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Chapter 7 Linear and angular momentum

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**7-1 impulse and linear momentum**

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**Impulse (change in momentum)**

The rate of change of momentum of a body is equal to the net force acting on it : During a collision, a force F acts on an object, thus causing a change in momentum of the object: The product of Force and time is known as IMPULSE units of impulse are Ns A change in momentum is called “impulse”: Think of hitting a soccer ball: A force F acting over a time Dt causes a change Dp in the momentum (velocity) of the ball.

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**Linear Momentum and Collisions**

The linear momentum of a particle of mass m and velocity v is defined as The linear momentum is a vector quantity. It’s direction is along v. The components of the momentum of a particle:

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**momentum also depends on the mass.**

So changing the mass of an object will also change the momentum vector. Therefore to change momentum one must change the mass or velocity or both. Regardless of what changes, the momentum vector is always in the same direction as the velocity vector. units of momentum are kg m/s

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m1 v1 p1 = m1v1 m2 v2 p2 = -m2v2 Note that particles moving in opposite directions have momentum which are opposite sign!

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example A 1200 kg car drives west at 25 m/s for 3 hours. What is the car’s momentum? 1200 kg = mass 25m/s, west = velocity 3 hours = time P = mv = 1200 x 25 = kg m/s,west

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example a car of mass kg with a velocity of 8 m/s. The momentum of the car is . p = m・ v = (1000)(8) = 8000 kg ・ m / s a motor cycle of mass 250 kg travelling at 32 m/s. The momentum of the motor cycle is p = m・ v = (250)(32)= 8000 kg ・ m / s

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**Both objects experience the same magnitude of force **

If the boulder and the boy have the same momentum, will the boulder crush the boy? Hint: Which would have the larger speed? When two objects with different masses collide, the force on the less massive object is larger than the force on the more massive object. Both objects experience the same magnitude of force but in opposite directions. However, greater damage is suffered by the less massive object because of the greater momentum of the more massive one.

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example 2 cars are heading east, car A is traveling 30m/s, car B is traveling 60m/s. Each car weighs 2000Kg. What is the momentum of car A? What is the momentum of car B? If car B crashes into car A, what is the total momentum? p=mv Car (A) momentum = 30m/s x 2000Kg pX = 60,000 mi-lbs/hr east Car (B)momentum = 60m/s x 2000Kg pY = 120,000 m-Kg/s west Total momentum = pA - pB = 120, ,000 = 60,000 m-Kg/s west

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**Impulse (change in momentum)**

The rate of change of momentum of a body is equal to the net force acting on it : During a collision, a force F acts on an object, thus causing a change in momentum of the object: The product of Force and time is known as IMPULSE units of impulse are Ns A change in momentum is called “impulse”: Think of hitting a soccer ball: A force F acting over a time Dt causes a change Dp in the momentum (velocity) of the ball.

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**How do the safety features of a car protect you in a collision?**

In an accident, the Dp stays the same, but you can change F and t of impact. When you increase the time that the collision lasts, the force of impact decreases Ft All safety features in a car are designed to reduce the force by increasing the time you are in contact with that force. Watch the video! Remember: Ft = Dp

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**the momentum change of an object is the mass X velocity change **

the impulse equals the momentum change Impulse is a vector quantity and has the same direction as the average force. Final momentum Initial momentum Impulse

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**Washing a Car: Momentum Change and Force. **

Water leaves a hose at a rate of 1.5 kg/s with a speed of 20 m/s and is aimed at the side of a car, which stops it. (That is, we ignore an splashing back.) What is the force exerted by the water on the car? If the water splashes back from the car, the change in momentum will be greater in magnitude, and so will the force on the car be greater in magnitude. Note that pfinal will now point in the negative x direction. So the result for F will be minus something of magnitude depending on the water’s rebound speed.). To put it simply, the car exerts not only a force to stop the water, but also an additional force to give it momentum in the opposite direction.

