stick together after the collision (maximum KE loss for the given momentum).

Impulse and its Real-World Implications

• Impulse definition:
  $I = \Delta p = \int \mathbf{F}<em>{\text{net}} \; dt = F</em>{\text{net}} \Delta t \quad (\text{for constant } F)$
• Impulse is a vector quantity with the same direction as the net force.
• Relationship to momentum: impulse equals the change in momentum: $I = \Delta \mathbf{p}$
• Practical implication: increasing the time of impact reduces the average force during a collision (I = F_{net} Δt).
• Examples:
  • Seatbelts and airbags increase the time over which the head/torso change momentum, reducing peak forces during crashes.
  • Crumple zones and padding serve a similar purpose by extending collision time.

Impulse: Worked Examples (Selected)

• Example 1: Golf ball and club
  • Golf ball mass m = 0.050 kg; initial velocity vi = 0; final velocity vf = +50 m/s after impact with club.
  • Impulse: I = m(vf - vi) = 0.050(50 - 0) = 2.5\text{ kg·m/s}.
  • If impact time Δt = 0.0005 s, average force: $F_{net} = \frac{I}{\Delta t} = \frac{2.5}{5\times 10^{-4}} = 5000\text{ N}.$
• Example 2: Ball dropping and bouncing
  • Ball mass m = 0.010 kg; initial downward velocity vi = -15 m/s; rebound velocity vf = +10 m/s; Δt = 0.0007 s.
  • Impulse: I = m(vf - vi) = 0.010(10 - (-15)) = 0.25\text{ kg·m/s}.
  • Average force: $F_{net} = \frac{I}{\Delta t} = \frac{0.25}{7\times 10^{-4}} \approx 357\text{ N}.$
• Seatbelt and airbag example emphasizes increasing the time of impact to minimize force.

Conservation of Momentum in Collisions (Summary)

• In any collision, the total momentum of the system is conserved if external forces are negligible (isolated system):
  $P<em>{sys,i} = P</em>{sys,f}.$
• Collision types recap:
  • Elastic collision: KE conserved; momentum conserved.
  • Inelastic collision: KE not conserved; momentum conserved.
  • Perfectly inelastic collision: objects stick together after collision; momentum conserved; KE lost.

Additional Worked Collision Examples (From Slides)

• Example: Elastic collision with equal masses (1D)
  • Initial: v1 = 6 m/s, v2 = 0; After collision: v1' = 0, v2' = 6 m/s.
• Example: Perfectly inelastic collision (two masses that stick)
  • Initial: mA = 85 kg with uA = 7 m/s; mB = 105 kg at rest. Final velocity vf ≈ 3.13 m/s; KE loss ≈ 1.15 × 10^3 J.
• Example: Rifle-bullet recoil (again for emphasis)
  • Confirmed: v_R = -0.5 m/s for given masses and bullet velocity.

Quick Revisit: Key Formulas to Remember

• Center of Mass (2-particle system):
  $X<em>{CM} = \frac{m</em>1 x<em>1 + m</em>2 x<em>2}{m</em>1 + m_2}.$
• General Center of Mass (N particles):
  $X<em>{CM} = \frac{\sum</em>{i=1}^N m<em>i x</em>i}{M}, \quad Y<em>{CM} = \frac{\sum</em>{i=1}^N m<em>i y</em>i}{M}, \quad Z<em>{CM} = \frac{\sum</em>{i=1}^N m<em>i z</em>i}{M}.$
• Total Momentum:
  $\mathbf{P}<em>{tot} = \sum</em>i \mathbf{p}<em>i = \sum</em>i m<em>i \mathbf{v}</em>i.$
• Momentum change (Newton's second law in momentum form):
  $\mathbf{F}_{\text{net}} = \frac{d\mathbf{p}}{dt}.$
• Impulse: $I = \Delta \mathbf{p} = \mathbf{F}_{\text{net}} \Delta t.$
• Elastic collision (1D):
  $v<em>1' = \frac{(m</em>1 - m<em>2) u</em>1 + 2 m<em>2 u</em>2}{m<em>1 + m</em>2},\quad v<em>2' = \frac{2 m</em>1 u<em>1 + (m</em>2 - m<em>1) u</em>2}{m<em>1 + m</em>2}.$
• Perfectly inelastic collision (1D):
  $v<em>f = \frac{m</em>1 u<em>1 + m</em>2 u<em>2}{m</em>1 + m_2}.$

Real-World Connections and Ethical/Practical Implications

• Understanding CoM helps in designing stable structures and vehicles (e.g., seatbelts and airbags reduce injury by increasing collision time).
• Momentum conservation principles underlie safety designs in sports, automotive engineering, and disaster-response simulations.
• Accurate CoM calculations are essential in robotics, aerospace, and biomechanics to ensure balance, control, and safety.

Notes on Key Concepts from the Slides

• The geometric center is shape-based; the CoM accounts for density distribution.
• A non-uniform density (e.g., a disk with a hole) shifts the CoM away from the geometric center.
• In many problems, 2D or 3D CoM calculations require summing mass-weighted coordinates and dividing by total mass.
• Parabolic paths of CoM can arise in certain rotating or projectile-like contexts (illustrative of how CoM motion can resemble projectile motion).
• Impulse provides a practical link between force and momentum: long contact times reduce peak forces during impacts.