M.I.T. C.P. PHYSICS MOMENTUM

Momentum = mass of an object x its velocity P = mv

Momentum is a vector quality!

Sample Problem What is the value of the momentum of a 10-kg bowling ball as it rolls down the alley at a velocity of 5 m/s? Step 1 P = mv Step 2 P = (10)(5) Step 3 P = 50 kg•m/s

This change in momentum is called IMPULSE! In football, a defensive player applies a force for a given amount of time to stop the momentum of the offensive player. This change in momentum is called IMPULSE!

If the impulse force is large, the object’s momentum is changed quickly! If the impulse force is small, the object’s momentum is changed slowly!

Deriving the Impulse Formula

Deriving the Impulse Formula Impulse = Change in Momentum F = ma a = Δv/t F = m•Δv/t F • t = m •Δv

In boxing, riding a punch increases Time and reduces the Force of the collision. FT = mΔv Ft = mΔv

Introduction to Engineering

Lab: Eggstatic Over Impulse

Lab: Eggstatic Over Impulse Data Chart Mass of Zip-lock Bag & Egg 1 kilograms Mass of Zip-lock Bag & Egg 2 & Test Material Breakage Height (Egg) meters (Egg & Test Material)

Lab: Eggstatic Over Impulse Calculation Table Calculation # Egg Egg & Test Material 1 2 3 4 5 6 7

Lab: Eggstatic Over Impulse Egg Calculations Calculate the time of free fall of the egg when it broke in the zip-lock bag. T = √2d/g Calculate the velocity of the egg right before it broke in the zip-lock bag. v = gt

Calculate the momentum of the egg and the zip-lock bag before impact. Egg Calculations Calculate the momentum of the egg and the zip-lock bag before impact. p = mv Calculate the momentum of the egg and the zip-lock bag after impact. p = mv Calculate the change in momentum of the egg and the zip-lock bag. Δp = mΔv

Calculate the impulse of the system. Egg Calculations Calculate the impulse of the system. Impulse = Δp Assuming that the impact time lasted 0.025 seconds, calculate the force the egg encountered when it broke. F = Impulse/t

Egg & Test Material Calculations Calculate the time of free fall of the egg & test material when it broke in the zip-lock bag. T = √2d/g Calculate the velocity of the egg & test material right before it broke in the zip-lock bag. v = gt

Egg & Test Material Calculations Calculate the momentum of the egg & test material and the zip-lock bag before impact. p = mv Calculate the momentum of the egg & test material and the zip-lock bag after impact. p = mv Calculate the change in momentum of the egg & test material and the zip-lock bag. Δp = mΔv

Egg & Test Material Calculations Calculate the impulse of the system. Impulse = Δp Using the force value you calculated for the egg, calculate the time the egg endured the impact in the test material. t = Impulse/F

Lab: How Heavy Was the Rider?

Lab: How Heavy Was the Rider? Given Data Coefficient of Static Friction (u) = 0.4 Breaking Force = 1128.4 N Bike Mass = 175kg

Lab: How Heavy Was the Rider? Given Formulas d = v^2/2ug d = ½ (v1 + v2) Δt F • t = Δp Δp = m•Δv

Law of Conservation of Energy Law on Conservation of Mass Mass can not be created nor destroyed, only changed. Law of Conservation of Energy Law of Conservation of Momentum The total momentum in a system remains constant.

Momentum Conservation

Lab: Validating the Law of Conservation of Momentum Interactive

Lab: Validating the Law of Conservation of Momentum Data Chart: Cart Trial # Mass (kg) Velocity (m/s) 1 100 2 3 200 4 5 400 6 7 500

Lab: Validating the Law of Conservation of Momentum Data Chart: Jumper Trial # Mass (kg) Velocity (m/s) 1 50 2 100 3 4 5 6 7

Lab: Validating the Law of Conservation of Momentum Calculation Chart Trial # p(cart) (kg•m/s) p(jumper) 1 2 3 4 5 6 7

Sample Problem 1 Consistent with Newton's third law of motion, when a rifle is fired, the bullet pushes backwards upon the rifle. The acceleration of the recoiling rifle is ... more than the bullet. less the bullet. the same as the bullet.

Sample Problem 2 A 50.0-kg cart is rolling at a constant velocity of 25.0m/s. How much mass should an additional brick have in order to slow its velocity to 15.0m/s. Step 1 m1 x v1 = m2 x v2 Step 2 (50.0)(25.0) = m2 (15.0) Step 3 m2 = 83.3kg Step 4 83.3kg – 50.0kg = 33.3kg

Explosions Even during an explosion, the Law of Conservation of Momentum takes effect!

Explosions p(before) = p(after) p(before) = m1v1 + m2v2 p1 = p2 p(after) = m1v1 + m2v2 Since velocity & momentum is a vector quality, make one of the velocities negative!

