Impulse and Momentum

Let’s start with everyday language What do you say when a sports team is on a roll? They may not have the lead but they may have ___________ MOMENTUM A team that has momentum is hard to stop.

What is Momentum? An object with a lot of momentum is also hard to stop Momentum = p = mv Units: kg∙m/s m=mass v=velocity Momentum is also a vector (it has direction)

Let’s practice Identify the variables: A 1200 kg car drives west at 25 m/s for 3 hours. What is the car’s momentum? Identify the variables: 1200 kg = mass 25m/s, west = velocity 3 hours = time P = mv = 1200 x 25 = 30000 kg m/s, west

How hard is it to stop a moving object? To stop an object, we have to apply a force over a period of time. This is called Impulse Impulse = FΔt Units: N∙s F = force (N) Δt = time elapsed (s)

Deriving Impulse-Momentum Theorem

Impulse = change in Momentum Consider Newton’s 2nd Law and the definition of acceleration Units of Impulse: Units of Momentum: Ns Kg x m/s Momentum is defined as “Inertia in Motion”

Impulse – Momentum Theorem CHANGE IN MOMENTUM This theorem reveals some interesting relationships such as the INVERSE relationship between FORCE and TIME

Impulse – Momentum Relationships

Impulse – Momentum Relationships Constant Since TIME is directly related to the VELOCITY when the force and mass are constant, the LONGER the cannonball is in the barrel the greater the velocity. Also, you could say that the force acts over a larger displacement, thus there is more WORK. The work done on the cannonball turns into kinetic energy.

How about a collision? Consider 2 objects speeding toward each other. When they collide...... Due to Newton’s 3rd Law the FORCE they exert on each other are EQUAL and OPPOSITE. The TIMES of impact are also equal. Therefore, the IMPULSES of the 2 objects colliding are also EQUAL

How about a collision? If the Impulses are equal then the MOMENTUMS are also equal!

Momentum is conserved! The Law of Conservation of Momentum: “In the absence of an external force (gravity, friction), the total momentum before the collision is equal to the total momentum after the collision.”

Ice Skaters 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.

F12 F21 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

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.

Sample Problem 1 continued on next slide 35 g 7 kg 700 m/s v = 0 A rifle fires a bullet into a giant slab of butter on a frictionless surface. The bullet penetrates the butter, but while passing through it, the bullet pushes the butter to the left, and the butter pushes the bullet just as hard to the right, slowing the bullet down. If the butter skids off at 4 cm/s after the bullet passes through it, what is the final speed of the bullet? (The mass of the rifle matters not.) 35 g 7 kg v = ? 4 cm/s continued on next slide

Sample Problem 1 (cont.) Let’s choose left to be the + direction & use conservation of momentum, converting all units to meters and kilograms. 35 g 7 kg p before = 7 (0) + (0.035) (700) = 24.5 kg · m /s 700 m/s v = 0 35 g 7 kg p after = 7 (0.04) + 0.035 v = 0.28 + 0.035 v v = ? 4 cm/s p before = p after 24.5 = 0.28 + 0.035 v v = 692 m/s v came out positive. This means we chose the correct direction of the bullet in the “after” picture.

Sample Problem 2 (0.035) (700) = 7.035 v v = 3.48 m/s 35 g 7 kg Same as the last problem except this time it’s a block of wood rather than butter, and the bullet does not pass all the way through it. How fast do they move together after impact? v 7. 035 kg (0.035) (700) = 7.035 v v = 3.48 m/s Note: Once again we’re assuming a frictionless surface, otherwise there would be a frictional force on the wood in addition to that of the bullet, and the “system” would have to include the table as well.

To Be Continued…