MOMENTUM! Momentum Impulse Conservation of Momentum in 1 Dimension Conservation of Momentum in 2 Dimensions Angular Momentum Torque Moment of Inertia

p = m v Momentum Defined p = momentum vector m = mass v = velocity vector

Momentum Facts p = m v Momentum is a vector quantity! Velocity and momentum vectors point in the same direction. SI unit for momentum: kg · m /s (no special name). Momentum is a conserved quantity (this will be proven later). A net force is required to change a body’s momentum. Momentum is directly proportional to both mass and speed. Something big and slow could have the same momentum as something small and fast.

Momentum Examples p = 45 kg · m /s at 26º N of E 3 m /s 30 kg · m /s Note: The momentum vector does not have to be drawn 10 times longer than the velocity vector, since only vectors of the same quantity can be compared in this way. 9 km /s 26º p = 45 kg · m /s at 26º N of E 5 g

Equivalent Momenta Car: m = 1800 kg; v = 80 m /s p = 1.44 ·105 kg · m /s Bus: m = 9000 kg; v = 16 m /s p = 1.44 ·105 kg · m /s Train: m = 3.6 ·104 kg; v = 4 m /s p = 1.44 ·105 kg · m /s continued on next slide

Equivalent Momenta (cont.) The train, bus, and car all have different masses and speeds, but their momenta are the same in magnitude. The massive train has a slow speed; the low-mass car has a great speed; and the bus has moderate mass and speed. Note: We can only say that the magnitudes of their momenta are equal since they’re aren’t moving in the same direction. The difficulty in bringing each vehicle to rest--in terms of a combination of the force and time required--would be the same, since they each have the same momentum.

Impulse Defined Impulse is defined as the product force acting on an object and the time during which the force acts. The symbol for impulse is J. So, by definition: J = F t Example: A 50 N force is applied to a 100 kg boulder for 3 s. The impulse of this force is J = (50 N) (3 s) = 150 N · s. Note that we didn’t need to know the mass of the object in the above example.

{ Impulse Units proof: 1 N · s = 1 (kg · m /s2) (s) = 1 kg · m /s J = F t shows why the SI unit for impulse is the Newton · second. There is no special name for this unit, but it is equivalent to a kg · m /s. proof: 1 N · s = 1 (kg · m /s2) (s) = 1 kg · m /s { Fnet = m a shows this is equivalent to a newton. Therefore, impulse and momentum have the same units, which leads to a useful theorem.

Impulse - Momentum Theorem The impulse due to all forces acting on an object (the net force) is equal to the change in momentum of the object: Fnet t =  p We know the units on both sides of the equation are the same (last slide), but let’s prove the theorem formally: Fnet t = m a t = m ( v / t) t = m  v =  p

Stopping Time F t = F t Imagine a car hitting a wall and coming to rest. The force on the car due to the wall is large (big F ), but that force only acts for a small amount of time (little t ). Now imagine the same car moving at the same speed but this time hitting a giant haystack and coming to rest. The force on the car is much smaller now (little F ), but it acts for a much longer time (big t ). In each case the impulse involved is the same since the change in momentum of the car is the same. Any net force, no matter how small, can bring an object to rest if it has enough time. A pole vaulter can fall from a great height without getting hurt because the mat applies a smaller force over a longer period of time than the ground alone would.

Impulse - Momentum Example A 1.3 kg ball is coming straight at a 75 kg soccer player at 13 m/s who kicks it in the exact opposite direction at 22 m/s with an average force of 1200 N. How long are his foot and the ball in contact? answer: We’ll use Fnet t =  p. Since the ball changes direction,  p = m  v = m (vf - v0) = 1.3 [22 - (-13)] = (1.3 kg) (35 m/s) = 45.5 kg · m /s. Thus, t = 45.5 / 1200 = 0.0379 s, which is just under 40 ms. During this contact time the ball compresses substantially and then decompresses. This happens too quickly for us to see, though. This compression occurs in many cases, such as hitting a baseball or golf ball.