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Force & Impulse Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each box for exactly 1 second. Which box has the most momentum after the force acts? (a) heavier (b) lighter (c) same We know so In this problem F and Dt are the same for both boxes! The boxes will have the same final momentum. F F F F heavy heavy light light

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**Why does an egg break or not break?**

An egg dropped on a tile floor breaks, but an egg dropped on a pillow does not. Why? FΔt= mΔv In both cases, m and Δv are the same. If Δt goes up, what happens to F, the force? Right! Force goes down. When dropped on a pillow, the egg starts to slow down as soon as it touches it. A pillow increases the time the egg takes to stops.

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**Impulse and contact time:**

Impulses and Contact Time

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Example The baseball has a mass of 0.14kg and an initial speed of 30 ms-1. It rebounds from the bat with a speed of 40 ms-1 in the opposite direction and is in contact with the bat for s. Determine the average force on a baseball hit by a bat. Example : The engine of a model rocket is rated with a total impulse of 5.00 Ns and a thrust duration 0f 1.20 s. what is the average force exerted by the engine?

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Example A 68 kg soccer player kicks a stationary 0.425kg ball giving it a speed 0f 13.7 ms-1. The player foot is in contact with the ball for s. What is the average force on the ball?

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**The cart’s change of momentum is**

30 kg m/s. 10 kg m/s. –10 kg m/s. –20 kg m/s. –30 kg m/s.

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**example Δvelocity = +1.2 – (-1.2) = 2.4 m/s**

A 57 gram tennis ball falls on a tile floor. The ball changes velocity from -1.2 m/s to +1.2 m/s in 0.02 s. What is the average force on the ball? Mass = 57 g = kg Δvelocity = +1.2 – (-1.2) = 2.4 m/s Time = 0.02 s FΔt= mΔv F x (0.02 s) = (0.057 kg)(2.4 m/s) F= 6.8 N

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Car Crash Would you rather be in a head on collision with an identical car, traveling at the same speed as you, or a brick wall? Assume in both situations you come to a complete stop.

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It Does Not Matter! Look at FΔt= mΔv In both situations, Δt, m, and Δv are the same! The time it takes you to stop depends on your car, m is the mass of your car, and Δv depends on how fast you were initially traveling.

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example If the halfback experienced a force of 800 N for 0.9 seconds to the north, determine the impulse F (Δ t ) = m D v 800N ( 0.9s ) = 720 N . s the impulse was 720 N . s or a momentum change of 720 kg . m/s Given: F = 800 N t = 0.9 s Find : Impulse (F Δt)

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example A 0.10 Kg model rocket’s engine is designed to deliver an impulse of 6.0 N*s. If the rocket engine burns for 0.75 s, what is the average force does the engine produce? F (Δ t ) = m D v 6.0 N . s = F ( 0.75s ) 6.0 N . s / 0.75s = F 8.0 N = F Given: F = 800 N t = 0.9 s Find : Average Force

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example A Bullet traveling at 500 m/s is brought to rest by an impulse of 50 N*s. What is the mass of the bullet? F ( Δt ) = m D v 50 N . s = m ( 500 m/s – 0 m/s ) 50 kg-m/s 2 . s / 500 m/s = m 0.1 kg = m Given: v = 500 m/s F Δ t= 50 N . s Find : m = ?

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Example A baseball (m=0.14kg) has initial velocity of v0=-38m/s as it approaches a bat. The bat applies an average force that is much larger than the weight of the ball, and the ball departs from the bat with a final velocity of vf=+58m. Determine the impulse applied to the ball by the bat. Assuming time of contact is =1.6*10-3s, find the average force exerted on the ball by the bat.

