Taylor 3.1-3.2 (today) and Monday we will review for the Midterm
The equations of motion for a charge in a magnetic field are:
The z equation is easily solved, but the x and y equations are coupled. Changing to a new variable η
the x and y equations of motion become
a linear first order differential equation with solution
With the solution for the velocity as a function of time, we can now solve for the position. To do that, it is convenient to define another complex variable, ξ, pronounced xi,
By taking the derivative of ξ we see that it equals η
so we can solve for ξ, and hence the positions x and y, by integrating η
Since x is the real part of ξ and y is the imaginary part, we can write, dropping the constant since it basically corresponds to a displacement of the motion:
When t=0, the exponential is 1, so C corresponds to the position of the particle at that moment,
In general, the charge moves in a helix determined by the constant velocity in z and the circular motion in x-y. If the z velocity is zero, the motion reduces to a circle, centered on the origin. The radius of the circular motion is equal to the tangential velocity (x-y speed) divided by the angular velocity, ω (recall that v = rω)
This motion is the basis for many devices, in particular, the cyclotron.
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
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.
If the net external force Fext on an N-particle system is zero, the system's total mechanical momentum P = ∑ pα = ∑ mαvα is constant.
The following examlple demonstrates the power of this principle.
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
and solve for the final velocity to find
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 with mass m moving at velocity v at an instant of time t. A small amount of mass -Δm is expelled with velocity vex 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 vex is in the -x direction. The momentum of the rocket at time t is
and a short time later the momentum of the rocket (counting the expelled mass) is
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
The change in momentum must be zero when we account for both the rocket and the exhaust, so
or
If we divide both side by Δt, and take the limit as Δt goes to zero, we get
That is, the expelled exhaust produces a change of momentum of the remaining mass (rocket housing, payload, and fuel). The quantity -vex dtm is called the thrust; it is a measure of how much push the exhaust gives to the rest of the rocket, and is proportional to the mass expelled by the rocket and its velocity relative to the rocket. 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.