Impulse/Momentum And Its Conservation

Which is easier to stop? Truck, car, bowling ball, or baseball all moving at 30 mph. Baseball -it is the least massive. Baseball at 30 mph or a baseball at 90 mph. Baseball at 30 mph - it is moving the slowest.

Baseball at 90 mph or a bowling ball at 20 mph. Hard to make a decision here. The least massive object is also moving the fastest. Past judgements were made by utilizing the concept of momentum.

A slow moving train and a high speed bullet can both have large amounts of momentum.

Momentum at any point in time is a combination of an objects mass and velocity. Momentum = Mass X Velocity p = mv Units p = kg m s Momentum is a vector quantity. Momentum is sometimes referred to as moving inertia.

If the velocity of an object is changing, the object is accelerating. Therefore, there must be a net force acting on the object. Net forces then also cause momentum changes.

p = m V F net = ma A = V t F net = m V t F net t = mv Ft = impulse m V = change in momentum

Impulses cause momentum changes. Units F t = N s m V = kg m s N s = kg m The units are equivalent. s

A net force of 20 N acts on a 2 kg object for 10 s. A. What is the change in velocity of the object? V = F net t m F net t = m V V = 20 N ( 10 s) 2 kg V = 100 m/s

B. What is the change in momentum of the object? m V = change in momentum m V = 2 kg ( 100 m/s ) m V = 200 kg m s

C. What impulse caused the change in momentum of the object? Ft = impulse Ft = 20 N ( 10 s) Impulse = 200 N s

A car weighs 7840 N. It is accelerated from rest to a velocity of 25 m/s by a net force of 1000 N. How long did the force act?

d =? m =? t =? F Net = 1000 N V i = 0 m/s F W = 7840 N V f = 25 m/s F net t = m V A =? t = m V F Net t = m( V f - V i ) F Net

F W = mg m = 7840 N 9.8 m/s 2 m = 800 kg t = 800 kg( 25 m/s - 0 m/s ) 1000 N t = 20 s

x x x x m V = F t Stopping in a short period of time requires a large force.

x x m V = F t Stopping in a long period of time requires a small force.

What is the purpose of a follow-through?

Apply the greatest force possible for the longest time possible. Accelerates the ball from 0 to high speed in a very short time.

Newton s 3 rd Law: Every force is accompanied by an equal and opposite force. When an object applies a net force on a second object, the second object accelerates and the first object also accelerates. (decelerates slows down)

The impulse of the bat decelerates the ball and accelerates it in the opposite direction very quickly.

Bouncing Falling: Speed = v p = mv Striking the ground: Speed = 0 p = 0 Rising: Speed = v p = mv Impulse needed to stop the ball = mv Total Impulse = 2mv Impulse needed to accelerate the ball upward = mv Important point: It only takes an impulse of mv to stop the ball. It takes twice that much (2mv) to make it bounce.

Law of Conservation of Momentum The total momentum of an isolated system can not change. Gain in momentum of 1 object occurs only through the loss of the same amount of momentum by a 2 nd object.

Types of Collisions Elastic Collisions: Collision where objects collide and bounce apart. Inelastic Collisions: Collision where objects collide and stick together.

Elastic Collisions

Elastic Collisions P A + P B = P A + P B indicates after the collision m A V A + m B V B = m A V A + m B V B Two objects before the collision and two objects after the collision.

Elastic Collision

Notice KE before KE car = ½ 1000 kg(20 m/s) 2 = 200000 J KE truck = 0 J KE total = 200000 J KE after KE car = ½ 1000 kg(-10 m/s) 2 = 50000 J KE truck = ½ 3000 kg(10 m/s) 2 = 150000 J KE total = 200000 J

Note: A perfectly elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter. Perfectly elastic collisions occur only if there is no net conversion of kinetic energy into other forms.