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(a) = kg.m/s (b)

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example 2 cars are heading east, car A is traveling 30m/s, car B is traveling 60m/s. Each car weighs 2000Kg. What is the momentum of car A? What is the momentum of car B? If car B crashes into car A, what is the total momentum? p=mv Car (A) momentum = 30m/s x 2000Kg pX = 60,000 mi-lbs/hr east Car (B)momentum = 60m/s x 2000Kg pY = 120,000 m-Kg/s west Total momentum = pA - pB = 120, ,000 = 60,000 m-Kg/s west

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**In a crash test, a car of mass 1**

In a crash test, a car of mass 1.5103 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

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**How do the safety features of a car protect you in a collision?**

In an accident, the Dp stays the same, but you can change F and t of impact. When you increase the time that the collision lasts, the force of impact decreases Ft All safety features in a car are designed to reduce the force by increasing the time you are in contact with that force. Watch the video! Remember: Ft = Dp

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A 10 g rubber ball and a 10 g clay ball are thrown at a wall with equal speeds. The rubber ball bounces, the clay ball sticks. Which ball exerts a larger impulse on the wall? They exert equal impulses because they have equal momentum. The clay ball exerts a larger impulse because it sticks. Neither exerts an impulse on the wall because the wall doesn’t move. The rubber ball exerts a larger impulse because it bounces.

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**vf is greater than v2, but less than v1. vf = v1. **

The two particles are both moving to the right. Particle 1 catches up with particle 2 and collides with it. The particles stick together and continue on with velocity vf. Which of these statements is true? vf = v2. vf is less than v2. vf is greater than v2, but less than v1. vf = v1. vf is greater than v1.

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example A 2000 kg truck traveling west at 12 m/s collides with a 1200 kg car traveling east at 16 m/s. They collide and remain stuck together. What is their final velocity? 1.5 m/s west

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Example Rain comes straight down with velocity of v0=-15m/s and hits the roof of a car perpendicularly. Mass of rain per second that strikes the car roof is 0.06kg/s. Assuming the rain comes to rest upon striking the car (vf=0m/s), find the average force exerted by the raindrop.

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-(0.06kg/s)(-15m/s) = 0.9 N According to action-reaction law, the force exerted on the roof also has a magnitude of 0.9 N points downward: N

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example A 10 g rubber ball and a 10 g clay ball are thrown at a wall with equal speeds. The rubber ball bounces, the clay ball sticks. Which ball exerts a larger impulse on the wall? They exert equal impulses because they have equal momenta. The clay ball exerts a larger impulse because it sticks. Neither exerts an impulse on the wall because the wall doesn’t move. The rubber ball exerts a larger impulse because it bounces.

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**In which case is the impulse greater when You drop an egg onto **

a) the floor b) a thick piece of foam rubber In both cases, the egg does not bounce. A) case 1 B) case 2 C) the same In which case is the average force greater T. Norah Ali Almoneef

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**Example: suppose m=1 kg, v(initial)=-1 m/s mv(initial)= -1 kg-m/s **

Two identical balls are dropped from the same height onto the floor.In case 1 the ball bounces back up, and in case 2 the ball sticks to the floor without bouncing. In which case is theimpulse given to the ball by the floor larger? 1. Case 1 2. Case 2 3. The same the impulse is greater for case 1 because: the change in momentum of the object is proportional to the change in velocity which is greater in case 1 because it has a greater final velocity (down then up) than case 2 (which is only from down to zero). Impulse must be greater for case 1. Example: suppose m=1 kg, v(initial)=-1 m/s mv(initial)= -1 kg-m/s Case 1 mv(final)= +1 kg-m/s Impulse = 1- (-1)=2 Case 2 mv(final)= 0 Impulse = = 1 Note: the direction (upward) important. T. Norah Ali Almoneef

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Is it possible for a system of two objects to have zero total momentum while having a non-zero total kinetic energy? 1. YES 2. NO in an isolated system, two ice skaters starting at rest and pushing on one another will move in opposite directions thus the momentum of the two are equal and opposite and total momentum is zero. but they are moving apart after the push and therefore the KE is non-zero. two hockey pucks moving towards each other with the same speed on a collision course have zero total momentum, but a non zero total kinetic energy T. Norah Ali Almoneef

