Chapter 7 - Physics 121

Chapter Seven Lecture

Who has the Momentum?

Formulas:

Demonstrations:

Newton's spheres
Ballistic Pendulum
Big ball bouncing off little ball.
Elastic and inelastic balls.
Center of Mass.

Main Ideas:

Momentum and Impulse
Conservation of Momentum.
Elastic Collisions
Inelastic Collisions
Center of Mass and motion.

Momentum & Impulse

Last chapter we talked about the fundamental law of the Conservation of Energy. Energy is always conserved in any process. This chapter will introduce another conservation law (which is not applicable in all circumstances, but in many important ones). That is the Conservation of Linear Momentum. First we have to define what linear momentum is;

Linear momentum is defined as the mass of an object times the velocity of that object. It is a vector, having the same direction as the velocity of the object. We use momentum in this way in everyday language. A 300 lb lineman has more momentum than a 180 lb flanker when they are both moving at the same speed. A car going 60 mph has more momentum than one going 10 mph.

Now consider Newton's Second Law:

So Newton's Second Law can be re-stated as the rate of change of momentum of a body is proportional to the net force applied to it.

I can rewrite this equation as . This says that the net force acting on an object over a period of time (Dt) produces a change in momentum (Dp). Often when the momentum of an object is changed, it is because there is a very large force acting on the object for a very brief time or the force of impact is in a different direction from other forces acting on an object. This happens when you hit a ball, or bounce a ball, or hammer a nail, or in many collisions. In such cases, we neglect all forces except the major force changing the momentum and write;

where the quantity F is the major force changing the momentum, and FDt is called the impulse. Impulse is the quantity that gives the impetus necessary to change the momentum.

Problem : A 50-g golf ball is struck with a club. The ball is deformed by about 2.0cm during the time of collision and the ball leaves the club-face with a velocity of 44 m/s. (a) What is the impulse during collision? (b) How long is the collision? (c) What is the average force during the collision?
(a) To solve for the impulse =

(b)
(c)

Problem: A 100-g ball is dropped from 2.00 m above the ground. It rebounds to a height of 1.50 m. What was the average force exerted by the floor if the ball was in contact with the floor for 1.00´10^-2 s.
Since , and we know Dt, we must find Dp.

From conservation of energy:

Now the force acting on the golf ball can be determined from;

Conservation of Momentum

Suppose we look at a collection of objects called a "system". If we have two billiard balls, that could be the system. A ball bouncing off the floor would consist of the ball and the floor as the "system." Let's define two classes of forces. The first are forces internal to the system. Since our system consists only of two balls, the only forces internal to the system would be the equal and opposite force between the two balls when they collide or the force of gravity between the two balls. Other forces, like the force of gravity from the earth, or the normal force of the table on the balls are external to the system. That is they require agents that are not a part of the system. They require the earth to make gravity, or the table to impart the forces to the system. If there are no external forces, we say that the system is isolated. In many cases where a collision occurs, the internal forces are much greater than the external forces. If I hit a baseball or two cars collide, the internal forces between the ball and the bat, or between the two cars, are much greater over a brief period of time than the external forces like gravity. In such a case the system acts as if it is isolated during the collision. So what happens in an isolated (or nearly isolated) collision?

Look at two billiard balls colliding:

Before collision:

After collision:

Looking at billiard ball 1 and at billiard ball 2

From Newton's Third Law we know that the force of ball 2 on ball 1 (F1) is equal to the force of ball 1 on ball 2 (F2), and the time of the collision is the same for each ball, so we have;

This is the law of the conservation of momentum. It states:

The Total Momentum of an isolated system of bodies remains constant.

Problem: A 90-kg fullback attempts to dive over the goal line with a velocity of 6.00 m/s. He is met at the goal line by a 110-kg linebacker moving at 4.00 m/s in the opposite direction. The linebacker holds on to the fullback. Does the fullback cross the goal line?
During the collision between the players the total momentum is conserved, so;

Where M is the mass of the players during the tackle.

The direction is positive so the two move forward and the fullback scores! Touchdown!

Since p_i = p_f is a vector equation, if we have a collision in two dimensions, we must conserve momentum in each direction. So the conservation of momentum becomes p_xi = p_xf and p_yi = p_yf. We must make the momentum initially and finally equal in both the x-direction, and the y-direction.

Problem: A firecracker with a mass of 100g, initially at rest, explodes into 3 parts. One part with a mass of 25g moves along the x-axis at 75m/s. One part with mass of 34g moves along the y-axis at 52m/s. What is the velocity of the third part?
The third part has a mass of 100-34-25 = 41g.

Along x-direction:

Along y-direction:

So the magnitude of the velocity of the third piece is;

And the direction of v₃ is given by

below the neg. x-axis.

Elastic Collisions

In a certain class of collisions, both the momentum and the kinetic energy are conserved. That is, the total momentum and the total kinetic energy are the same before and after the collision. Remember that momentum is a vector, so the total momentum must be summed like a vector. Kinetic energy is a scalar, so the total energy is summed like regular numbers. An example of a nearly elastic collision is a super ball. An example of a nearly inelastic collision is a lump of clay colliding with a wall and sticking to it.

Problem: A ball with a mass of 1.2 kg moving to the right at 2.0 m/s collides with a ball of mass 1.8 kg moving at 1.5 m/s to the left. If the collision is an elastic collision, what are the velocities of the balls after the collision?

Conservation of Momentum says;

                       Equation (A)

Conservation of Kinetic Energy says;

     Equation (B)

We have two equations and two unknowns, so the physics stops here, and the algebra begins.   The algebra is quite involved.

