9-1 - 9-3   9-4 - 9-6   9-7 - 9-8   next        

Chapter Review

---

In chapters 7 and 8 we saw how defining concepts beyond just force, velocity, acceleration, and displacement made analyzing some situations much easier, especially the conservation of energy. In this chapter we do more of the same by introducing the concept of linear momentum. In a way similar to work and energy the idea of linear momentum often provides a more direct path to understanding physical interactions than working with forces only. Furthermore, as with energy, there is a conservation principle for linear momentum, which is of tremendous value in physics.

9-1 Linear Momentum

The linear momentum of an object is defined as the product of its mass and its velocity

p = mv,

where p is the symbol for linear momentum. Often, we just refer to p as the momentum and leave off the word "linear." Notice that momentum is a vector quantity whose direction is that of the velocity v and whose units are just the mass unit times velocity units, . In one sense you can think of linear momentum as a measure of the effect the motion of an object has when that object interacts with other objects (you'll see this more clearly in the discussion of collisions). In another sense, linear momentum can be viewed as an alternative measure of an object's inertia, i.e., its tendency to maintain a constant state of motion (you'll see this more clearly in the discussion of momentum conservation).

For a system of particles we often speak of the total momentum of the system. This total momentum is the vector sum of the momenta of every particle in the system

We will see later that the total momentum of a system tells us a lot about how a system behaves as a whole.

---

Example 9.1 The Total Momentum: Three balls of mass 2.0 kg, 3.0 kg, and 4.0 kg move with the velocities shown below. Determine the magnitude and direction of the total momentum of the balls.

Picture the Problem The picture shows the 3 balls indicating the magnitudes and directions of their velocities.

Strategy We need to determine the momentum of each particle, then form a vector sum to get the total momentum. Let's take m₁ = 2.0 kg, v₁ = 3.0 m/s, m₂ = 3.0 kg,
v₂ = 1.5 m/s, m₃ = 4.0 kg, and v₃ = 2.0 m/s.

Solution

1. Determine the components of p1:
2. Determine the components of p2:
3. Determine the components of p3:
4. Determine the components of ptotal:
5. Determine the magnitude of ptotal:
6. Determine the direction of ptotal:

Insights Take note of the fact that for p₂ the signs of the components were put in from our knowledge of their directions instead of using the angle from the +x axis. This way of doing it is common so be sure you understand it. Also notice that since both components of p_total are positive, the 3.8^o is the angle from the +x axis and nothing else needs to be said.

---

Practice Quiz

If two objects have the same velocity but one has twice the mass of the other, the object with larger mass has half the momentum of the other. has twice the momentum of the other. has the same momentum of the other. must move slower than the other. [none of the above]

If two objects have the same mass and speed but move in opposite directions, then they have momenta of equal magnitude and opposite directions. they have equal momenta. they each have zero momenta. one must move faster than the other. [none of the above]

A 3.2 kg object has a velocity of 1.8 m/s in the x-direction. What is its linear momentum? 5.0 1.8 0.56 5.8 1.4

9-2 Momentum and Newton's Second Law

The form of Newton's second law that we have been using until now, F_net = ma, only applies to circumstances in which the mass remains constant. However, in the most general cases, as with rockets, the mass may change during the motion. The most general form of Newton's second law is expressed in terms of momentum,

.

So, force equals the rate at which momentum changes whether the change is in the mass, the velocity, or both. For cases when mass is constant, this reduces to F_net = ma.

---

Example 9.2 Newton's Second Law: A 0.25 kg object moves due east at 2.1 m/s. What force is needed to cause it to move due north at 3.6 m/s in 1.52 s?

Solution: Even though this is a case of constant mass, let's work it in terms of momentum to illustrate that approach.

Given: m = 0.25 kg, v_i = 2.1 m/s, v_f = 3.6 m/s, t = 1.52 s. Find: F

The intial and final momenta of the object are

p_i = mv_i = 0.25 kg(2.1 m/s) = 0.525 ,
p_f = mv_f = 0.25 kg(3.6 m/s) = 0.900 .

The change in momentum, therefore, is

.

The force, then, is given by

.

Insight If you are curious, work this out as F = ma and check that you get the same result.

---

Practice Quiz

When does the equation F = ma not accurately describe the dynamics of an object? when gravity is not present when more than one force acts on the object when kinetic frictions does work on the object when the mass of the object is constant [none of the above]

If the linear momentum of a 1.6 kg object is decreasing at a rate of , what is the magnitude of the force on the object? 8.0 N 3.4 N 5.0 N 6.6 N 3.1 N

9-3 Impulse

As mentioned above, the concept of momentum is important when two objects interact. In this chapter, the interaction we focus on is called a collision. A collision occurs when the forces of interaction between two objects are large for a finite period of time. The average force applied to an object times the amount of time this force is applied is called the impulse I

I = F_avDt.

The SI unit of impulse is the which has no special name. The same impulse can be delivered by a weak force acting for a long period of time or a strong force acting for a short period of time. The most common usage of impulse is for the latter case. The concept of impulse is closely related to momentum by what is often called the impulse-momentum theorem

I = Dp.

The above expression is really just a restatement of Newton's second law in a form that is convenient to describe interactions, like collisions, for which it is difficult to know precise values of the forces involved.

---

Physlet Illustration: Momentum and Impulse
		Interactive Help on      off
Two 100-gm balls are thrown at the same wall. The blue ball rebounds elastically from the wall, while the red ball sticks to the wall upon impact. The distance grid is in meters and the times are shown in seconds. What is the change in momentum that each ball undergoes upon impact with the wall? What can you conclude about the impulse delivered by the wall to each ball? Start
Hints What is the blue ball's velocity before it hits the wall? Its momentum? (Remember: these are vector quantities!) What is its velocity after it hits the wall? Its momentum? At impact, then, what is the blue ball's change in momentum? What is the red ball's velocity before it hits the wall? Its momentum? What is its velocity after it hits the wall? Its momentum? At impact, then, what is the red ball's change in momentum? Recall that the impulse equals the change in momentum.

Example 9.3 Ricochet: The velocity of a rock of mass 0.24 kg moving with a speed of 3.33 m/s makes a 60^o angle with the normal to a brick wall. The rock is in contact with the wall for only 0.032 s. If the velocity of the rock makes an angle of 40^o with the normal to the wall after it strikes and has a magnitude of 2.68 m/s, (a) what impulse does the wall apply to the rock, and (b) what average force causes this impulse?

Picture the Problem The picture shows the initial and final momenta of the rock as it bounces off the wall.

Strategy To solve for the impulse, we can find the change in momentum that results from the bounce. Once the impulse is known, we'll use it to get the average force.

Solution
Part (a)

1. Determine the components of the initial and final momenta of the rock:
2. Determine the impulse as the change in momentum:

Part (b)

1. Use the definition of impulse to get an expression for the average force:
2. Calculate the numerical value of the force:

Insight Be sure that the signs and relative magnitudes of the components of I and F_av make sense to you.

---

Practice Quiz

During one trial, a force F is applied to a ball for an amount of time t. During a second trial a force of 3F is applied for a time of t/2. Which of the following is true concerning the magnitude of the change in momentum of the ball? there's a greater change for the second trial there's a smaller change for the second trial we get the same nonzero momentum change in each trial the momentum doesn't change in either trial [none of the above]

An object of mass 2.8 kg moves at 1.1 m/s in the +x-direction. If an impulse of is applied, what is its final momentum? 3.1 4.0 5.2 4.4 4.2

9-1 - 9-3   9-4 - 9-6   9-7 - 9-8   next