- •Preface
- •Contents
- •1 Introduction
- •1.1 Physics
- •1.2 Mechanics
- •1.3 Integrating Numerical Methods
- •1.4 Problems and Exercises
- •1.5 How to Learn Physics
- •1.5.1 Advice for How to Succeed
- •1.6 How to Use This Book
- •2 Getting Started with Programming
- •2.1 A Python Calculator
- •2.2 Scripts and Functions
- •2.3 Plotting Data-Sets
- •2.4 Plotting a Function
- •2.5 Random Numbers
- •2.6 Conditions
- •2.7 Reading Real Data
- •2.7.1 Example: Plot of Function and Derivative
- •3 Units and Measurement
- •3.1 Standardized Units
- •3.2 Changing Units
- •3.4 Numerical Representation
- •4 Motion in One Dimension
- •4.1 Description of Motion
- •4.1.1 Example: Motion of a Falling Tennis Ball
- •4.2 Calculation of Motion
- •4.2.1 Example: Modeling the Motion of a Falling Tennis Ball
- •5 Forces in One Dimension
- •5.1 What Is a Force?
- •5.2 Identifying Forces
- •5.3.1 Example: Acceleration and Forces on a Lunar Lander
- •5.4 Force Models
- •5.5 Force Model: Gravitational Force
- •5.6 Force Model: Viscous Force
- •5.6.1 Example: Falling Raindrops
- •5.7 Force Model: Spring Force
- •5.7.1 Example: Motion of a Hanging Block
- •5.9.1 Example: Weight in an Elevator
- •6 Motion in Two and Three Dimensions
- •6.1 Vectors
- •6.2 Description of Motion
- •6.2.1 Example: Mars Express
- •6.3 Calculation of Motion
- •6.3.1 Example: Feather in the Wind
- •6.4 Frames of Reference
- •6.4.1 Example: Motion of a Boat on a Flowing River
- •7 Forces in Two and Three Dimensions
- •7.1 Identifying Forces
- •7.3.1 Example: Motion of a Ball with Gravity
- •7.4.1 Example: Path Through a Tornado
- •7.5.1 Example: Motion of a Bouncing Ball with Air Resistance
- •7.6.1 Example: Comet Trajectory
- •8 Constrained Motion
- •8.1 Linear Motion
- •8.2 Curved Motion
- •8.2.1 Example: Acceleration of a Matchbox Car
- •8.2.2 Example: Acceleration of a Rotating Rod
- •8.2.3 Example: Normal Acceleration in Circular Motion
- •9 Forces and Constrained Motion
- •9.1 Linear Constraints
- •9.1.1 Example: A Bead in the Wind
- •9.2.1 Example: Static Friction Forces
- •9.2.2 Example: Dynamic Friction of a Block Sliding up a Hill
- •9.2.3 Example: Oscillations During an Earthquake
- •9.3 Circular Motion
- •9.3.1 Example: A Car Driving Through a Curve
- •9.3.2 Example: Pendulum with Air Resistance
- •10 Work
- •10.1 Integration Methods
- •10.2 Work-Energy Theorem
- •10.3 Work Done by One-Dimensional Force Models
- •10.3.1 Example: Jumping from the Roof
- •10.3.2 Example: Stopping in a Cushion
- •10.4.1 Example: Work of Gravity
- •10.4.2 Example: Roller-Coaster Motion
- •10.4.3 Example: Work on a Block Sliding Down a Plane
- •10.5 Power
- •10.5.1 Example: Power Exerted When Climbing the Stairs
- •10.5.2 Example: Power of Small Bacterium
- •11 Energy
- •11.1 Motivating Examples
- •11.2 Potential Energy in One Dimension
- •11.2.1 Example: Falling Faster
- •11.2.2 Example: Roller-Coaster Motion
- •11.2.3 Example: Pendulum
- •11.2.4 Example: Spring Cannon
- •11.3 Energy Diagrams
- •11.3.1 Example: Energy Diagram for the Vertical Bow-Shot
- •11.3.2 Example: Atomic Motion Along a Surface
- •11.4 The Energy Principle
- •11.4.1 Example: Lift and Release
- •11.4.2 Example: Sliding Block
- •11.5 Potential Energy in Three Dimensions
- •11.5.1 Example: Constant Gravity in Three Dimensions
- •11.5.2 Example: Gravity in Three Dimensions
- •11.5.3 Example: Non-conservative Force Field
- •11.6 Energy Conservation as a Test of Numerical Solutions
- •12 Momentum, Impulse, and Collisions
- •12.2 Translational Momentum
- •12.3 Impulse and Change in Momentum
- •12.3.1 Example: Ball Colliding with Wall