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The diagram depicts Before and After velocities of an 800-kg car in two different collisions with a wall. In case A, the car rebounds upon collision. In case B, the car hits the wall, crumples up and stops. Assume that the collision time for each collision are the same. 1. In which case does the car experience the greatest momentum change? a. Case A b. Case B c. Both the same d. Insufficient information 2. In which case does the car experience the greatest impulse? a. Case A b. Case B c. Both the same d. Insufficient information 3. The impulse encountered by the 800-kg car in case A has a magnitude of ___ N•s. a b c d. 4000 e f. Not enough information to determine. T. Norah Ali Almoneef

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**(a) The impulse was experienced by the wall is. **

A 320 g ball with a speed v of 6.22 m/s strikes a wall at angle θ of 33o and then rebounds with the same speed and angle (Fig). It is in contact with the wall for 10.4 m s. (a) The impulse was experienced by the wall is. (b) The average force exerted by the ball on the wall is According to the impulse-momentum theorem: Δp =F Δt The impulse was experienced by the wall is Δp = m v sin − (−m v sin ) = 2m v sin = 2× 0.32× 6.22× sin 33o = 2.17 N⋅ s The average force exerted by the ball on the wall is T. Norah Ali Almoneef

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Example A baseball (m=0.14kg) has initial velocity of v0=-38m/s as it approaches a bat. The bat applies an average force that is much larger than the weight of the ball, and the ball departs from the bat with a final velocity of vf=+58m. Determine the impulse applied to the ball by the bat. Assuming time of contact is =1.6*10-3s, find the average force exerted on the ball by the bat. T. Norah Ali Almoneef

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**Conservation of momentum 2 particle system**

Before colliding W 2 N 1 W 1 N2 m 1 m 2 F21 F12 during collision W 1 W 2 N 1 N2 m 1 m 2 After collision W 1 W 2 N 1 N2

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**F12 is force of 1 on 2 F21 is force of 2 on 1**

From Newton’s 3rd Law F12 = - F OR F12 + F21 = 0 F12 is force of 1 on F is force of 2 on 1 F12 Δ t = p 1 f - p 1i F21 Δ t = p2f - p 2i P 1 f - P 1i + P2f - P 2i = 0

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**Conservation of linear momentum**

If Fext = 0 then the total momentum remains constant. The total momentum of an isolated system remains constant momentum before = momentum after

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Example Starting from rest, two skaters push off against each other on smooth level ice (friction is negligible). One is a woman (m1=54kg), and one is a man(m2=88kg). The woman moves away with a velocity of vf1=2.5m/s. Find the recoil velocity vf2 of the man.

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**For the two skater system, what are the forces?**

In horizontal direction F12 F21 internal forces system taken together No external forces isolated system conservation of momentum

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m1vf1 + m2vf2 = 0 after pushing before pushing It is important to realize that the total linear momentum may be conserved even when the kinetic energies of the individual parts of a system change.

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example Assume that an 8kg mass m1 moving to the right at 4ms-1 collides with a 6kg mass m2 moving to the left at 5ms-1. what is the total momentum before and after the collision? (Ans: 2kgms-1) example A 1400kg cannon mounted on wheels fires a 60kg ball in a horizontal direction with a velocity of 50ms-1. Assume that the cannon can move freely, what will be its recoil velocity? (Ans: vc=-2.14ms-1)

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example What is the recoil velocity of a 4.0-kg rifle that shoots a kg bullet at a speed of 280 m/s? the total momentum is conserved. The initial momentum of the rifle-bullet system is 0 because the riffle and the bullet were at rest at the beginning. By the Law of Conservation of Momentum,

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example A 3-kg mass is moving with an initial velocity v1. The mass collides with a 5-kg mass m2, which is initially at rest. Find the final velocity of the masses after the collision if it is perfectly inelastic. According to the Law of Conservation of Momentum

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example 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?