A glass ball has a mass of 5.0 g and is moving with a velocity of 20 cm/s. The ball collides with a second glass ball that has a mass of 10.0 grams moving in the opposite direction at 5.0 cm/s. After the collision, the 5.0 gram ball continues rolling in the same direction with a velocity of 2.0 cm/s. What is the velocity of the second ball?

BEFORE 5 g 20 cm/s 5 cm/s 10 g AFTER 2 cm/s? 5 g 10 g Why this way?

m A V A + m B V B = m A V A + m B V B Must mathematically show directions. 5 g ( 20 cm/s) + 10 g ( 5 cm/s) = V B = + 4 cm/s 5 g ( 2 cm/s) + 10 g ( V B ) + indicates to the right - as expected

A 20 kg projectile leaves a 1200 kg launcher with a velocity of 600 m/s. What is the recoil velocity of the launcher? A = Launcher B = Projectile m A V A + m B V B = m A V A + m B V B

1200 kg ( 0 m/s) + 20 kg ( 0 m/s) = 1200 kg (V A ) + 20 kg ( + 600 m/s) Must mathematically V A = - 10 m/s show directions. Indicates recoil - velocity opposite the projectile.

Inelastic Collisions

Inelastic Collisions P A + P B = P A,B m A V A + m B V B = m (A+B) V (A&B) Two objects before the collision and one larger object after the collision or vice-versa.

Inelastic Collision

Notice KE before KE car = ½ 1000 kg(20 m/s) 2 = 200000 J KE truck = 0 J KE total = 200000 J KE after KE car/truck = ½ 4000 kg(5 m/s) 2 = 50000 J KE total = 50000 J

Note: An inelastic collision, in contrast to an elastic collision, is a collision in which kinetic energy is not conserved. Only momentum is conserved.

Reminder: A perfectly elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter. Perfectly elastic collisions occur only if there is no net conversion of kinetic energy into other forms.

Inelastic Collision

Make a statement about the mass of the two cars.

A VW has a mass of 1000 kg and is traveling 50 m/s E. It runs into a Mack truck having a mass of 10000 kg which is moving 50 m/s W. What is the outcome of the inelastic collision? 50 m/s 50 m/s

m A V A + m B V B = m (A+B) V (A&B) 1000 kg(50 m/s) + 10000 kg( 50 m/s) = ( 1000 kg + 10000 kg ) ( V (A&B) ) V (A&B) = - 40.9 m/s Indicates m A+B is moving in the original direction of the Mack truck.

Frictional Effects on Momentum A brick has a mass of 5.0 kg. It is released on a very steep frictionless inclined plane at a point 10.0 m above a horizontal wooden table. The brick slides down the plane and strikes a second brick that weighs 98 N.

5 kg 10 m µ = 0 A. If the collision is inelastic, what is the initial speed of the two brick system? V (A&B) =? 98 N

m A V A + m B V B = m (A+B) V (A&B) V A =? V f 2 = V i2 + 2Ad d = t = V i = V f = - 10 m? 0 m/s? V f 2 = 2 (- 9.8 m ) (- 10 m) s 2 V f = 14.0 m/s A = g = - 9.8 m/s 2

5 kg (14.0 m/s) + 10 kg (0m/s) = F Wg (5 kg + 10 kg) V (A&B) V (A&B) = 4.67 m/s

B. If the coefficient of friction is 0.4, how long will the bricks slide? F f F N µ = F f F N 49 N 98 N F W Σ F y = O N = F N + F w

F N = F w F N = 147 N F f = µf N F f =.4 ( 147 N) F f = 58.8 N F net t = m V Σ F X = F net = F A + F f F net = F f

F Net t = m V t = m V F Net t = 15 kg (0 m/s - 4.67 m/s) Frictional 58.8 N Force t = 1.2 s

C. How far will the bricks slide before stopping? d H = V f + V i (t) 2 d H = 0 m/s + 4.67 m/s ( 1.2 s ) 2 d H = 2.80 m

The End