CHAPTER 3 LINEAR MOMENTUM COLLISIONS Semester 1 2015

CHAPTER 3: LINEAR MOMENTUM & COLLISIONS Semester 1 2015 / 2016

Linear Momentum and Collisions

OBJECTIVES • Ability to compute linear momentum and components of momentum • Ability to relate impulse and momentum & kinetic energy and momentum • Ability to describe and solve the conditions on kinetic energy and momentum in elastic and inelastic collisions.

TOPICS: • Linear Momentum • Impulse • Conservation of Linear Momentum • Elastic and Inelastic Collisions

Linear Momentum • Definition The linear momentum of an object is the product of its mass and velocity • Note that momentum is a vector – it has both a magnitude and a direction. • SI unit of momentum: kg • m/s. This unit has no special name.

Linear Momentum • For a system of objects, the total momentum is the vector sum of each.

Example 1: A 100 kg football player run with a velocity of 4. 0 m/s straight down the field. A 1. 0 kg artillery shell leaves the barrel of a gun with a muzzle velocity of 500 m/s. Which has the greater momentum (magnitude), the football player or the shell?

Example 2: What is the total momentum for each of the systems of particles illustrated in Fig (a)?

What is the total momentum for each of the systems of particles illustrated in Fig (b)?

Linear Momentum • The change in momentum is the difference between the momentum vectors. • Here, the vector sum is zero, but the vector difference, or change in momentum, is not. (The particles are displaced for convenience. )

7. 1 A car weighing 12 k. N is driving due north at 30. 0 m/s. After driving around a sharp curve, the car is moving east at 13. 6 m/s. What is the change in momentum of the car? Slide 11

7. 1 Strategy There are two potential pitfalls: 1. momentum depends not on weight but on mass, and 2. momentum is a vector, so we must take its direction into consideration as well as its magnitude. To find the change in momentum, we need to do a vector subtraction. Slide 12

7. 1 Solution Slide 13

7. 1 Solution The change in momentum of the car as 4. 0 × 104 kg·m/s directed 24° east of south. Slide 14

Linear Momentum • The change in momentum is found by computing the change in the components.

Linear Momentum • If an object’s momentum changes, a force must have acted on it. • The net force is equal to the rate of change of the momentum.

Impulse • Impulse is the change in momentum: • SI units of impulse : newton-second (N. s)/ 1 kgm/s= 1 N. s • Dimension of impulse MLT-1 • Impulse is not property of the particle, but is a measure of the change in momentum of the particle

THE IMPULSE-MOMENTUM THEOREM The area under the Fx(t) graph for a variable force is the impulse. Slide 19

THE IMPULSE-MOMENTUM THEOREM The average force for a given time interval is the constant force that would produce the same impulse. Slide 20

Impulse-Momentum Theorem Impulse When Forces Are Not Constant: Slide 21

Impulse When a moving object stops, its impulse depends only on its change in momentum. This can be accomplished by a large force acting for a short time, or a smaller force acting for a longer time.

Example 3: A golfer drives a 0. 1 kg ball from an elevated tee, the ball an initial horizontal speed of 40 m/s. The club and the ball are in contact for 1 ms. Calculate average force exerted by the club on the ball during his time.

Conservation of Linear Momentum • If there is no net force acting on a system, its total momentum cannot change. • This is the law of conservation of momentum. • If there are internal forces, the momentum of individual parts of the system can change, but the overall momentum stays the same.

Conservation of Linear Momentum • For the linear momentum of an object to be conserved, its follow Newton’s Second Law • If the net force acting on a particle is zero, hence Fnet = ∆p / ∆t = 0 ∆p = 0 = p - p 0 • Therefore conservation of linear momentum:

Conservation of Linear Momentum • In this example, there is no external force, but the individual components of the system do change their momentum. • The spring force is an internal force, so the momentum of the system is conserved.

Types of Collisions: Elastic Collisions: i. Total kinetic energy is conversed ii. the total kinetic energy of all the objects of the system after the collision is the same as the total kinetic energy before the collision

Types of Collisions: Inelastic collision: i. total kinetic energy is not conversed

Energy and Momentum in Inelastic Collisions • Figure above, m 1=m 2 and V 10=-V 20 • Total momentum before collision is zero. • After collisions, the balls are stuck together and stationary, so total momentum is unchanged, still zero.

• One ball is initially at rest, and the ball stick together after collisions. • Both balls have same velocity • Assume balls have different mass; • Since the momentum is conserved even in inelastic collisions,

before after Lets us consider how much kinetic energy has been lost;

Substitute v to equation Kf and simplify the results;

• (m 2 initially at rest, completely inelastic collision only) • In a completely inelastic collision, the maximum amount of kinetic energy is lost, consistent with the conservation of momentum.

