Momentum with Examples (Online)

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Chapter 7 Momentum Momentum n The linear momentum p of an object of mass m moving with a velocity v is defined as the product of the mass and the velocity

n p = mv

n SI Units are kg m / s

n Vector quantity, the direction of the momentum is the same as the velocity’s n In order to change the momentum of an object, a force must be applied. n The time rate of change of momentum of an object is equal to the net force acting on it

n ∆� = = � ∆ ∆ net Momentum and Its Relation to Force

Momentum is a vector symbolized by the symbol p, and is defined as

The rate of change of momentum is equal to the net force:

This can be shown using Newton’s second law:

F = ma = m (Dv/Dt) = D(mv)/Dt F = Dp/Dt Impulse n When a single, constant force acts on the object, there is an impulse delivered to the object

n J = Ft r n IJ is defined as the impulse

n Vector quantity, the direction is the same as the direction of the force Impulse-Momentum Theorem n The theorem states that the impulse acting on the object is equal to the change in momentum of the object

n J = FDt = Dp = mvf - mvi

n If the force is not constant, use the average force applied Collisions and Impulse The impulse tells us that we can get the same change in momentum with a large force acting for a short time, or a small force acting for a longer time. This is why you should bend your knees when you land; why airbags work; and why landing on a pillow hurts less than landing on concrete. Impulse Applied to Auto Collisions n The most important factor is the collision time or the time it takes the person to come to a rest

n This will reduce the chance of dying in a car crash n Ways to increase the time

n Seat belts

n Air bags Air Bags n The air bag increases the time of the collision n It will also absorb some of the energy from the body n It will spread out the area of contact

n Decreases the pressure

n Helps prevent penetration wounds Example 1: How Good Are the Bumpers?

In a crash test, a car of mass 1.5 x 103 kg collides with a wall and rebounds as in figure. The initial and final velocities of the car are vi=-15 m/s and vf = 2.6 m/s, respectively. If the collision lasts for 0.15 s, find (a) the impulse delivered to the car due to the collision (b) the size and direction of the average force exerted on the car Example 1: How Good Are the Bumpers? In a crash test, a car of mass 1.5 x 103 kg collides with a wall and rebounds as in figure. The initial and final velocities of the car are vi=-15 m/s and vf = 2.6 m/s, respectively. If the collision lasts for 0.15 s, find (a) the impulse delivered to the car due to the collision (b) the size and direction of the average force exerted on the car (a) Find the impulse delivered to the car.

J = FDt = Dp = mvf - mvi Calculate the initial and final momenta of the car. 3 4 pi = mvi = (1.50 x 10 kg)(-15.0 m/s) = -2.25 x 10 kg·m/s 3 3 pf = mvf = (1.50 x 10 kg)(+2.60 m/s) = + 3.90 x 10 kg·m/s The impulse is just the difference between the final and initial momenta. 3 4 J = pf - pi = +3.90 x 10 kg·m/s - (-2.25 x 10 kg·m/s) J = 2.64 x104 kg · m/s

(b) Find the average force exerted on the car.

Apply the impulse-momentum theorem.

4 5 Fav = Δp = 2.64 x 10 kg · m/s = +1.76 x 10 N Δt 0.150 s Conservation of Momentum n Momentum in an isolated system in which a collision occurs is conserved

n A “collision” happens anytime physical contact is made between two objects Conservation of Momentum, cont n The principle of conservation of momentum states: n When no external forces act on a system consisting of two or more objects that collide with each other, the total momentum of the system remains constant in time

n Specifically, the total momentum before the collision will equal the total momentum after the collision Conservation of Momentum, Example n The momentum of each object will change n The total momentum of the system remains constant Forces in a Collision n The force with which object 1 acts on object 2 is equal and opposite to the force with which object 2 acts on object 1 n Impulses are also equal and opposite Conservation of Momentum, cont. n Mathematically: m1v1i + m2v2i = m1v1f + m2v2f n Momentum is conserved for the system of objects

n The system includes all the objects interacting with each other

n Assumes only internal forces are acting during the collision

n Can be generalized to any number of objects The Archer

2. An archer stands at rest on frictionless ice and fires a 0.5-kg arrow horizontally at 50.0 m/s. The combined mass of the archer and bow is 60.0 kg. With what velocity does the archer move across the ice after firing the arrow? vAf = - mBvBf / mA = - (0.5 kg)(50 m/s) / (60 kg)

= - 0.417 m/s Conservation of Momentum Example 3 : A 10,000-kg railroad car, A, traveling at a speed of 24.0 m/s strikes an identical car, B, at rest. If the cars lock together as a result of the collision, what is their common speed immediately after the collision?

mAvAi = (mA + mB) vf

(a) An empty sled is sliding on frictionless ice when Susan drops vertically from a tree above onto the sled. When she lands, does the sled speed up, slow down, or keep the same speed?

