MOMENTUM Momentum Conservation of Momentum Impulse

Momentum A body travelling at 100m/s is harder to stop than the same body travelling at 10m/s. If two bodies were travelling at the same speed the heavier one would be much harder to stop. The combination of the bodies velocity and mass seem to affect how difficult it is to stop. This combination is called Momentum.

Momentum Defined Momentum is the product of mass and velocity. p = m x v momentum = mass x velocity (units are kgms-1) p = momentum m = mass v = velocity

Example A car with mass 2000kg is travelling at 20m/s. Calculate the car’s momentum. p =mv = 2000 x 20 = 40,000 kgms -1 Note Momentum is a vector quantity! Velocity and momentum vectors point in the same direction. SI unit for momentum: kgm/s or kgms -1 (no special name).

Examples – Direction Matters! 10 kg 3 m /s 10 kg p =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. p = 45 kg m /s but remember…. at 26º N of E 9 km /s 5 g 26º

Equivalent Momentums Bus: m = 9000 kg; v = 16 m /s p = 1.44 x10 5 kg m /s Train: m = 3.6 x10 4 kg; v = 4 m /s p = 1.44 x10 5 kg m /s Car: m = 1800 kg; v = 80 m /s p = 1.44 x10 5 kg m /s

Equivalent Momentums (cont.) The train, bus, and car all have different masses and speeds, but their momentums 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. 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. Note: We can only say that the magnitudes of their momentums are equal since they’re aren’t moving in the same direction.

Momentum and Collisions There are many types of collisions: two cars in a crash or two billiard balls on a table. Collisions can be divided into two types: Elastic Inelastic

Inelastic Collisions Two vehicles were set to collide on a linear air track. The two vehicles are arranged to collide as shown in the table below.

Table of Results Momentum before the collision Mass M1 (kg) Velocity V1 (m/s) Mom 1 M1V1 (kgms -1 ) Mass M2 (kg) Velocity V2 (m/s) Mom 2 M2V2 (kgms -1 ) Momentum Before M1V1+M2 V2 (kgms -1 )

Table of Results Momentum after the collision Mass (M1+M2) (kg) Velocity V3 (m/s) Momentum After Collision (M1+M1)V3 (kgms -1 )

Conservation of Momentum in 1-D Whenever two objects collide the 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. before: p = m 1 v 1 + m 2 (-v 2 ) after: p = m 1 (-v 3 ) + m 2 v 4 v1v1 m1m1 m2m2 v2v2 (Choosing right as the + direction, m 2 has - momentum.) m1m1 m2m2 v3v3 v4v4 m 1 v 1 + m 2 (-v 2 ) = m 1 (-v 3 )+ m 2 v 4

Directions after a collision If we solve the conservation of momentum equation (red box) for v 4 and get a negative answer, it would mean that m 2 was moving to the left after the collision. If the answer was positive for v 4 then it would still be moving to the right. m1m1 m2m2 v1v1 v2v2 m1m1 m2m2 v3v3 v4v4 m 1 v 1 +m 2 (-v 2 ) = m 1 (-v 3 )+m 2 (-v 4 ) Note: the momentum of the system is conserved even when the objects exert forces on each other without colliding, such as an object falling due to gravity.

Inelastic Collisions 1 A body of mass 5kg is travelling at 10m/s to the right when it collides with a stationary body of mass 2kg. After the collision they stick together and move off to the right. Calculate the speed with which they move off. m 1 =5kgm 2 =2kg 10m/s What happens next? m 1 v 1 + m 2 v 2 = (m 1 +m 2 )v 3 5x = ( 5 + 2)v 3 50 = 7v 3 Thus v 3 =7.14 m/s (to the right!)