- •12.3.2 Example: Hitting a Tennis Ball
- •12.4 Isolated Systems and Conservation of Momentum
- •12.5 Collisions
- •12.5.1 Example: Ballistic Pendulum
- •12.5.2 Example: Super-Ball
- •12.6 Modeling and Visualization of Collisions
- •12.7 Rocket Equation
- •12.7.1 Example: Adding Mass to a Railway Car
- •12.7.2 Example: Rocket with Diminishing Mass
- •13 Multiparticle Systems
- •13.1 Motion of a Multiparticle System
- •13.2 The Center of Mass
- •13.2.1 Example: Points on a Line
- •13.2.2 Example: Center of Mass of Object with Hole
- •13.2.3 Example: Center of Mass by Integration
- •13.2.4 Example: Center of Mass from Image Analysis
- •13.3.1 Example: Ballistic Motion with an Explosion
- •13.4 Motion in the Center of Mass System
- •13.5 Energy Partitioning
- •13.5.1 Example: Bouncing Dumbbell
- •13.6 Energy Principle for Multi-particle Systems
- •14 Rotational Motion
- •14.2 Angular Velocity
- •14.3 Angular Acceleration
- •14.3.1 Example: Oscillating Antenna
- •14.4 Comparing Linear and Rotational Motion
- •14.5 Solving for the Rotational Motion
- •14.5.1 Example: Revolutions of an Accelerating Disc
- •14.5.2 Example: Angular Velocities of Two Objects in Contact
- •14.6 Rotational Motion in Three Dimensions
- •14.6.1 Example: Velocity and Acceleration of a Conical Pendulum
- •15 Rotation of Rigid Bodies
- •15.1 Rigid Bodies
- •15.2 Kinetic Energy of a Rotating Rigid Body
- •15.3 Calculating the Moment of Inertia
- •15.3.1 Example: Moment of Inertia of Two-Particle System
- •15.3.2 Example: Moment of Inertia of a Plate
- •15.4 Conservation of Energy for Rigid Bodies
- •15.4.1 Example: Rotating Rod
- •15.5 Relating Rotational and Translational Motion
- •15.5.1 Example: Weight and Spinning Wheel
- •15.5.2 Example: Rolling Down a Hill
- •16 Dynamics of Rigid Bodies
- •16.2.1 Example: Torque and Vector Decomposition
- •16.2.2 Example: Pulling at a Wheel
- •16.2.3 Example: Blowing at a Pendulum
- •16.3 Rotational Motion Around a Moving Center of Mass
- •16.3.1 Example: Kicking a Ball
- •16.3.2 Example: Rolling down an Inclined Plane
- •16.3.3 Example: Bouncing Rod
- •16.4 Collisions and Conservation Laws
- •16.4.1 Example: Block on a Frictionless Table
- •16.4.2 Example: Changing Your Angular Velocity
- •16.4.3 Example: Conservation of Rotational Momentum
- •16.4.4 Example: Ballistic Pendulum
- •16.4.5 Example: Rotating Rod
- •16.5 General Rotational Motion
- •Index
12.3 Impulse and Change in Momentum |
361 |
The change in momentum is therefore |
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p = mv1 − mv0 = m(−v0) − mv0 = −2mv0 = −2 p0 . |
(12.27) |
If we know the duration, t , of the collision, we can find the average net force on the ball during the collision from
F net |
= |
−2mv0 |
. |
(12.28) |
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avg |
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t |
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12.3.2 Example: Hitting a Tennis Ball
Problem: A tennis ball of mass 57 g is approaching you with a horizontal velocity v0 = 20 m/s. You hit the ball, returning it with a horizontal velocity v1 = 20 m/s, now in the opposite direction. (a) What is the impulse J on the ball while it is in contact with the racket during the collision? (b) The ball and racket are in contact for 2.0 ms. What is the average net force on the racket during the collision? (c) You want to return the ball as a high lob and give the ball a velocity v1 = 15 m/s at angle of 45◦ upward. What is now the impulse on the ball and the net force from the racket on the ball?