Example 5: A 30 g bullet with speed 400 m/s strikes a glancing blow to a target brick of mass 1. 0 kg. The brick breaks into two fragments. The bullet deflect at an angle of 30 degree and has a reduced speed of 100 m/s. One piece of the brick with mass 0. 75 kg goes off to the right or in the initial direction of the bullet with speed 5 m/s. (i)Sketches the situations (ii)Calculate the speed and direction of 2 nd piece

Energy and Momentum in Inelastic Collisions • For general elastic collision of two objects, Conservation of kinetic energy Conservation of momentum

• Consider this situation, if the ball m 2 is stationary, the equation for elastic collision; • Rearrange these equations gives; m 1(V 1 o 2 - V 12) = m 2 V 22 ……(1) and m 1(V 1 o-V 1)=m 2 V 2 …………(2)

• Using algebraic relationship x 2 + y 2 = (x+y) (xy) and dividing equation (1) by equation (2), we get; m 1(V 1 o +V 1) (V 1 o-V 1)/m 1(V 1 o-V 1) = m 2 V 22/m 2 V 2 • V 1 o + V 1 = V 2 ……. (3) • Equation 3 can be used to eliminate V 1 or V 2 from Equation total momentum

Final velocities for elastic, head on collision with m 2 initially stationary

7. 9 A krypton atom (mass 83. 9 u) moving with a velocity of 0. 80 km/s to the right and a water molecule (mass 18. 0 u) moving with a velocity of 0. 40 km/s to the left collide head-on. The water molecule has a velocity of 0. 60 km/s to the right after the collision. What is the velocity of the krypton atom after the collision? (The symbol “u” stands for the atomic mass unit. ) Slide 40

7. 9 Strategy Slide 41

7. 9 Solution For simplicity we drop the “x” subscripts from the xcomponents , remembering that all quantities refer to x components: Slide 42

7. 9 Solution After the collision, the krypton atom moves to the right with a speed of 0. 59 km/s. Slide 43

7. 7 COLLISIONS IN ONE DIMENSION Elastic and Inelastic Collisions A collision in which the total kinetic energy is the same before and after is called elastic. There is no conservation law for kinetic energy by itself. The elastic collision is just a special kind of collision in which no kinetic energy is changed into other forms of energy. When the final kinetic energy is less than the initial kinetic energy, the collision is said to be inelastic. Slide 44

7. 7 COLLISIONS IN ONE DIMENSION Elastic and Inelastic Collisions When a collision results in two objects sticking together, the collision is perfectly inelastic. The decrease of kinetic energy in a perfectly inelastic collision is as large as possible (consistent with the conservation of momentum). Slide 45

Problem-Solving Strategy for Collisions Involving Two Objects 1. Draw before and after diagrams of the collision. 2. Collect and organize information on the masses and velocities of the two objects before and after the collision. Express the velocities in component form (with correct algebraic signs). 3. Set the sum of the momenta of the two before the collision equal to the sum of the momenta after the collision. Write one equation for each direction: Slide 46

Problem-Solving Strategy for Collisions Involving Two Objects 4. If the collision is known to be perfectly inelastic, set the final velocities equal: 5. If the collision is known to be perfectly elastic, then either set the final kinetic energy equal to the initial kinetic energy: or, for a one-dimensional collision, set the relative speeds equal: Slide 47

Problem-Solving Strategy for Collisions Involving Two Objects 6. Solve for the unknown quantities. Slide 48

7. 10 At a Route 3 highway on-ramp, a car of mass 1. 50 × 103 kg is stopped at a stop sign, waiting for a break in traffic before merging with the cars on the highway. A pickup of mass 2. 00 × 103 kg comes up from behind and hits the stopped car. Assuming the collision is elastic, how fast was the pickup going just before the collision if the car is pushed straight ahead onto the highway at 20. 0 m/s just after the collision? Slide 49

7. 10 Strategy Slide 50

7. 10 Solution Slide 51

Example 6: A 3. 0 kg object with speed of 2. 0 m/s in the positive x-direction has head on elastic collision with a stationary 0. 70 kg object located at x=0. Calculate the distance separating the objects 2. 5 after the collision.

Example 7: A 7. 1 kg bowling ball with a speed of 6. 0 m/s has head on elastic collision with a stationary 1. 6 kg pin. (i) Calculate velocity of each object after collision (ii) Total momentum after collision.

SUMMARY • The linear momentum of an object is the product of its mass and velocity • For a system of objects, the total momentum is the vector sum of each

SUMMARY • Impulse is the change in momentum • conservation of linear momentum • Energy and Momentum in Inelastic Collisions

SUMMARY • Energy and Momentum in Inelastic Collisions

SUMMARY • Final velocities for elastic, head on collision with m 2 initially stationary