Inelastic Collisions 2 A body of mass 20kg is travelling to the right at 10m/s when it collides with a stationary mass of 15kg. If the 20kg mass remains at rest after the collision with what speed and in what direction does the 15kg mass move? 10m/s 20kg 15kg

Inelastic Collisions 2 -Solution 00 Momentum before = Momentum after (20x10)+(15x0) = (20x0)+(15 x v 4 ) 200 = 15v 4 v 4 =200/15 v 4 = 13.3 ms -1 (positive so to the right) V 4 =? 20kg 15kg V 3 =0

Inelastic Collisions 3 7 kg v = m/s A rifle fires a bullet into a giant slab of butter on a frictionless surface. The bullet penetrates the butter 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 does not matter.) 35 g 7 kg v = ? 35 g 4 cm/s continued on next slide

Inelastic Collisions 3 (contd) 7 kg v = m/s 35 g 7 kg v = ? 35 g 4 cm/s p before =M1v1+m2v2 7 (0) + (0.035) (700) = 24.5 kg · m /s Choose left to be the + direction & use conservation of momentum, converting all units to metres and kilograms. p after = 7 (0.04) v = v p before = p after 24.5 = v v = 692 m/s v came out positive. This means the direction of the bullet in the “after” picture is to the left.

Inelastic Collisions 4 7 kg v = m/s 35 g 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 kg (0.035) (700) = 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 3 rd 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. M m F F 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. Force and the contact time is the same for each mass therefore they each undergo the same change in momentum, but in opposite directions.  p for the system is zero. (Note  p =Ft) The momentum one mass gains, the other loses. Hence, the momentum of the system before equals the momentum of the system after.

Exploding Bomb A c m eA c m e before after A c m e A bomb, which was originally at rest (and therefore with zero momentum), explodes and shrapnel flies every which way, each piece with a different mass and speed. The momentum vectors are shown in the after picture. From the law of conservation of momentum the net momentum after the collision must also be zero

Co-linear Explosion Example 7kg of explosives explodes into two fragments, one of 5kg and one of 2kg. After the explosion the 5kg piece flies off to the right at 2.6m/s. The 2kg pieces flies off in the same plane, what is its speed and direction? Mom Before = Mom after =0 =M 1 V 1 +M 2 V 2 = 0 = 2xV 1 + 5x2.6 = 0 -2V 1 =13 V 1 = -6.5m/s (To the left) 2kg 5kg 2.6m/s 6.5m/s

Collisions and Energy During a game of squash the ball warms up as the game progresses. Heat is generated by the collisions of the ball with the wall. A collision where energy is lost is called INELASTIC. A collision where energy is conserved is called ELASTIC

Force in Collisions From N2 we saw F= ma. Now a=(v-u)/t =  v/t Thus F= m  v/t or rearranging m  v= Ft This quantity is known as Impulse. Impulse = F t = m  v =  p Impulse is defined as the product of the force acting on an object and the time during which the force acts. Example: A 50 N force is applied to a 100 kg boulder for 3 s. The impulse of this force is = (50 N) (3 s) = 150 Ns. Note that we didn’t need to know the mass of the object in the above example.

Impulse Units Impule = 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 /s 2 ) (s) = 1 kg m /s { F net = ma shows this is equivalent to 1 N. 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: F t =  momentum We know the units on both sides of the equation are the same (last slide), but let’s prove the theorem formally: F t = m a t = m (  v / t) t = m  v =  p

Now imagine the same car moving at the same speed but this time hitting a giant haystack and coming to rest. The momentum of the car is the same but the force car is much smaller now since it acts for a much longer time. 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. Stopping Time Ft = m  v 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).

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? F t =  p p = m  v = m (u - v) = 1.3 [22 - (-13)] = (1.3 kg) (35 m/s) = 45.5 kg m /s. Thus, t =  p/F T= 45.5 / 1200 = s (= 40ms)

What Happens to the ball? 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.

F net vs. t graph F net (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 F net 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 - the area below the curve. This is just like a v vs. t graph, in which net displacement is given area under the curve. Area =  p

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 Earth M apple 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? m F F 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, i.e. include the 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: V v m V = M v

Sample Problem 4 before after 3 kg 15 kg 10 m/s 6 m/s 3 kg 15 kg 4.5 m/s v A crate of raspberry jelly collides with a tub of lime jelly on a frictionless surface. Which way and how fast does the lime jelly rebound? 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 jelly is moving 3.1 m/s to the left.