Fnet vs. t graph Net area =  p Fnet (N) t (s) 6 A variable strength net force acts on an object in the positive direction for 6 s, thereafter in the opposite direction. Since impulse is Fnet t, the area under the curve is equal to the impulse, which is the change in momentum. The net change in momentum is the area above the curve minus the area below the curve. This is just like a v vs. t graph, in which net displacement is given area under the curve.

Conservation of Momentum in 1-D Whenever two objects collide (or when they exert forces on each other without colliding, such as gravity) momentum of the system (both objects together) is conserved. This mean the total momentum of the objects is the same before and after the collision. (Choosing right as the + direction, m2 has - momentum.) before: p = m1 v1 - m2 v2 v1 v2 m1 m2 m1 v1 - m2 v2 = - m1 va + m2 vb after: p = - m1 va + m2 vb va vb m1 m2

Directions after a collision On the last slide the boxes were drawn going in the opposite direction after colliding. This isn’t always the case. For example, when a bat hits a ball, the ball changes direction, but the bat doesn’t. It doesn’t really matter, though, which way we draw the velocity vectors in “after” picture. If we solved the conservation of momentum equation (red box) for vb and got a negative answer, it would mean that m2 was still moving to the left after the collision. As long as we interpret our answers correctly, it matters not how the velocity vectors are drawn. v1 v2 m1 m2 m1 v1 - m2 v2 = - m1 va + m2 vb va vb m1 m2

Sample Problem 1 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.

Proof of Conservation of Momentum The proof is based on Newton’s 3rd Law. Whenever two objects collide (or exert forces on each other from a distance), the forces involved are an action-reaction pair, equal in strength, opposite in direction. This means the net force on the system (the two objects together) is zero, since these forces cancel out. F M F m force on M due to m force on m due to M For each object, F = (mass) (a) = (mass) (v / t ) = (mass v) / t = p / t. Since the force applied and the contact time is the same for each mass, they each undergo the same change in momentum, but in opposite directions. The result is that even though the momenta of the individual objects changes, p for the system is zero. The momentum that one mass gains, the other loses. Hence, the momentum of the system before equals the momentum of the system after.

Conservation of Momentum applies only in the absence of external forces! In the first two sample problems, we dealt with a frictionless surface. We couldn’t simply conserve momentum if friction had been present because, as the proof on the last slide shows, there would be another force (friction) in addition to the contact forces. Friction wouldn’t cancel out, and it would be a net force on the system. The only way to conserve momentum with an external force like friction is to make it internal by including the tabletop, floor, or the entire Earth as part of the system. For example, if a rubber ball hits a brick wall, p for the ball is not conserved, neither is p for the ball-wall system, since the wall is connected to the ground and subject to force by it. However, p for the ball-Earth system is conserved!

Sample Problem 3 An apple is originally at rest and then dropped. After falling a short time, it’s moving pretty fast, say at a speed V. Obviously, momentum is not conserved for the apple, since it didn’t have any at first. How can this be? answer: Gravity is an external force on the apple, so momentum for it alone is not conserved. To make gravity “internal,” we must define a system that includes the other object responsible for the gravitational force--Earth. The net force on the apple-Earth system is zero, and momentum is conserved for it. During the fall the Earth attains a very small speed v. So, by conservation of momentum: apple m V F v Earth M F m V = M v

Sample Problem 4 A crate of raspberry donut filling collides with a tub of lime Kool Aid on a frictionless surface. Which way on how fast does the Kool Aid rebound? answer: Let’s draw v to the right in the after picture. 3 (10) - 6 (15) = -3 (4.5) + 15 v v = -3.1 m/s Since v came out negative, we guessed wrong in drawing v to the right, but that’s OK as long as we interpret our answer correctly. After the collision the lime Kool Aid is moving 3.1 m/s to the left. before 6 m/s 10 m/s 3 kg 15 kg after 4.5 m/s v 3 kg 15 kg