CHAPTER 3 LINEAR MOMENTUM COLLISIONS Edited by MOHD

CHAPTER 3: LINEAR MOMENTUM & COLLISIONS Edited by MOHD ROSYDI ZAKARIA Prepared by NURJULIANA JUHARI

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 Momentum Elastic of Linear and Inelastic Collisions

Linear Momentum Definition of linear momentum: • the quantity of motion • how much stuff is moving (mass) • how fast the stuff is moving(velocity) The linear momentum of an object is the product of its mass and velocity • Momentum = mass (kg) velocity (ms-1) p = m v • Linear momentum is a vector quantity – it has both a magnitude and a direction. (Its direction is the same as the direction of the velocity) • SI unit of momentum: kg • m/s. dimensions of momentum are ML/T.

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

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. )

Linear Momentum and Collisions

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

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.

7. 1 Example 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

Impulse Impulse is the change in momentum: ◦ The product of the force and the time over which the force acts on an object is called impulse. 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 states that when a net force is applied to an object over a certain time, the force will cause a change in the object’s momentum. F∆t = ∆p = mvf – mvi force time interval = change in momentum

Chapter 6 Section 1 Momentum and Impulse-Momentum Theorem

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.

Momentum is Conserved The Law of Conservation of Momentum: The total momentum of all objects interacting with one another remains constant regardless of the nature of the forces between the objects. If there are internal forces, the momenta of individual parts of the system can change, but the overall momentum stays the same. If there is no net force acting on a system, its total momentum cannot change. m 1 v 1, i + m 2 v 2, i = m 1 v 1, f + m 2 v 2, f total initial momentum = total final momentum

Momentum is Conserved Newton’s third law leads to conservation of momentum During the collision, the force exerted on each bumper causes a change in momentum for each car. The total momentum is the same before and after the collision.

Conservation of Linear Momentum For the linear momentum of an object to be conserved, its follow Newtons 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: In this example, there is no external force, but the individual components of the system do change their momenta. The spring force is an internal force, so the momentum of the system is conserved

Chapter 6 Sample Problem Conservation of Momentum A 76 kg boater, initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock. If the boater moves out of the boat with a velocity of 2. 5 m/s to the right, what is the final velocity of the boat?

Given: m 1 = 76 kg m 2 = 45 kg v 1, i = 0 v 2, i = 0 v 1, f = 2. 5 m/s to the right Unknown: v 2, f = ? m 1 v 1, i + m 2 v 2, i = m 1 v 1, f + m 2 v 2, f = 0 v 2, f = 4. 2 m/s to the left negative sign for v 2, f indicates that the boat is moving to the left

Types of Collisions: Elastic Collisions i. Momentum and Kinetic Energy Are Conserved 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 between two objects is one in which the total kinetic energy (as well as total momentum) of the system is the same before and after collision. In an elastic collision, both momentum and kinetic energy are conserved.

Sample Problem, continued Elastic Collisions A 0. 015 kg marble moving to the right at 0. 225 m/s makes an elastic head-on collision with a 0. 030 kg shooter marble moving to the left at 0. 180 m/s. After the collision, the smaller marble moves to the left at 0. 315 m/s. Assume that neither marble rotates before or after the collision and that both marbles are moving on a frictionless surface. What is the velocity of the 0. 030 kg marble after the collision?

Elastic Collisions 1. Define Given: m 1 = 0. 015 kg m 2 = 0. 030 kg v 1, i = 0. 225 m/s to the right, v 1, i = +0. 225 m/s v 2, i = 0. 180 m/s to the left, v 2, i = – 0. 180 m/s v 1, f = 0. 315 m/s to the left, v 1, i = – 0. 315 m/s Unknown: v 2, f = ? 1. Choose an equation or situation: Use the equation for the conservation of momentum to find the final velocity of m 2, the 0. 030 kg marble. m 1 v 1, i + m 2 v 2, i = m 1 v 1, f + m 2 v 2, f 2. Rearrange the equation to isolate the final velocity of m 2

Substitute the values into the equation and solve: Evaluate Confirm your answer by making sure kinetic energy is also conserved using these values.

Types of Collisions: i. Inelastic collision total kinetic energy is not conversed q q is one in which the total kinetic energy of the system is not the same before and after the collision (even thought the momentum is conserved) TWO types: perfectly inelastic (stick together after collide) and inelastic (do not stick together but some kinetic energy is transformed or transferred away) m 1 v 1, i + m 2 v 2, i = (m 1 + m 2)vf total initial momentum = total final momentum

In a non-perfect inelastic collision, momentum is conserved but kinetic energy is not. Moreover, the objects do not stick together In a perfectly inelastic collision, momentum is conserved, kinetic energy is not, and the two objects stick together after the collision, so their final velocities are the same. Elastic and perfectly inelastic collisions are limiting cases, most actual collisions fall in between these two types. Momentum is conserved in all collisions

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: An SUV with mass 1. 80 x 10 -3 kg is travelling eastbound at +15. 0 m/s, while a compact car with mass 9. 00 x 102 kg is travelling westbound at – 15. 0 m/s. The cars collide headon, becoming rentangled. (a) Find the speed of the entangled cars after the collision. (b) Find the change in the velocity of each car. (c) Find the change in the kinetic energy of the system consisting of both cars.

Energy and Momentum in Elastic 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) m 1(V 1 o-V 1)=m 2 V 2 …………(2) and

Using algebraic relationship x 2 + y 2 = (x+y) (x-y) 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

Sample Problem Kinetic Energy in Perfectly Inelastic Collisions Two clay balls collide head-on in a perfectly inelastic collision. The first ball has a mass of 0. 500 kg and an initial velocity of 4. 00 m/s to the right. The second ball has a mass of 0. 250 kg and an initial velocity of 3. 00 m/s to the left. What is the decrease in kinetic energy during the collision?

Sample Problem, continued Kinetic Energy in Perfectly Inelastic Collisions 1. Define Given: m 1= 0. 500 kg m 2 = 0. 250 kg m/s v 1, i = 4. 00 m/s to the right, v 1, i = +4. 00 v 2, i = 3. 00 m/s to the left, v 2, i = – 3. 00 m/s Unknown: ∆KE = ?

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