Notation: the underscore symbol indicates a subscript ("p_i") and the caret indicates a superscript ("v^2").

If possible, please keep your textbook nearby while reading this web lecture.

Definition: We introduced the definition of linear momentum last week: (momentum vector) = mass * (velocity vector) [eqn 6-11] (We will introduce angular momentum on Wednesday March 7, so everything today will just be simple linear momentum.)

Experiments: We saw momentum in action in two experiments we conducted on the frictionless air track with carts of equal mass m. In experiment 1, initially cart 1 has v1, cart 2 at rest, so the initial momentum is m*v1; after the collision, cart 1 was at rest and cart 2 had velocity v1, so the final momentum was m*v1. In experiment 2, initially cart 1 has v1, cart 2 at rest, so the initial momentum is m*v1; after the collision, cart 1 and cart 2 stuck together to form a single object of mass 2m which had velocity 0.5*v1, so the final momentum was (2m)*(0.5*v1) = m*v1 again.

Conservation: We can very simply obtain an equation that describes how momentum changes, leading us to conservation of momentum in some situations and a definition of impulse. From Newton's second law, F=ma which by the definition of acceleration gives F= m (dv/dt). Using the definition of momentum as mass times velocity (p=mv) we get F = d/dt (mv) = dp/dt [eqn 6-12]. The first consequence is that if the external force is zero, the change in momentum is zero, meaning the momentum is conserved. This relationship can also be used even in the presence of external forces for brief collisions. (A "collision" means that the objects in our system interact with each other in a short period of time, then move along their own way again. The external forces are finite, so if (delta t) gets very small the change in momentum approaches zero, so we can consider the momentum to have the same value immediately before and after the collision.) The second consequence is that if you look at finite changes and multiply both sides by a time interval (delta t), then the change in momentum is equal to the average net external force times the time interval, which is further defined as the "impulse". (delta p) = (average net force)*(delta t) = I [eqn 8-20]. Please read the Problem-Solving Guide on page 245.

Example: Let's re-examine some homework problems in terms of conservation of momentum. For Chapter 8, Problem 31, a girl of mass 55kg jumps off the bow of a 75kg canoe that is initially at rest. If her velocity is 2.5m/s to the right, what is the velocity of the canoe after she jumps? We can treat the girl and the canoe as a system that experiences no external forces, so momentum will be conserved. Initially both masses are at rest so the initial momentum is zero. We set up an x-axis to the right and evaluate the momentum after she jumps. p_x = (55kg)*(2.5m/s) + (75kg)*(v_canoe), but since this equals the initial momentum which is zero, we solve for v_canoe = -(55kg/75kg)*(2.5m/s) = -1.83m/s. The concept of "recoil" is a handy way to think about conservation of momentum for system. When you throw an object in one direction, you experience an impulse in the opposite direction.

Example 8-7: A railroad car on a frictionless track fills up with rain, so the mass of the system changes. Initially the momentum was (14000kg)*(4m/s). After the downpour, the momentum was (16000kg)*(v_final). Equating the initial and final momentum yields v_final = 3.5m/s.

Example: Given: a bullet of mass 0.01kg with an initial velocity of 200m/s and a box of mass 1kg initially at rest on a frictionless surface. The bullet passes through the box of mass and emerges with a final velocity of 100m/s. Determine the velocity of the box after the bullet passes through it. ... Answer: Since the interaction between the bullet and the box takes such a short time and the external forces are finite, the momentum of the system of bullet+box is the same before and after the collision. Before: p=(0.01kg)*(200m/s)=2kgm/s. After: p=(0.01kg)*(100m/s)+(1kg)*(v_box). Equating p_i=p_f yields v_box=1m/s.