Calculate the velocity of the pink cart after the explosion. Sample Problem 1 Calculate the velocity of the pink cart after the explosion. Step 1 m1v(initial) + m2v(initial) = m1v(final) + m2v(final) Step 2 0 = (0.5)(-40) + (1.0)v(final) Step 3 v(final) = 20m/s

Calculate the velocity of the pink cart after the explosion. Sample Problem 2 Calculate the velocity of the pink cart after the explosion. Step 1 m1v(initial) + m2v(initial) = m1v(final) + m2v(final) Step 2 0 = (1.0)(-20) + (2.0)v(final) Step 3 v(final) = 10m/s

Calculate the velocity of the pink cart after the explosion. Sample Problem 3 Calculate the velocity of the pink cart after the explosion. Step 1 m1v(initial) + m2v(initial) = m1v(final) + m2v(final) Step 2 0 = (0.5)(-40) + (2.0)v(final) Step 3 v(final) = 10m/s

Lab: Explosions

Lab: Explosions Data Chart 1 Cart Mass (kg) Distance (m) Time (s) Left Right

Lab: Explosions Data Chart 2 Cart Mass (kg) Distance (m) Time (s) Left Right

Lab: Explosions Calculation Chart 1 Cart Mass (kg) Velocity (m/s) Momentum (kg•m/s) Left Right

Lab: Explosions Calculation Chart 2 Cart Mass (kg) Velocity (m/s) Momentum (kg•m/s) Left Right

Collisions During a collision, the Law of Conservation of Momentum still takes effect!

In an inelastic collision, two bodies collide and act as one! Inelastic Collisions In an inelastic collision, two bodies collide and act as one! The total momentum of the system before the collision must equal the total momentum of the system after the collision.

Inelastic Collisions The total momentum of the system before the collision must equal the total momentum of the system after the collision.

Inelastic Collisions

Inelastic Collisions

Determine the initial momentum of each object. Sample Problem Calculate the final velocity of the girl on frictionless ice after she catches the ball. Step 1 Determine the initial momentum of each object. p = mv Ball = 300kg•m/hr Girl = 0

Determine the total momentum before impact. Sample Problem Step 2 Determine the total momentum before impact. p(before) = m1v1 + m2v2 p(before) = 300kg•m/hr

Determine the total mass after impact. Sample Problem Step 3 Determine the total mass after impact. m(total) = (m1 + m2) m(total) = (15 + 60) m(total) = 75kg

Determine the velocity after impact. Sample Problem Step 4 Determine the velocity after impact. p(after) = m(total) v2 300kg•m/hr = (75) v2 v2 = 4m/s

Lab: Inelastic Collisions

Lab: Inelastic Collisions Red Car Data Chart Trial # m(kg) vi (m/s) vf (m/s) 1 10.0 2 5.0 3 4 5 8.0 6 6.0 7 3.0

Lab: Inelastic Collisions Blue Car Data Chart Trial # m(kg) vi (m/s) vf (m/s) 1 3 2 4 5 3.0 6 2.0 7 1.0

Lab: Inelastic Collisions Red Car Calculation Chart Trial # p(before) p(after) 1 2 3 4 5 6 7

Lab: Inelastic Collisions Blue Car Calculation Chart Trial # p(before) p(after) 1 2 3 4 5 6 7

Lab: Inelastic Collisions Final Calculation Chart Trial # Total p(before) Total p(after) Total KE(before) Total KE(after) 1 2 3 4 5 6 7

Sample Problem A 5.0g bullet is fired upward at a velocity of 700. m/s. It hits and embeds itself into a 5.0-kg block of wood. What is the momentum of the bullet before impact? Step 1 p = mv Step 2 p = (.0050)(700.) Step 3 p = 3.5kg•m/s

What is the momentum of the bullet after impact? Sample Problem What is the momentum of the bullet after impact? Step 1 p(before) = p(after) Step 2 3.5kg•m/s = p(after) Step 3 p(after) = 3.5kg•m/s

What is the velocity of the bullet after impact? Sample Problem What is the velocity of the bullet after impact? Step 1 p(after) = m(total) v(final) Step 2 3.5kg•m/s = (5.005)v(final) Step 3 v(final) = .70m/s

What is the kinetic energy of the bullet before impact? Sample Problem What is the kinetic energy of the bullet before impact? Step 1 KE = ½ mv^2 Step 2 KE = (.5)(.0050)(700.)^2 Step 3 KE = 12225J

What is the kinetic energy of the bullet after impact? Sample Problem What is the kinetic energy of the bullet after impact? Step 1 KE = ½ mv^2 Step 2 KE = (.5)(5.0 + .0050)(.70)^2 Step 3 KE = 1.2224J

How high does the block rise after impact? Sample Problem How high does the block rise after impact? Step 1 KE = PE = mgh Step 2 1.2224J = (5.005)(9.8)h Step 3 h = 0.025m

In an inelastic collision, there is no loss of momentum! In an inelastic collision, there is a loss of kinetic energy.