We will first use the Conservation of Momentum Equation (A) and do a little bit of algebraic manipulation.

Now we can plug in the values that we know. (m₁=1.2kg, m₂=1.8kg, v_1i=2.0 m/s, and v_2i=-1.5 m/s)

We now have this relation derived from the Conservation of Momentum Equation (A). We will now plug this into the Conservation of Kinetic Energy Equation (B).

Notice, the only number we do not have in this equation is v_2f, so we can plug in all the known quantities.

We need to take care of the multiplication of blue term above.

So we plug this in, again.

And do some more multiplication. Isn't this fun? You know it is.

And we can begin to combine like terms together based on the units.

Now we finally have a simpler mathematical relation that we can solve for one of the final velocities after the collision. Unfortunately, it is a quadratic equation. A quadratic equation of the form;

With "a" = 4.5kg, "b" = 0.90 kg×m/s, "c" = -8.775kg×m²/s², and "x" = v_2f --- we get;

There are two possible answers! We will have to do a bit of careful problem analyzing to decide what the answers really mean.

or

We are almost to the end!! Thank heavens. We have some possibilities for an answer to v_2f and we can go back to the relation we got from Conservation of Momentum;

or

Finally, what do we do with all of these answers!? There is only two of them that are what we are looking for. We have;

Why did I cross out two of the possible answers? Well, we have seen them before!! Where? Those numbers were the initial velocities of the balls before the collision! The two answers are the ones that we haven't seen before – or – the final velocities of the two balls after the collision.

What does all this work and fun mean? We have got our final picture.

Note, that there is a built-in way to determine if you do a problem like this correctly, if two of your possible four answers come back matching the initial velocities then you have done all the algebra correctly. If they do not, then you need to do the algebra again.

Inelastic Collisions

An Inelastic Collision is one in which kinetic energy is not conserved. A collision is called a completely inelastic collision if the objects stick together after colliding. In an inelastic collision, kinetic energy is not conserved. However, linear momentum is conserved even in inelastic collisions.

Problem: A 3kg sphere makes a perfectly inelastic collision with a second sphere initially at rest. The composite system moves with a speed equal to 1/3^rd the original speed of the 3 kg sphere. What is the mass of the second sphere?

First we need to write out the velocity relationship:

Now we can apply conservation of momentum:

Demonstration/Problem: The ballistic pendulum is used to determine the velocity of a projectile. Suppose we have a projectile with mass m and velocity v₁, and a pendulum with mass M. The projectile hits the stationary (v₂=0m/s) pendulum and sticks to it, so that the projectile and pendulum have a mass (m + M) and final velocity V. The pendulum then raises a distance h from its original position. From conservation of momentum during the collision we know that;

with v₂ = 0m/s (pendulum stationary to start)

After the collision the kinetic energy of the projectile and pendulum swings upward transforming to gravitational potential energy (mechanical energy is conserved) and we have;

plugging the previous equation in;

Now we can solve this equation for v₁ which is the velocity of the projectile.

Now the height ("h") is not easy to measure but it is given by h+ l = R

Thus, we can plug this geometric result into the equation for the projectile's velocity;

If the gun is horizontal, then from the Kinematic Equations of Motion we can do a bit more mathematical tinkering and get the following;

and

Substituting these into the equation for the projectile velocity, we get;

Which is a neat little formula for the horizontal position of the projectile as it collides and swings up as part of the ballistic pendulum.

Center of Mass

We have talked about the motion of an object a lot in this class. We have usually used simple objects and described their movement. Sometimes the movement of an object is not so simple, (like a spinning, wobbling, tossed Frisbee). How do we describe the motion of that object? The motion of most of the points on the object is quite complicated. However, there is one point of the object that behaves in the same way as a single particle would move subject to the same forces. That point of the body is called the center of mass. Even with rotating, and spinning, the center of mass moves in the same way that a single particle would move.

Demonstration: Finding the center of mass.

We find the center of mass of an object along a certain axis by using the equations:

Problem: Three masses are located in the xy-plane have the following coordinates: a 2kg mass has coordinates given by (3,-2)m; a 3 kg mass has coordinates (-2,4)m, a 1kg mass has coordinates (2,2)m. Find the coordinates of the center of mass.

So, the center of mass, , has coordinates (1/3,5/3)m.

Problem: If the mass distribution of a person sitting down with his legs outstretched can be approximated by the two rectangles, where is the person's center of mass?

The center of mass has coordinates of (24, 34)cm and is marked with an "x" in the diagram.

Note a few things about this example:

The CM may be outside the body.
The CM depends on the shape of the body.

If this person stands up, the center of mass will change location relative to his body. In fact, it will then be somewhere in his torso.

A high jumper can actually clear the bar without the center of mass clearing the bar. A basketball player or gymnast in flight has their center of mass follow the same path that a single particle would follow (parabolic near the earth if we neglect air resistance) although their extremities may be changing direction, and they may be twisting and tumbling.

This means that even if there are internal forces, like muscles pulling, the motion of the center of mass of the system is governed by the external forces only. Suppose, I shoot fireworks up in the air and they explode. What is the motion of the center of mass of all the particles? What is the motion of each particle? Look at the rocket example in the book.

Case where m_II = m_I

Problem: Suppose that m_II = 3m_I. Where would m_I land? (m_I would still land straight down, and now the Center of Mass must be at 2D, so;

Note that every one of the individual objects, as well as the center of mass of all the objects follow a parabolic path. (A straight line down is a type of parabola).

Text Box: A man's worst difficulties begin when he is able to do as he likes. -- Thomas Huxley