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**In an inelastic collision**

1. both kinetic energy and momentum are conserved. 2. only kinetic energy is conserved. 3. only momentum is conserved. 4. neither kinetic energy nor momentum are conserved. 1: False by definition of inelastic collision 2: False by definition of inelastic collision 4: False by definition of collision T. Norah Ali Almoneef T. Norah Ali Almoneef

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Collisions Term collision is used to represent an event during which 2 particles come close to each other and interact by means of forces. Time interval during which the velocities of the particles change from initial to final values is assumed to be short. The interaction forces are assumed to be much greater than any external forces present. When two particles of masses m1 and m2 collide, the impulsive forces may vary in time in complicated ways. The two particles form an isolated system, and the momentum must be conserved. The total momentum of an isolated system just before a collision equals to the momentum of the system just after the collision. The total kinetic energy of the system, may or may not be conserved, depending on the type of the collision.

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T. Norah Ali Almoneef

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**Elastic and inelastic collisions in one dimension**

Momentum is conserved in any collision, elastic and inelastic. Mechanical Energy is only conserved in elastic collisions. Perfectly inelastic collision: After colliding, particles stick together. There is a loss of energy (deformation). Elastic collision: Particles bounce off each other without loss of energy. Inelastic collision: Particles collide with some loss of energy, but don’t stick together.

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**Perfectly inelastic collision of two particles **

OR completely inelastic collision (Particles stick together) thy moves as one object Notice that p and v are vectors and, thus have a direction (+/-)

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**( totally in elastic collision)**

There is a loss in energy Eloss

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**There is a loss in energy Eloss**

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**Totally Inelastic collision:**

All masses are equal. vf 2m After m Before vi T. Norah Ali Almoneef

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**Calculate the final velocity**

T. Norah Ali Almoneef

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Example Car 1 has a mass of m1=65*103kg and moves at a velocity of v01= vi1=+0.8m/s. Car 2 has a mass of m2=92*103kg and a velocity of v02= vi2= +1.3m/s. Neglecting friction, find the common velocity vf of the cars after they become coupled.

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(m1+m2) vf = m1v01 + m2v02 After collision Before collision =+1.1 m/s

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**Inelastic Collisions v = 10 v = 0 M M Before Collision p = Mv vf = 5 M**

After Collision p = 2Mvf Mv = 2Mvf vf = ½ v

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**example Find the speed of the entangled cars after the collision.**

An SUV with mass 1.80103 kg is travelling eastbound at m/s, while a compact car with mass 9.00102 kg is travelling westbound at m/s. The cars collide head-on, becoming entangled. Find the speed of the entangled cars after the collision. Find the change in the velocity of each car. Find the change in the kinetic energy of the system consisting of both cars.

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**Find the speed of the entangled cars after the collision.**

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**Find the change in the velocity of each car.**

March 24, 2009

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**Find the change in the kinetic energy of the system consisting of both cars.**

March 24, 2009

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example Two blocks are traveling toward each other. The first has a speed of 10cm/sec and the second a speed of 60 cm/sec. After the collision, the second block is moving with a speed of 20 cm/sec in a direction opposite to its initial velocity. If the mass of the first block is twice that of the second, determine the velocity of the first block after the collision. Since the surface is frictionless, the net force acting on the system is 0, and from the Law of Conservation of Momentum P1i + P2i = P1f + P2f Adding the x-components we have m1v1i - m2v2i = m1v1f + m2v2f Since m1 = 2m2, we find that 2v1i - v2i = 2v1f + v2f (2)(10) - (60) = 2v1f + 20 Thus, 2v1f = - 60 and v1f = - 30 cm/sec ('-' means to left).

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**The two particles are both moving to the right**

The two particles are both moving to the right. Particle 1 catches up with particle 2 and collides with it. The particles stick together and continue on with velocity vf. Which of these statements is true? vf = v2. vf is less than v2. vf is greater than v2, but less than v1. vf = v1. vf is greater than v1.