RESEARCH PROJECT #3

MIDDLESEX

Calculate the initial momentum. Warm Up Problem Calculate the initial momentum. Step 1 p (initial) = (m1 v1(initial)) + (m2 v2(initial) ) Step 2 p(initial) = (0.045)(400.) + (16) x (0) Step 3 p(initial) = 18 kg•m/s

Calculate the final velocity. Warm Up Problem Calculate the final velocity. Step 1 p(final) = (m1 + m2) v(final) Step 2 18 = (0.045 +16) v(final) Step 3 v(final) = 1.1 m/s

Elastic Collisions In an elastic collision, both objects travel independently after the collision.

As with an inelastic collision, there is no loss of momentum in an elastic collision.

But unlike an inelastic collision, there is no loss of kinetic energy during an elastic collision.

Lab: Elastic Collisions Interactive

Lab: Elastic Collisions Red Car Data Chart Trial # m(kg) vi (m/s) vf (m/s) 1 10.0 2 5.0 3 4 5 8.0 6 6.0 7 3.0

Lab: Elastic Collisions Blue Car Data Chart Trial # m(kg) vi (m/s) vf (m/s) 1 3 2 4 5 3.0 6 2.0 7 1.0

Lab: Elastic Collisions Red Car Calculation Chart Trial # p(before) p(after) KE(before) KE(after) 1 2 3 4 5 6 7

Lab: Elastic Collisions Blue Car Calculation Chart Trial # p(before) p(after) KE(before) KE(after) 1 2 3 4 5 6 7

Lab: Elastic Collisions Final Calculation Chart Trial # Total p(before) Total p(after) Total KE(before) Total KE(after) 1 2 3 4 5 6 7

Which ball(s) would be moving after #1 & #2 make contact? Sample Problem 1 Which ball(s) would be moving after #1 & #2 make contact?

Sample Problem 2 What would be the final velocity of ball 2 after an elastic collision if ball 1’s final velocity is 0 m/s?

Jeff Laws Step 1: Calculate Initial Momentum

Sample Problem 2 What would be the final velocity of ball 2 after an elastic collision if ball 1’s final velocity is 0 m/s? Step 1 Calculate the total initial momentum. Total p(initial) = m1 v1(initial) + m2 v2(inital) Total p(initial) = (0.25)(+2.0) + (0.25)(-0.5) Total p(initial) = (.50) + (-.125) Total p(initial) = .375kg•m/s

Step 2: Calculate Final Velocity of Ball #2

Sample Problem 2 What would be the final velocity of ball 2 after an elastic collision if ball 1’s final velocity is 0 m/s? Step 2 Calculate the final velocity of ball #2. Total p(final) = m1 v1(final) + m2 v2(final) .375 = (0.25)(0) + (0.25)v2(final) .375 = (0.25)v2(final) v(final) = 1.5m/s

Sample Problem 3 What would be the final velocity of ball 2 after an elastic collision if ball 1’s final velocity is -1.0 m/s? Step 1 Calculate the total initial momentum. Total p(initial) = m1 v1(initial) + m2 v2(inital) Total p(initial) = (0.25)(+2.0) + (0.25)(-0.5) Total p(initial) = (.5) + (-.125) Total p(initial) = .375 kg•m/s

Sample Problem 3 What would be the final velocity of ball 2 after an elastic collision if ball 1’s final velocity is -1.0 m/s? Step 2 Calculate the final velocity of ball #2. Total p(final) = m1 v1(final) + m2 v2(final) .375= (0.25)(-1.0) + (0.25)v2(final) .375 = (-0.25) + (0.25)v2(final) v(final) = 2.5 m/s

Nice Job!!!!!!!!

Inelastic Collisions

Inelastic Collisions

2-Dimentional Inelastic Collisions

Sketch the before & after systems Sample Problem A 2165 kg car moving east at 17.0 m/s collides with a 1325 kg car moving north at 27.0m/s. They collide and stick together. In what direction and at what velocity do they move after the collision? Step 1 Sketch the before & after systems

Calculate the initial momentum of each car. Sample Problem Step 2 Calculate the initial momentum of each car. Car A (east) p = mv p = (2165)(17.0) p = 36800 kg•m/s

Calculate the initial momentum of each car. Sample Problem Step 2 Calculate the initial momentum of each car. Car B (north) p = mv p = (1325)(27.0) p = 35800 kg•m/s

Establish the coordinate axis. tanΘ = opposite/adjacent Sample Problem Step 3 Establish the coordinate axis. Car A = opposite Car B = adjacent tanΘ = opposite/adjacent

Calculate the direction of the collision. tanΘ = opposite/adjacent Sample Problem Step 4 Calculate the direction of the collision. tanΘ = opposite/adjacent tanΘ = 35800/36800 tanΘ = .973 Θ = 44.2°

Calculate the total momentum after the collision. Sample Problem Step 5 Calculate the total momentum after the collision. a^2 + b^2 = c^2 (36800)^2 + (35800)^2 = c^2 2630000000 = c^2 c = 51300kg•m/s

Sample Problem Step 6 Calculate the final velocity after the collision. p(final) = (m1 + m2)v(final) 51300 = (2165 + 1325)v(final) 51300 = 3490v(final) v(final) = 14.7m/s