(a) Because Susan falls vertically onto the sled, she has no initial horizontal momentum. Thus, the total horizontal momentum afterward equals the momentum of the sled initially. Since the mass of the system (sled + person) has increased, the speed of the sled must decrease.

(b) Later: Susan falls sideways off the sled. When she drops off, does the sled speed up, slow down, or keep the same speed?

(b) At the instant Susan falls off, she is moving with the same horizontal speed as she was while on the sled. At the moment she leaves the sled, she has the same momentum she had an instant before. Because momentum is conserved, the sled keeps the same speed.

(m1+m2)V= m1v1 + m2v2

where m1 is Susan So…

m1V+m2V = m1v1 + m2v2

m1V+m2V = m1V + m2v2

The only way for this to be true is if v2 = V

Copyright © 2009 Pearson Education, Inc. Types of Collisions • Momentum is conserved in any collision • Regarding Energy: • Inelastic collisions – Kinetic energy is not conserved • Some of the kinetic energy is converted into other types of energy such as heat, sound, work to permanently deform an object – Perfectly inelastic collisions occur when the objects stick together • A lot of KE is lost • Not all of the KE is necessarily lost • Elastic collision – Both momentum and kinetic energy are conserved

Most collisions fall between elastic and perfectly inelastic collisions

Momentum is conserved in all collisions.

Collisions in which kinetic energy is conserved as well are called elastic collisions, and those in which it is not are called inelastic. Elastic Collisions in One Dimension

Example: A nuclear collision. Study this example in half-sheet handout…. A proton (p) of mass 1.01 u (unified atomic mass units) traveling with a speed of 3.60 x 104 m/s has an elastic head-on collision with a helium (He)

nucleus (mHe = 4.00 u) initially at rest. What are the velocities of the proton and helium nucleus after the collision? Assume the collision takes place in nearly empty space.

Copyright © 2009 Pearson Education, Inc. Inelastic Collisions Example: Railroad cars again. A 10,000-kg railroad car, A, traveling at a speed of 24.0 m/s strikes an identical car, B, at rest. If the cars lock together as a result of the collision, how much of the initial kinetic energy is transformed to thermal or other forms of energy?

Before collision

After collision Copyright © 2009 Pearson Education, Inc. Solution: Momentum is conserved; the initial kinetic energy is 2.88 x 106 J and the final kinetic energy is 1.44 x 106 J, so 1.44 x 106 J of kinetic energy

Copyright © 2009 Pearson Education, Inc. have been transformed to other forms. An SUV Versus a Compact q An SUV with mass 1.80 x 103 kg is travelling eastbound at +15.0 m/s, while a compact car with mass 9.00x102 kg is travelling westbound at -15.0 m/s. The cars collide head- on, becoming entangled.

(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. To follow are two versions of this solution. The first is brief, the second explains more… An SUV Versus a Compact

(a) Find the speed of the entangled cars after the collision. An SUV Versus a Compact

(a) Find the speed of the entangled cars after the collision.

5 m/s An SUV Versus a Compact

(b) Find the change in the velocity of each car. An SUV Versus a Compact

(b) Find the change in the velocity of each car.

Dv1 = vf – v1i = -10 m/s

Dv2 = vf – v2i = 20 m/s An SUV Versus a Compact

!"# $%&'()*+("*,&-+(%&()*+(.%&+)%"( m 1.80103 kg,v 15m / s +&+/-0(12()*+(303)+4("1&3%3)%&-( 1 1i 2 12(51)*(",/36( m2 9.0010 kg,v2i 15m / s

v f 5.00m/ s 1 1 KE m v2 m v2 3.04105 J i 2 1 1i 2 2 2i 1 1 KE m v2 m v2 3.38104 J f 2 1 1 f 2 2 2 f

5 KE KE f KEi 2.7010 J Here are the solutions again…. pAi + pBi = pBf + pBf

mAvAi + mBvBi = (mA + mB) vf

1800kg (15 m/s) + 900kg (-15m/s) = (1800kg + 900kg)vf

vf = +5 m/s or

m v 2 + m v 2 K = 1 1 2 2 i 2

2 2 Ki = m1v1 /2+ m2v2 /2

= 3.04 x 105 J