Section 8.4 Stretching and Compressing Materials © 2015 Pearson Education, Inc.

Stretching and Compressing Materials We can model most solid materials as being made up of particle-like atoms connected by spring-like bonds. Pulling on a steel rod will slightly stretch the bonds between particles, and the rod will stretch. © 2015 Pearson Education, Inc.

Stretching and Compressing Materials Steel is elastic, but under normal forces it experiences only small changes in dimension. Materials of this sort are called rigid. Rubber bands and other materials that can be stretched easily or show large deformations with small forces are called pliant. © 2015 Pearson Education, Inc.

Stretching and Compressing Materials For a rigid rod, the spring constant depends on the cross-sectional area A, the length of the rod, L, and the material from which it is made. The constant Y is called Young’s modulus and is a property of the material from which the rod is made. © 2015 Pearson Education, Inc.

QuickCheck 8.4 Bars A and B are attached to a wall on the left and pulled with equal forces to the right. Bar B, with twice the radius, is stretched half as far as bar A. Which has the larger value of Young’s modulus Y? YA > YB YA = YB YA < YB Not enough information to tell Answer: A © 2015 Pearson Education, Inc. 5 5

QuickCheck 8.4 Bars A and B are attached to a wall on the left and pulled with equal forces to the right. Bar B, with twice the radius, is stretched half as far as bar A. Which has the larger value of Young’s modulus Y? YA > YB YA = YB YA < YB Not enough information to tell F A = Y D L Area of B increases by 4. If YB = YA, stretch would be only L/4. Stretch of L/2 means B is “softer” than A. © 2015 Pearson Education, Inc. 6 6

Stretching and Compressing Materials © 2015 Pearson Education, Inc.

Stretching and Compressing Materials The restoring force can be written in terms of the change in length ΔL: It is useful to rearrange the equation in terms of two ratios, the stress and the strain: The unit of stress is N/m2 If stress is due to stretching, we call it tensile stress. © 2015 Pearson Education, Inc.

Example 8.8 Finding the stretch of a wire A Foucault pendulum in a physics department (used to prove that the earth rotates) consists of a 120 kg steel ball that swings at the end of a 6.0 m long steel cable. The cable has a diameter of 2.5 mm. When the ball was first hung from the cable, by how much did the cable stretch? © 2015 Pearson Education, Inc.

Example 8.8 Finding the stretch of a wire (cont.) prepare The amount by which the cable stretches depends on the elasticity of the steel cable. Young’s modulus for steel is given in Table 8.1 as Y = 20  1010 N/m2. solve Equation 8.6 relates the stretch of the cable ∆L to the restoring force F and to the properties of the cable. Rearranging terms, we find that the cable stretches by The cross-section area of the cable is © 2015 Pearson Education, Inc.

Example 8.8 Finding the stretch of a wire (cont.) The restoring force of the cable is equal to the ball’s weight: The change in length is thus © 2015 Pearson Education, Inc.

Example 8.8 Finding the stretch of a wire (cont.) assess If you’ve ever strung a guitar with steel strings, you know that the strings stretch several millimeters with the force you can apply by turning the tuning pegs. So a stretch of 7 mm under a 120 kg load seems reasonable. © 2015 Pearson Education, Inc.

Beyond the Elastic Limit As long as the stretch stays within the linear region, a solid rod acts like a spring and obeys Hooke’s law. As long as the stretch is less than the elastic limit, the rod returns to its initial length when force is removed. The elastic limit is the end of the elastic region. © 2015 Pearson Education, Inc.

Beyond the Elastic Limit For a rod or cable of a particular material, there is an ultimate stress. The ultimate stress, or tensile strength, is the largest stress the material can sustain before breaking. © 2015 Pearson Education, Inc.

Biological Materials Most bones in your body are made of two different kinds of bony material: dense and rigid cortical (or compact bone) on the outside, and porous, flexible cancellous (or spongy) bone on the inside. © 2015 Pearson Education, Inc.

Biological Materials © 2015 Pearson Education, Inc.