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**p1 = ( (1000 kg)(13.4 m/s)) = (1.34 X 104 kg m/s)**

A 1000 kg car moving north at (13.4m/s) collides with a 2000 kg moving east at (22.4m/s). The cars get entangled and moved together. What is their final speed? How much energy is converted to other forms? Answer Say we define our coordinates so that north is the y direction and east is the x -direction. Then the initial momentum of the first car is p1 = ( (1000 kg)(13.4 m/s)) = (1.34 X 104 kg m/s) T. Norah Ali Almoneef

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T. Norah Ali Almoneef

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example T. Norah Ali Almoneef

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**Example (m1+m2) vf = m1v01 + m2v02 =+1.1 m/s**

Car 1 has a mass of m1=65X103kg and moves at a velocity of v01=+0.8m/s. Car 2 has a mass of m2=92X103kg and a velocity of v02=+1.3m/s. Neglecting friction, find the common velocity vf of the cars after they become coupled. (m1+m2) vf = m1v01 + m2v02 After collision Before collision =+1.1 m/s T. Norah Ali Almoneef

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T. Norah Ali Almoneef

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T. Norah Ali Almoneef

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example Two cars travelling, one from west to east and the other from south to north got stuck and move together of the first one is 45 km /h and that of the second is 30 km/ h . The masses of the first and second cars are 1000 g and 1500 respectively, What is the velocity of the wreck after collision ? T. Norah Ali Almoneef

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**In elastic collision after the collision they moves as tow objects**

Total momentum after collision Total momentum before collision Total kinetic energy after collision Total kinetic energy before collision

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**Elastic collision of two particles**

(Particles bounce off each other without loss of energy. Momentum is conserved: Energy is conserved:

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If V2 not equal zero

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**Elastic collision: Collision of Two Large Balls: All masses are equal.**

v m Before v m After V1= V2= V V1 = V V2 = 0 2 2 T. Norah Ali Almoneef

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T. Norah Ali Almoneef

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**Elastic Collisions Case 1**

Objects A and B are of the same mass, the objects exchange velocities If m1=m2, then v1=u2 and v2 = u1 If 2nd object is initially at rest If u2=0, then v1=u2=0 and v2 = u1 Case 2 If m2>>m1, u2=0 A small object collides with a big object which is at rest. then v – u1 and v When a small/light object such as a ball collides with a very large/heavy object such as bus, the ball rebounds with the same speed as that before collision while the bus remains at rest. Case 3 If m1>>m2, u2=0 A heavy object collides with a light object which is at rest. If u2=0, then v u1 and v u1 After collision, the heavy object continues its motion with the same velocity. The light object moves off with a velocity which is twice the initial velocity of the heavy object.

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example A 2000 kg truck traveling west at 12 m/s collides with a 1200 kg car traveling east at 16 m/s. They collide and remain stuck together. What is their final velocity (include direction)? 1.5 m/s west

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**In an inelastic collision**

1. both kinetic energy and momentum are conserved. 2. only kinetic energy is conserved. 3. only momentum is conserved. 4. neither kinetic energy nor momentum are conserved. 1: False by definition of inelastic collision 2: False by definition of inelastic collision 4: False by definition of collision T. Norah Ali Almoneef T. Norah Ali Almoneef

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example The version shown in Fig. consists of a large block of wood of mass M = 5.4 kg, hanging from two long cords. A bullet of mass m = 9.5 g is fired into the block, coming quickly to rest. The block + bullet then swing upward, their center of mass rising a vertical distance h = 6.3 cm before the block comes momentarily to rest end. What is the speed of the bullet just before the collision?

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Example In a game of billiard, a player wishes to sink the target ball in the corner pocket as shown in the figure. If the angle of the corner pocket is 350, at what angel θ is the cue ball deflected? Assume collision elastic. solution

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Example A 1500 kg car traveling east with a speed of 25.0 ms-1 collides at an intersection with a 2500 kg van traveling north at a speed of 20.0 ms-1 as shown in the figure. Find the direction and magnitude of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is they stick together). solution

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**Simple Examples of Head-On Collisions**

(Energy and Momentum are Both Conserved) Collision between two objects of the same mass. One mass is at rest. Collision between two objects. One at rest initially has twice the mass. Collision between two objects. One not at rest initially has twice the mass.

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