Conservation of Momentum in 2-D m1m1 v1v1 11 m2m2 v2v2 22 m1m1 vava aa m2m2 vbvb  b b To handle a collision in 2-D, we conserve momentum in each dimension separately. Choosing down & right as positive: before: p x = m 1 v 1 cos  1 - m 2 v 2 cos  2 p y = m 1 v 1 sin  1 + m 2 v 2 sin  2 after: p x = -m 1 v a cos  a + m 2 v b cos  b p y = m 1 v a sin  a + m 2 v b sin  b Conservation of momentum equations: x) m 1 v 1 cos  1 - m 2 v 2 cos  2 = -m 1 v a cos  a + m 2 v b cos  b y) m 1 v 1 sin  1 + m 2 v 2 sin  2 = m 1 v a sin  a + m 2 v b sin  b

Conserving Momentum - Vectors m1m1 p ap a aa m2m2 p bp b  b b m1m1 m2m2 p1p1 p 2p 2 11 22 BEFOREBEFORE AFTERAFTER Note p 1 + p 2 = p a + p b and p before = p after as conservation of momentum demands. p1p1 p 2p 2 p before p ap a p bp b p after

Exploding Bomb A c m eA c m e before after A c m e A bomb, which was originally at rest, explodes and shrapnel flies every which way, each piece with a different mass and speed. The momentum vectors are shown in the after picture. continued on next slide

Exploding Bomb (cont.) Since the momentum of the bomb was zero before the explosion, it must be zero after it as well. Each piece does have momentum, but the total momentum of the exploded bomb must be zero afterwards. This means that it must be possible to place the momentum vectors tip to tail and form a closed polygon, which means the vector sum is zero. If the original momentum of the bomb were not zero, these vectors would add up to the original momentum vector.

2D Problems Two objects, one of mass 2.0kg and moving at 30.0m/s whilst the other of mass 3.0kg moving at 20.0m/s are travelling at right angles to each other. The 2.0kg mass is moving in an easterly direction whilst the 3.0kg is heading North. They collide and remain coupled together. Find the velocity of the coupled mass after the collision.

Kinetic Energy and Momentum THEY ARE NOT THE SAME! Dependence on Velocity: The momentum of an object is proportional to the object's velocity - if you double its velocity, you double its momentum. p=mv The kinetic energy of an object is proportional to the square of the object's velocity - if you double its velocity, you quadruple its kinetic energy. Ek = ½mv 2 This has important consequences...

A Thought Experiment: Suppose that you were captured by an evil physicist who gave you the following choice: You must either: Stand in front of a 1000 kg. truck moving at 1 m/s, or Stand in front of a 1 kg. hotdog moving at 1000 m/s. What's your choice?

Your Choice? Hopefully, you picked the truck! It's a big truck, but it is moving rather slowly (about walking speed), so assuming you don't fall down when it hits you (That would be bad...) the truck is just going to bump into you and move you out of the way. On the other hand the hotdog is a very dangerous object. It isn't that massive, but it is moving very fast (about 10 football fields per second) - and if it were to hit you, it would do considerable damage to you, and keep going! Splat!

Energy & Momentum Calculations Consider the momentum and kinetic energy of the truck and the hotdog: Truck's momentum = mv = (1000 kg)(1 m/s) = 1000 kg m/s Truck's kinetic energy = ½mv 2 = (0.5)(1000 kg)(1 m/s) 2 = 500 Joules Hotdog's momentum = mv = (1 kg)(1000 m/s) = 1000 kg m/s Hotdogs's kinetic energy = ½mv 2 = (0.5)(1 kg)(1000 m/s) 2 = Joules

Killer Hotdog! We know intuitively that the hotdog is more dangerous than the truck, yet the momentum of the truck and the hotdog are the same. On the other hand, the hotdog has times the kinetic energy of the truck! Clearly, momentum and kinetic energy tell different things about an object!