Biological Materials and Others Note: A N/m2 is called a Pascal (Pa), so 106 N/m2 = 1 MPa Compare these to some other materials such as Steel : 841 MPa Glass: 33 MPa Concrete: 3 MPa Wood: 40 MPa Diamond: 2800 MPa Carbon Nanotubes: 63,000 MPa © 2015 Pearson Education, Inc.

© 2015 Pearson Education, Inc.

Chapter 9 Momentum Chapter Goal: To learn about impulse, momentum, and a new problem-solving strategy based on conservation laws. © 2015 Pearson Education, Inc.

Chapter 9 Preview Looking Ahead: Impulse This golf club delivers an impulse to the ball as the club strikes it. You’ll learn that a longer-lasting, stronger force delivers a greater impulse to an object. © 2015 Pearson Education, Inc.

Chapter 9 Preview Looking Ahead: Momentum and Impulse The impulse delivered by the player’s head changes the ball’s momentum. You’ll learn how to calculate this momentum change using the impulse-momentum theorem. © 2015 Pearson Education, Inc.

Chapter 9 Preview Looking Ahead: Conservation of Momentum The momentum of these pool balls before and after they collide is the same—it is conserved. You’ll learn a powerful new before-and-after problem-solving strategy using this law of conservation of momentum. © 2015 Pearson Education, Inc.

Chapter 9 Preview Looking Back: Newton’s Third Law In Section 4.7, you learned about Newton’s third law. In this chapter, you’ll apply this law in order to understand the conservation of momentum. Newton’s third law states that the force that object B exerts on A has equal magnitude but is directed opposite to the force that A exerts on B. © 2015 Pearson Education, Inc.

Chapter 9 Preview Stop to Think A hammer hits a nail. The force of the nail on the hammer is Greater than the force of the hammer on the nail. Less than the force of the hammer on the nail. Equal to the force of the hammer on the nail. Zero. Stop to Think Answer: C © 2015 Pearson Education, Inc.

Chapter 9 Preview Stop to Think A hammer hits a nail. The force of the nail on the hammer is Greater than the force of the hammer on the nail. Less than the force of the hammer on the nail. Equal to the force of the hammer on the nail. Zero. Stop to Think Answer: C © 2015 Pearson Education, Inc.

Section 9.1 Impulse © 2015 Pearson Education, Inc.

Impulse A collision is a short-duration interaction between two objects. During a collision, it takes time to compress the object, and it takes time for the object to re-expand. The duration of a collision depends on the materials. © 2015 Pearson Education, Inc.

Impulse When kicking a soccer ball, the amount by which the ball is compressed is a measure of the magnitude of the force the foot exerts on the ball. The force is applied only while the ball is in contact with the foot. The impulse force is a large force exerted during a short interval of time. © 2015 Pearson Education, Inc.

Impulse The effect of an impulsive force is proportional to the area under the force-versus-time curve. The area is called the impulse J of the force. © 2015 Pearson Education, Inc.

Impulse It is useful to think of the collision in terms of an average force Favg. Favg is defined as the constant force that has the same duration Δt and the same area under the force curve as the real force. © 2015 Pearson Education, Inc.

Impulse Impulse has units of N  s, but N  s are equivalent to kg  m/s. The latter are the preferred units for impulse. The impulse is a vector quantity, pointing in the direction of the average force vector: © 2015 Pearson Education, Inc.

Example 9.1 Finding the impulse on a bouncing ball A rubber ball experiences the force shown in the figure to the right as it bounces off the floor. What is the impulse on the ball? What is the average force on the ball? © 2015 Pearson Education, Inc.

Example 9.1 Finding the impulse on a bouncing ball (cont.) prepare The impulse is the area under the force curve. Here the shape of the graph is triangular, so we use the area of a triangle:  height  base. solve a. The impulse is Since J = Favg ∆t, we compute © 2015 Pearson Education, Inc.

Example 9.1 Finding the impulse on a bouncing ball (cont.) assess In this particular example, the average value of the force is half the maximum value. This is not surprising for a triangular force because the area of a triangle is half the base times the height. © 2015 Pearson Education, Inc.