Approach: We may solve this problem by determining the motion of the ball from Newton’s laws of motion, but this would require a detailed force model for the force from the tennis racket on the ball. In this case, we do not have such a model. Instead, we want to use the measured change in velocity to determine the average force on the ball.
Identify: In this problem we address the motion of the tennis ball, described by the position r(t ). The ball starts with the velocity v0 = −v0 i at the time t0 (before the collision), and gets the velocity v1 after the collision.
Model: The ball is affected by a force, F(t ), from the racket on the ball, and by gravity. However, we assume that gravity is small compared with the typical force from the racket, and ignore the effects of gravity.
Solve: The impulse is defined as the integral of the net force on the ball, and it is equal to the change in momentum of the ball:
J = p = p1 − p0 = mv1 − mv0 , |
(12.29) |
Solution part a: In part (a) of the problem, the final velocity is v1 = v1 i. The key idea is that momentum is a vector quantity. The ball therefore experiences a change in momentum, even though the magnitude of the momentum does not change, because the direction of the momentum changes.
362 |
12 Momentum, Impulse, and Collisions |
|
The impulse on the ball in the collision is: |
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|
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J = m (v1 i − (−v0 i)) = m(v1 + v0) i |
(12.30) |
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= 0.057 kg (20.0 m/s + 20.0 m/s) i = 2.28 kg m/s i . |
(12.31) |
The impulse is positive, since the force acting on the ball is in the positive x - direction—this is also the direction of the acceleration of the ball.
Solution of part b: In part (b) of the problem, we find the average force from the change in momentum during the collision:
Favg = |
1t |
t0 |
Fdt = |
t |
= 2 |
2 10−3 s |
= 1140N , |
(12.32) |
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t1 |
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p |
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.28 kg m/s |
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This is the average net force on the ball. We recall from Fig. 12.3 that the net force is smaller than the maximum force, although they are typically of comparable magnitude.
Dicussion: What about gravity? We neglected gravity because we assumed it to be small compared with the force from the racket. We could check this assumption in two ways. First, we could check that the impulse of gravity is much smaller than the total impulse on the ball—we can do this without calculating the average force. The magnitude of gravity is:
W = mg = 0.057 kg 9.8 m/s2 = 0.56 N . |
(12.33) |
The impulse of gravity is therefore:
Jg = mg t = 0.56 M 2 10−3 s = 1.1 10−2 kg m/s , |
(12.34) |
which is much smaller than the impulse of the net force.
From this calculation we also found the force from gravity, which is much smaller than the average net force on the ball. We were therefore right in neglecting the effect of gravity.
Solution to part c: Finally, we address question (c), where the collision is not head on, and the ball leaves the racket at an angle α, as illustrated in Fig. 12.6. In this case, we need to treat the collision as two-dimensional. First, we introduce the velocity of the ball after the collision as:
v1 = v1 (cos α i + sin α j) , |
(12.35) |
where v1 = 15 m/s and α = 45◦ = π/4. The impulse on the ball is still given as the change in momentum:
J = p1 − p0 = m (v1 − v0) . |
(12.36) |