Chapter 9 Momentum and Collisions Linear Momentum is

Linear Momentum • is a measure of how hard it is to stop a

Sample problems What is the momentum of • a 0. 45 Caliber bullet (m

• c. … a 360, 000 -kg passenger plane taxiing down a runway

Impulse • equals the change in momentum of an object • Also a vector

• A boy falls and hits his head with an impulse of 20

Linear Momentum and Impulse – – – the change in momentum of an object

• A 5. 0 kg mass moving with a velocity of 8. 0

• J = mΔv = (5. 0 kg)(12. m/s East) • • =

Conservation of Momentum • Def: – in a system of objects on which no

Conservation of Momentum • Momentum is conserved in all cases. • When objects collide

When objects push away from each other they have the same momentum but equal

Perfectly Inelastic Collisions • This is when two objects collide and then remain stuck

• Ima Rilla Saari rushes to her car in order to hurry home

Elastic Collision • This occurs when two objects collide and bounce off each other

Collisions • It is very important to remember that momentum is a vector quantity

Sample Problems (Momentum) • A 2 kg ball is traveling at a constant speed

Sample Problems Impulse • If you exert a force of 30 N (E) on

Sample Problem Cons of p (inelastic) • A 300 kg box is sliding across

Sample Problem Cons of p (inelastic) • A 4 kg snowball is thrown to

Sample Problems Cons of p (elastic) • A 2 kg ball is moving right

Sample Problem Collision – A 2 kg ball is moving right at 4 m/s

Sample Problem – A 2 kg ball is moving right at 4 m/s when

Sample Problem Collision • In the following problem determine the final velocity A 60

Sample Problem Collision (Cons of E) • In the following problem determine the final

Sample Problem – Stopping Distance • A 2300 kg car is traveling north at

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Chapter 9 Momentum and Collisions

Linear Momentum • is a measure of how hard it is to stop a moving object • is equal to the product of an object’s mass and velocity • Formula: p = mv p = momentum; m = mass in kilograms and v = velocity • Units are: kg * m/s • is a vector quantity • can derive formula from Newton’s 2 nd Law of Motion

Sample problems What is the momentum of • a 0. 45 Caliber bullet (m = 0. 162 kg) leaving the muzzle of a gun at 860 m/s. • Answer 139. 32 kg • m/s) b. … a 110 -kg professional fullback running across the line at 9. 2 m/s. • Answer 1012 kg • m/s

• c. … a 360, 000 -kg passenger plane taxiing down a runway at 1. 5 m/s • Answer: 5. 4 x 105 kg • m/s

Impulse • equals the change in momentum of an object • Also a vector • product of the force and time – Formula: J = Fnet * t or = Δ p = m Δv = m (vf – vi) – J = Impulse, Fnet = Force in newtons ; t = time in sec Units: Newton seconds or N*s; Can also = kg* m/s – Practice Problem What is the impulse of a bat on a ball that is hit for 0. 050 s with a force of 25 N?

• 1. 25 N×s

• A boy falls and hits his head with an impulse of 20 N·s. On cement, the boy would hit for. 010 s. What force did the floor exert on the boy?

• 2000 N

Linear Momentum and Impulse – – – the change in momentum of an object produced by a force acting on the object is in the same direction as the force and is directly proportional to the product of the force and the time that it acts upon the object It is this relationship between momentum and impulse that protects us in car crashes and allows spaceships to achieve great speeds. How you say?

• J = Ft = mΔv = Δp • A 5. 0 kg mass has its velocity change from 8. 0 m/s east to 2. 0 m/s east. • Find the objects change in momentum. • What was the impulse applied?

• • Δp = mΔV = (5. 0 kg)(2. 0 m/s - 8. 0 m/s) = -30. kgm/s East or +30. kg m/s West

• A 5. 0 kg mass moving with a velocity of 8. 0 m/s east has an impulse applied to it which causes its velocity to change to 20. m/s East. Find Impulse:

• J = mΔv = (5. 0 kg)(12. m/s East) • • = 60. kg m/s east or 60 Ns east • Find the force if the impulse was applied for 3. 0 sec. • J = Ft = 60. Ns East • F(3. 0 sec) = 60. Ns East • F = 20. N East

Conservation of Momentum • Def: – in a system of objects on which no outside forces act, the total momentum of the system remains constant. • Because momentum is a vector, we can say that is conserved vectorially. – in all directions – SO momentum before = momentum after – pbefore = pafter

Conservation of Momentum • Momentum is conserved in all cases. • When objects collide the total momentum of the system is constant even though the momentums of the individual objects may change • Formula: m 1 vi + m 2 vi = m 1 vf + m 2 vf • this isn’t in reference table but goes with • pbefore = pafter

Formula: m 1 vi + m 2 vi = m 1 vf + m 2 vf • Rex (m=86 kg) and Tex (92 kg) board the bumper cars at the local carnival. Rex(86 kg + 125 kg car))are moving at a full speed of 2. 05 m/s when he rear-ends Tex who is at rest in his path. Tex(92 kg) and his 125 -kg car lunge forward at 1. 40 m/s. Determine the post-collision speed of Rex and his 125 -kg car.