Section 9.2 Momentum and the Impulse-Momentum Theorem © 2015 Pearson Education, Inc.

Momentum and the Impulse-Momentum Theorem Intuitively we know that giving a “kick” to a heavy object will change its velocity much less than giving the same “kick” to a light object. We can calculate how the final velocity is related to the initial velocity for a given “kick”. © 2015 Pearson Education, Inc.

Momentum and the Impulse-Momentum Theorem From Newton’s second law, the average acceleration of an object during the time the force is being applied is The average acceleration is related to the change in the velocity by We combine those two equations to find © 2015 Pearson Education, Inc.

Momentum and the Impulse-Momentum Theorem We can rearrange that equation in terms of impulse: Momentum is the product of the object’s mass and velocity. It has units of kg  m/s. © 2015 Pearson Education, Inc.

Momentum and the Impulse-Momentum Theorem Momentum is a vector quantity that points in the same direction as the velocity vector: The magnitude of an object’s momentum is simply the product of the object’s mass and speed. © 2015 Pearson Education, Inc.

Momentum and the Impulse-Momentum Theorem © 2015 Pearson Education, Inc.

The Impulse-Momentum Theorem Impulse and momentum are related as: The impulse-momentum theorem states that an impulse delivered to an object causes the object’s momentum to change. Impulse can be written in terms of its x- and y- components: © 2015 Pearson Education, Inc.

The Impulse-Momentum Theorem © 2015 Pearson Education, Inc.

QuickCheck 9.2 A 2.0 kg object moving to the right with speed 0.50 m/s experiences the force shown. What are the object’s speed and direction after the force ends? 0.50 m/s left At rest 0.50 m/s right 1.0 m/s right 2.0 m/s right Answer: D © 2015 Pearson Education, Inc. 43

QuickCheck 9.2 A 2.0 kg object moving to the right with speed 0.50 m/s experiences the force shown. What are the object’s speed and direction after the force ends? 0.50 m/s left At rest 0.50 m/s right 1.0 m/s right 2.0 m/s right px = Jx or pfx = pix + Jx © 2015 Pearson Education, Inc. 44

QuickCheck 9.4 A force pushes the cart for 1 s, starting from rest. To achieve the same speed with a force half as big, the force would need to push for s 1 s 2 s 4 s 1 4 1 2 Answer: D © 2015 Pearson Education, Inc. 45

QuickCheck 9.4 A force pushes the cart for 1 s, starting from rest. To achieve the same speed with a force half as big, the force would need to push for s 1 s 2 s 4 s 1 4 1 2 © 2015 Pearson Education, Inc. 46

Example 9.2 Calculating the change in momentum A ball of mass m = 0.25 kg rolling to the right at 1.3 m/s strikes a wall and rebounds to the left at 1.1 m/s. What is the change in the ball’s momentum? What is the impulse delivered to it by the wall? © 2015 Pearson Education, Inc.

Example 9.2 Calculating the change in momentum (cont.) prepare This is a new kind of visual overview, one in which we show the situation “before” and “after” the interaction. The change in momentum is the difference between the final and initial values of the momentum. By the impulse-momentum theorem, the impulse is equal to this change in momentum. [Insert Figure 9.8 (repeated)] © 2015 Pearson Education, Inc.

Example 9.2 Calculating the change in momentum (cont.) solve The x-component of the initial momentum is The y-component of the momentum is zero both before and after the bounce. After the ball rebounds, the x-component is [Insert Figure 9.8 (repeated)] © 2015 Pearson Education, Inc.

Example 9.2 Calculating the change in momentum (cont.) It is particularly important to notice that the x-component of the momentum, like that of the velocity, is negative. This indicates that the ball is moving to the left. The change in momentum is [Insert Figure 9.8 (repeated)] © 2015 Pearson Education, Inc.

Example 9.2 Calculating the change in momentum (cont.) By the impulse-momentum theorem, the impulse delivered to the ball by the wall is equal to this change, so assess The impulse is negative, indicating that the force causing the impulse is pointing to the left, which makes sense. © 2015 Pearson Education, Inc.