Centre of Mass, Linear Momentum, Collision

The centre of mass will be located at:

• Horizontally: exactly halfway across the square (at x = a/2, where a is the side length)
• Vertically: at 7/10 of the height (at y = 7a/10)

This position is biased toward the top-left corner where the 4m mass is located, and reflects how the centre of mass is weighted more heavily by larger masses.

Rocket equation derivation:

1. Initial rocket mass = M₀ (rocket + fuel)
2. Fuel ejection rate = r = -dM/dt (constant)
3. Gas ejection velocity relative to rocket = u (constant)
4. At time t: M = M₀ - rt

From momentum conservation: $\frac{dv}{dt} = \frac{ru}{M} = \frac{ru}{M_0-rt}$

Integrating gives: $v = u\ln\frac{M_0}{M_0-rt}$

Key insights:

• Acceleration increases over time as fuel is consumed (M decreases)
• The logarithmic relationship shows diminishing returns from additional fuel
• Final velocity depends on exhaust velocity (u) and mass ratio (M₀/final mass)

Example: If a rocket ejects half its initial mass, its final velocity is approximately 0.693u.

For elastic collision between two masses m₁ and m₂:

Given:

• Initial velocities: v₁, v₂
• Final velocities: v₁', v₂'
• Momentum conservation: m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
• Kinetic energy conservation: ½m₁v₁² + ½m₂v₂² = ½m₁v₁'² + ½m₂v₂'²

Proof:

1. From momentum: m₁(v₁-v₁') = m₂(v₂'-v₂)
2. From kinetic energy: m₁(v₁²-v₁'²) = m₂(v₂'²-v₂²)
3. Factoring: m₁(v₁-v₁')(v₁+v₁') = m₂(v₂'-v₂)(v₂'+v₂)
4. Using step 1: (v₁+v₁') = (v₂'+v₂)
5. Therefore: v₁-v₂ = -(v₁'-v₂')

This means: relative velocity after collision = -(relative velocity before collision)

Example: If two billiard balls approach at relative speed 10 m/s, they'll separate at relative speed 10 m/s.

Conservation of linear momentum for a system with only internal forces:

Key principles:

• Total linear momentum: P⃗ = ∑mᵢv⃗ᵢ = MV⃗ₒₘ
• When no external forces act (∑F⃗ₑₓₜ = 0):
  • P⃗ = constant (both magnitude and direction)
  • The centre of mass moves with constant velocity
  • Internal forces cannot change the total momentum of the system
  • If initially at rest, the centre of mass remains fixed

Applications:

1. Recoil of a gun when firing a bullet
2. Propulsion of rockets in space
3. Explosive separation of components
4. Collisions between particles in an isolated system

Example: When a person jumps from a boat, the boat moves in the opposite direction such that the total momentum remains zero.

• The centre of mass of a system moves according to Newton's laws as if all mass were concentrated there and all external forces were applied there
• Key principles in collisions:
  • The velocity of the centre of mass remains constant if no external forces act on the system
  • Internal forces (between the objects) cannot change the motion of the centre of mass
  • The system's total momentum is conserved in the absence of external forces
• In cricket terms:
  • When a ball hits a bat, the centre of mass of the ball-bat system moves according to external forces only
  • The distribution of momentum between ball and bat depends on internal forces during collision
  • The coefficient of restitution determines how much kinetic energy is conserved in the collision
• Practical example: When a cricket ball bounces off a stationary bat, the centre of mass of the system continues moving forward, but the ball may reverse direction relative to this centre of mass

Case (a): Heavy object ($m_1$) hits light object ($m_2$) from behind ($m_1 \gg m_2$)

• Heavy object continues with almost unchanged velocity: $v_1' \approx v_1$
• Light object's new velocity: $v_2' \approx 2v_1 - v_2$
• If light object was initially at rest ($v_2 = 0$): $v_2' \approx 2v_1$ (twice the velocity of the heavy object)

Case (b): Light object ($m_1$) hits heavy object ($m_2$) from behind ($m_2 \gg m_1$)

• Heavy object continues with almost unchanged velocity: $v_2' \approx v_2$
• Light object's velocity reverses: $v_1' \approx -v_1 + 2v_2$
• If heavy object was initially at rest ($v_2 = 0$): $v_1' \approx -v_1$ (rebounds with same speed)

This explains why balls bounce back when hitting walls (approximated as infinitely massive objects).

In an elastic collision between objects of equal mass ($m_1 = m_2$):

• Their velocities are completely exchanged: $v_1' = v_2$ and $v_2' = v_1$

This is derived by substituting $m_1 = m_2$ into the general elastic collision equations: $v_1' = \frac{(m_1-m_2)}{m_1+m_2}v_1 + \frac{2m_2}{m_1+m_2}v_2$ $v_2' = \frac{2m_1}{m_1+m_2}v_1 + \frac{(m_2-m_1)}{m_1+m_2}v_2$

Example: In billiards, when the cue ball hits a stationary target ball of the same mass in a direct hit, the cue ball stops and the target ball moves with the initial velocity of the cue ball.

For two-dimensional elastic collisions, you need:

1. Conservation of momentum in x-direction: $m_1u_1 = m_1v_1\cos\theta + m_2v_2\cos\phi$

2. Conservation of momentum in y-direction: $0 = m_1v_1\sin\theta - m_2v_2\sin\phi$

3. Conservation of kinetic energy: $\frac{1}{2}m_1u_1^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

Important notes:

• These three equations have four unknowns ($v_1$, $v_2$, $\theta$, $\phi$)
• Additional information needed - typically the impact parameter or angle
• Momentum is conserved in directions both parallel and perpendicular to the line of impact
• Final motion depends on the angle between the force line during collision and initial velocity

The system is underdetermined without additional information about the collision geometry.

Impulse is the change in momentum produced by a force acting over a time interval:

• Mathematical representation: $\vec{J} = \int_{t_i}^{t_f} \vec{F} dt$
• Units: N·s or kg·m/s
• Impulse equals the change in momentum: $\vec{J} = \Delta \vec{p} = \vec{p}_f - \vec{p}_i$

Example: When hitting a baseball, the bat applies a large force over a short time interval (~0.001s). The impulse delivered to the ball changes its momentum from moving toward the bat to moving away from it.

The area under a force-time curve represents impulse, which equals the change in momentum:

• Impulse = $\int_{t_i}^{t_f} F dt = p_f - p_i$
• For constant force: Impulse = $F \cdot \Delta t$
• For variable force: Calculate the area under the F-t curve
• Units: N·s = kg·m/s

Example: In a car crash test, the force exerted on a crash test dummy varies with time. Engineers calculate the total impulse by finding the area under the F-t curve, which equals the change in the dummy's momentum during impact.

Note: The shape of the F-t curve matters - a longer duration with lower peak force can deliver the same impulse as a shorter duration with higher peak force, which is why airbags extend impact time to reduce peak forces.