• 0. 61 m/s

When objects push away from each other they have the same momentum but equal and opposite in direction. Newton’s 3 rd law shows cons. of p • A 70. 9 -kg boy and a 43. 2 -kg girl, both wearing skates face each other at rest on a skating rink. The boy pushes the girl, sending her eastward with a speed of 4. 64 m/s. Neglecting friction, determine the subsequent velocity of the boy.

• 2. 83 m/s, West

Perfectly Inelastic Collisions • This is when two objects collide and then remain stuck together • Momentum is conserved – Note: during the collision the 2 objects become 1 • Formula: m 1 vi + m 2 vi = (m 1+m 2)vf

• Ima Rilla Saari rushes to her car in order to hurry home and get dressed for work. Failing to realize the dangers of driving under slick and icy conditions, she collides her 940 -kg Mazda Miata into the rear of a 2460 -kg pick-up truck which was at rest at the light on Lake Avenue. Ima's pre-collision speed was 12. 5 m/s. Determine the postcollision speed of the two entangled cars as they slide across the ice.

Elastic Collision • This occurs when two objects collide and bounce off each other with no loss in K. E. – Kinetic Energy is conserved – we will deal with this when we get to energy • Note: most collisions that occur are neither elastic or perfectly inelastic – To be elastic there can be no loss of energy either internally or to friction

Collisions • It is very important to remember that momentum is a vector quantity and as such the direction of the momentum must be taken into account when solving our problems.

Sample Problems (Momentum) • A 2 kg ball is traveling at a constant speed of 30 m/s east. What is its momentum? • A 30 kg object has a momentum of 100 kg*m/s. What is its speed? • A 10 kg object is traveling to the right at 20 m/s, when it hits a wall. If it bounces straight back at a speed of 10 m/s, what was its change in momentum?

Sample Problems (Momentum) • A 2 kg ball is traveling at a constant speed of 30 m/s. What is its momentum? 60 kg*m/s east • A 30 kg object has a momentum of 100 kg*m/s. What is its speed? 3. 33 m/s • A 10 kg object is traveling to the right at 20 m/s, when it hits a wall. If it bounces straight back at a speed of 10 m /s, what was its change in momentum? -300 kg*m/s

Sample Problems Impulse • If you exert a force of 30 N (E) on an object for 10 s what is the impulse imparted on the object? • If you have an impulse of 50 N*s (W) acting on a 4 kg object for 25 s… – What is the force acting on the object? – What is the change in momentum? – What is the change in velocity of the object?

Sample Problems Impulse • If you exert a force of 30 N (E) on an object for 10 s what is the impulse imparted on the object? 300 N*s • If you have an impulse of 50 N*s (W) acting on a 4 kg object for 25 s… – What is the force acting on the object? 2 N (W) – What is the change in momentum? 50 kg m/s (W) – What is the change in velocity of the object? 12. 5 m/s (W)

Sample Problem Cons of p (inelastic) • A 300 kg box is sliding across the floor with a velocity of 5 m/s (E) when it hits and sticks to a 200 kg box that is initially stationary. What is the velocity of the two after the collision?

Sample Problem Cons of p (inelastic) • A 4 kg snowball is thrown to the right at a speed of 7 m/s. If it hits a 60 kg person walking to the left at 1 m/s and sticks to them, what is the new velocity of the person after the snowball hits them?

Sample Problems Cons of p (elastic) • A 2 kg ball is moving right at 4 m/s when it collides elastically with a stationary 1 kg ball. If the first ball continues on after the collision with a speed of 1. 33 m/s, what is the speed of the second ball after the collision?

Sample Problem Collision – A 2 kg ball is moving right at 4 m/s when it collides elastically with a stationary 1 kg ball. If the first ball continues on after the collision with a speed of 1. 33 m/s, what is the speed of the second ball after the collision?

Sample Problem – A 2 kg ball is moving right at 4 m/s when it collides elastically with a stationary 1 kg ball. If the first ball continues on after the collision with a speed of 1. 33 m/s, what is the speed of the second ball after the collision?

Sample Problem Collision • In the following problem determine the final velocity A 60 kg football player running at 3 m/s (E) tackles another player (80 kg) running at 4 m/s (W).

Sample Problem Collision (Cons of E) • In the following problem determine the final velocity. • A 60 kg football player running at 3 m/s (E) tackles another player (80 kg) running at 4 m/s (W). -1 m/s , 840 J

Sample Problem – Stopping Distance • A 2300 kg car is traveling north at 23 m/s when a braking force of 4500 N is applied. – What is the car’s velocity after 2 s? – How far does the car move in the 2 s? – How long before the car stops? • A 30 kg ball initially traveling at 9 m/s stops after rolling 34 m. What is the coefficient of friction between the ball and the surface?