Unit 1 – Momentum and Impulse

Unit 1 – Momentum and Impulse
Physics 30 Unit 1 – Momentum and Impulse To accompany Pearson Physics

Momentum “quantity of motion” - Newton
is the product of mass and velocity Is a vector, has units of

Momentum Newton’s 2nd Law, , can be rewritten as
as your text shows on page We’ll use this later to determine a quantity called impulse Try Practice Problem 1 page 451

Momentum Practice Problem 1, page 451:
m = 65 kg kg = 600 kg Since velocity is a vector, direction must be given along with the numerical answer as shown on page 451

Momentum Momentum “varies directly as” or “is directly proportional to” both Study Example 9.2 page 452 Try Practice Problem 1 page 452

Momentum Practice Problem 1, page 452

Momentum Do questions 1 – 3, 5, 6, 8, 9, 11 page 453

Momentum and Impulse This equation can be rewritten as:
Quick Lab 9.2 page 455 (discuss or do and discuss) Earlier you saw that Newton’s 2nd Law could be rewritten as: This equation can be rewritten as:

Momentum and Impulse The change in momentum is equal to the net force times the time for the change, This quantity is called IMPULSE and is equal to the change in momentum

Momentum and Impulse Since , For a given change in momentum (or change in velocity if mass is constant), a fast change will mean a large force while a slower change will mean a smaller force

Momentum and Impulse Practical applications?
cushioned soles on runners • airbags • crumple zones in cars • others? Additional examples p

Momentum and Impulse Units of impulse: Units of momentum:
Since impulse = change in momentum, these units must be same. Your text shows this on page 457.

Momentum and Impulse Same as the direction of the change of momentum
Direction of Impulse??? Same as the direction of the change of momentum If you are travelling west in a car at constant speed, and hit the brakes, Impulse is east If you are travelling west in a car at constant speed, and step on the gas, Impulse is west

Momentum and Impulse Try Practice Problems 1 and 2 page 458

Momentum and Impulse Practice Problem 1, p. 458

Momentum and Impulse Practice Problem 2, p Since velocity is a vector, it is necessary to take direction into account when determining change in velocity

Momentum and Impulse When the net force is not constant, the impulse can be calculated by finding the area under an F vs t graph Your textbook has a number of examples on pages 459 – 461 Discuss golf ball graph page 462

Momentum Review Example 9.4, page 462 Do Practice Problem 1, page 462

Momentum and Impulse Practice Problem 1, page 462 c) a)
b) Impulse = total area = A1+A2 A1 A2

Momentum and Impulse Do questions 1, 2, 4, 7, 8, 9b, 10, page 467

Momentum: 1d Collisions
Note that in collisions friction is present as a constant force before, during, and after the collision Since collision times are relatively short, friction can be ignored in collision questions provided you are using only momentum of the objects immediately before or after the collision This will always be the case for you

Momentum Lab 9.5 page 471

Momentum : 1d Collisions
Law of Conservation of Momentum: When no external net force acts on a system, the momentum of the system remains constant Review pages 474 – 479 including examples 9.5, 9.6, 9.7, and 9.8

Try Practice Problem 1 on page 476, Practice Problem 2 on page 477, Practice Problem 1 on page 478, and Practice Problem 1 on page 479 Types of interactions: explosions including:???? hit and stick hit stationary and bounce back hit moving and bounce back

Practice Problem 1, page The negative sign on the velocity means it is opposite in direction to the astronaut, that is, towards the astronaut Final expression of answer???

Momentum:1d Collisions
Practice Problem 2, page 477 Since initial and final velocities are in the same direction, there is no need to assign positive and negative values

Practice Problem 1 page 478 1.2 m/s [ W ] Note the use of positive and negative on the velocities: + right, [E], up, [N] - left, [ W ], down, [S]

Practice Problem 1 page 479 0.62 m/s [E]

Momentum : Elastic and Inelastic Collisions
Elastic collisions: Ek conserved Since Ek is a scalar direction of v is unimportant No true elastic collisions in the macroscopic world: Collision between: % Elastic tennis ball and tile floor 66% golf ball and tile floor 87% basketball and tile floor 74%

Inelastic collisions: Ek not conserved Macroscopic collisions are always inelastic to some degree 100% inelastic – both objects are deformed and stick together Collisions between vehicles always involve deformation even when the vehicles bounce off each other Review examples 9.9 and 9.10 Do Practice Problem 1 on each of pages 484 and 485

Practice Problem 1, page 484 Energy conserved in pendulum swing: Momentum conserved in collision of bullet and block

Practice Problem 1 page 485

Do Check and Reflect Questions, page 486 1, 2a, 3, 6, 8, 10

Real collisions are most often 3 dimensional, but the techniques to solve them are the same as those for 2 dimensional Because momentum is a vector, vector analysis is required: Components!

Look at the information available on the Physics 30 Data Sheets: I recommend, that like in Physics 20, you measure all angles with respect to the positive x-axis

If you do this, x and y components of any vector R will be found by: The angle between the x axis positive or negative and the vector will always be determined by:

You can save work by drawing a good vector diagram and using the law of sines and law of cosines to solve even non-right angle triangles I don’t recommend this Review example 9.12 – I’ll redo it here

First change 12.0°[E of N] to 78.0° [N of E] (standard position) Since both masses are the same we can ignore mass in the calculation – work with velocities only vector x-component y-component 1.20 m/s ?

We now have everything needed to find !

Since x is negative and y is positive the resultant is in the 2nd quadrant – therefore 12.9° [N of W ] or 167°

Try Practice Problem 1, page 491

vector x-component Σx y-component Σy 189 kg•m/s 189 kg•m/s ?

Practice Problem 1, page 491, continued Since both x and y are positive, this is a 1st quadrant angle: [15.7° N of E]

Review Example 9.13, page 492 Try Practice Problem 2, page 492

Practice Problem 2, page 492 vector x-component Σx y-component Σy -150 kg•m/s -150 kg•m/s -454 -454 kg•m/s ?

Since x and y are both negative this is in a 3rd quadrant angle 71.7° [S of W ]

Review Example 9.14, page 494 Try Practice Problem 1, page 494

Practice Problem 1, page 494 Mass of missing 3rd fragment = 0.058 kg – kg – kg = kg

Practice Problem 2, page 494 Momentum of piece # x-component y-component 0.043 kg•m/s 0.034 kg•m/s ?

mass of 3rd piece Since both x and y components are negative the angle is in the 3rd quadrant: 52.1° [S of W ]

2d collisions can likewise be elastic or inelastic; the test is to check for conservation of kinetic energy Review Example 9.15, page 496 Try Practice Problem 1, page 497

Practice Problem 1, page 497 Since Ek is a scalar, it is not necessary to consider direction of velocity (nice) Ek puck: Ek goalie: Total Ek before:

Ek goalie and puck: Inelastic collision → percent elastic:

Subatomic particles are the only “objects” that can have true 100% elastic collisions Do Check and Reflect, page 499, questions 1, 5, 6, 7, 8, 10

Unit 1 – Momentum and Impulse

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