PHY5200 F06

Chapter 3: Momentum and Angular Momentum

Reading

Taylor 3.1-3.2 (today) and Monday we will review for the Midterm

Recall

The path of a charge in a magnetic field B=(0,0,B) is given by

x(t)+iy(t) = (x₀+iy₀)e^-iωt

where ω=qB/m. Expressing x₀+iy₀ as re^iδ the equation becomes

x(t)+iy(t) = re^iδe^-iωt = re^i(δ-ωt)

This path looks like a circle of radius r, centered on the origin, in the complex plane. One can show that the radius is given by

r = v/ω = (mv)/(qB) = p/(qB)

where v=(v_x0²+v_y0²)^½.

Conservation of Momentum

Using Newton's third law we showed that the time derivative of the total momentum of a system of N particles equals the total exteranl force on the system

P[dot] = F^ext

The concept of conservation of momentum is considered fundamental in physics. In quantum mechanics it is connected to translational invariance of the equations, that is, that physics is unchanged if the coordinates are displaced by a small amount. For this class, we formalize the concept with the following definition.

Principle of Conservation of Momentum

If the net external force F^ext on an N-particle system is zero, the system's total mechanical momentum P = ∑ p_α = ∑ m_αv_α is constant.

The following examlple demonstrates the power of this principle.

Example: Inelastic Collision of Two Bodies

An object of mass m₁ and velocity v₁ collides with a second object of mass m₂ and velocity v₂. The objects stick together. Find the final velocity of the stuck pair.

You did a similar problem as homework. It is easy to generalize to the situation where both objects are initially moving so that the initial momentum and final momenta are

P_i = m₁v₁ + m₂v₂ P_f = (m₁+m₂)v

and solve for the final velocity to find

v = (m₁v₁ + m₂v₂) / (m₁ + m₂) .

Rockets

Rockets are devices that expel an exhaust in order to move the bulk of the device, finessing the principle of conservation of momentum.

Imagine a rocket as an object m moving at velocity v at time t. A small amount of mass -Δm is expelled with velocity v_ex relative to the rest of the mass m during the time interval Δt. For simplicity, let's assume that v points in the +x direction and v_ex is in the -x direction. The momentum of the rocket at time t is

P(t) = mv

and a short time later the momentum of the rocket (counting the expelled mass) is

P(t+Δt) = (m+Δm)(v+Δv) + (-Δm)(v-v_ex)

where the first term, representing the rocket, includes a term to account for its change in velocity, and the second term represents the exhaust of mass -Δm (Δm is negative for convenience later in the calculation). Expanding this out, and dropping terms containing two $Delta;'s, yields

P(t+Δt) = mv + mΔv + v_exΔm

The change in momentum must be zero when we account for both the rocket and the exhaust, so

P(t+Δt) - P(t) = mΔv + v_exΔm = 0

mΔv = -v_exΔm .

If we divide both side by Δt, and take the limit as Δt goes to zero, we get

mv[dot] = -v_ex m[dot]

That is, the expelled exhaust produces a change of momentum of the remaining mass (rockethousing, payload, plus fuel). The quantity -v_ex m[dot] is called the thrust; it is a measure of how much mass is expelled by the rocket and with what velocity. The thrust increases if more mass is expelled, or if the same mass is expelled at a higher velocity. The thrust produces a change of momentum of the remaining rocket mass.