Two Dimensional example: Since momentum is a vector quantity, we often consider the conservation of momentum to apply separately to the x component and the y component of momentum. Example 8-18 discusses this type of collision -- please notice the need to keep the x and y components of momentum conserved separately. For another example, we will give values to Figure 8-37. Object 1 has mass m_1 = 1kg and initial velocity of v_1i to the east, while object 2 has mass m_2 = 2kg and is initially at rest. After the collision, object 1 has a velocity of v_1f in the direction theta_1 = 30 degrees north of east, while object 2 has a velocity of 10m/s in the direction theta_2 = 45 degrees south of east ... Answer: Initially, the x-component of the initial momentum for the system of two objects is p_ix = (1kg)(v_1i)+(2kg)(0m/s) while the y-component p_iy is zero. After the collision, the y-component of the final momentum is expressed as p_fy=(1kg)[(v_1f)(sin30)]-(2kg)[(10m/s)(sin45)]. Setting p_iy= p_fy yields v_1f=28.3m/s. This lets us express the x component as p_fy=(1kg)[(28.3m/s)(cos30)]+(2kg)[(10m/s)(cos45)]. Setting p_ix= p_fx yields v_1i=38.6m/s.

Energy considerations: For an "elastic" collision, the total kinetic energy of our system is the same before and after the collision. For our experiment 1 described at the beginning of the lecture, K_i=K_f= 0.5*m*(v1)^2. For an "inelastic" collision, the total kinetic energy of our system is NOT the same before and after the collision. For our experiment 2 described at the beginning of the lecture, K_f= 0.5*(2m)*(0.5*v1)^2, so half of the initial kinetic energy was lost during the collision. When the objects stick together during a collision, it is a "perfectly inelastic" collision.

You may have seen a desktop toy with five steel balls that can be made to collide with each other. It has the amazing property that if you pull back and release one ball, one ball flies off the opposite end, similarly if you launch two, two fly off etc. You can consider this as a consequence of the simultaneous conservation of momentum and energy. For momentum conservation alone, we could launch two and have only one fly off with twice the velocity of the initial balls, but then energy would not be conserved!

Another handy tool is to be able to express the kinetic energy of an object in terms of its momentum K=0.5*m*v^2=(p^2)/(2m).

Some handy results for two object ELASTIC collisions are that in one dimension, the velocity of object 2 relative to object 1 as it moves away after the collision is exactly equal and opposite to the velocity of object 2 relative to object 1 when it was approaching, before the collision. [eqn 8-29 and the discussion that continues on the following page]. For an elastic collision in two dimensions, if one object is initially at rest and if the masses are equal, the angle between the final velocities is 90 degrees. [paragraph after eqn 8-33 and caption of fig 8-38] Notice that we cannot use this clue for the two dimensional example we worked out above, since although one object was initially at rest, the objects did not have equal masses. So even though the final angle was 75 degrees, we cannot conclude from that alone whether the collision was elastic or inelastic. We would have to calculate the total kinetic energy and compare before and after.

Forces during a collision can be quite large, but can be estimated using the impulse. For example, suppose you are in your car travelling with an initial speed v_i, and hit a brick wall. Thanks to the "crumple zones" in your car, you are brought to a somewhat gentle stop (v_f=0) over a distance of 1m. You can estimate the time required using v_av = distance / time, so (delta t) = distance / v_av, but we can estimate v_av = (v_i + v_f)/2 = 0.5*v_i. Then the average force F_av=(delta p)/(delta t)={m*v_f - m*v_i}/{distance/0.5*v_i}=(K_i)/distance which also makes sense from an energy point of view.

Section 8-7 describes what happens when we change our point of view to observe collisions in a frame where the center of mass is at rest. Section 8-8 points out that the reason a rocket moves is conservation of momentum -- by giving the exhaust gas momentum backward, the rocket gains momentum forward.

HOMEWORK 14 will still be due on Wednesday, March 7, 2001.

Problem 40 uses the ideas of section 8-5. In part a) you find the straightforward kinetic energy of two masses moving and in part b) you calculate the velocity of the center of mass. In part c) you dredge equation 2-7a out of the dim recesses of your memory to find the velocity of each object relative to the center of mass. If you are adventurous, you can draw a figure like that in section 8-7.

Problem 48 uses basic ideas of impulse. Don't forget to pay attention to signs and your choice of the positive direction.

Problem 59 should be straightforward, Problem 68 is similar to a lecture example above, and Problem 87 has roots in both a textbook example and lecture.