Unit 4, Part 1---Momentum

Every object has mass. You also know that if an object is moving, it has velocity. And, if an object is moving, it also has momentum. Momentum is simply the vector product (multiplication) of an object’s mass and its velocity. The symbol for momentum is p (just a small “p”). To put momentum into an equation form:

p = mv

This says that an object’s momentum is a product of its mass and the velocity it’s moving with. You can get a large momentum (high value of p) in a couple of different ways:

  1. An object with a small mass (like a bullet) with a large velocity (moving really, really fast) can have a large momentum.
  2. An object with a large mass (like a train) can have a large momentum even if it’s moving very, very slow.

Objects with a lot of momentum can cause a lot of damage to other things (like a person). Think of how much damage a fast (but small) bullet can do to the flesh of a person. Or, think about how much damage a slow moving (but large) train can do to a person. Ouch.

The units of momentum are (kg)(m/s). The product of these two don’t form another unit (like the product of kg and m/s2 form a Newton). The unit of momentum is simply kg x m/s (kilogram-meter-per-second or kilogram “times” meter-per-second). Sometimes you might see the units of momentum as a “Newton-second” (Newton “times” second). If you take the real units that make up a Newton (Kg “times” m/s2) and multiply that by seconds, this will simplify down to Kg “times” m/s. So, you can use either way; both are acceptable.

One more general thing about momentum: Momentum is avector quantity with its direction being in the same direction as its velocity. Two objects with the same mass traveling at the same speed could actually have different momentums if they are traveling in opposite directions. In this unit we’ll still be using + or – to show the direction something is moving when we use equations.

Constant Momentum

If an object is moving, it has momentum. If it’s moving with constant velocity, then its momentum is constant (since

p= mv). In the momentum equation, if an object’s velocity changes, then its momentum must change as well. The only way to make an object’s momentum change is to either change its velocity or change its mass. In all of the physics problems that you will do in this unit, an object’s mass won’t be changing, but velocity is very easy to change---just speed up, slow down, or change direction. For example, if a car is moving along with some velocity and then it slows down (decreases its velocity), then its momentum will also decrease. Velocity and momentum are directly proportional to each other. If one goes up, the other goes up and vice versa.

1. Here’s an example problem: (This is how you would calculate the momentum of any object).

A baseball of mass 0.14 kg is moving at +35 m/s. a) What is the momentum of the ball? (b) Find the velocity at which a bowling ball (mass = 7.26Kg) would have the same momentum as the baseball.

a. m = 0.14Kgv = +35m/sp = ?p = mv +4.9Kgm/s

b. p = 4.9 Kgm/s m = 7.26 Kg v = ?p = mv0.67 m/s

2. A compact car (m = 725 Kg) is moving at 100 km/h. (a) What is its momentum? (b) What velocity will a larger car (m = 2175 Kg) have if it has the same momentum as the smaller car?

a) m = 725kgp = mv

v = 100km/h (needs to be converted to m/s)

p = ?

b)This part of the problem is saying that if a different, bigger (greater mass) car was moving and had the same momentum as the initial car from the above problem, how fast is it moving (what is its velocity?)?

p = “this is the momentum from part ‘a’ above”

m = 2175kg

v = ?p = mvv = p/m

Changing Momentum

If something is moving with constant velocity (and, hence, has constant momentum) its acceleration must be zero. And, if it isn’t accelerating, then the overall net force acting on it must also equal zero (Fnet = 0N). Conversely, if the velocity of an object changes (i.e., the object accelerates), there MUST be an overall net force acting on it. Soooooo, to get an object’s momentum to change, a net force must be applied to the object. The only way to get an object’s velocity, and hence its momentum, to change is by applying a net force to it.

  • Constant velocity = zero acceleration = Fnetis zero = constant momentum
  • Changing velocity = acceleration = some overall net force on the object = changing momentum

All of what was just said can be restated in an equation as:p = mv

The above equation shows that if the momentum of an object changes (“delta” p, Δp), the velocity had to have changed, too. If the velocity changes, it’s because there was some net force acting on the object for some period of time. So, we can also relate the size of the force acting on an object to the object’s change in momentum. It can be put into an equation as: F = ma  F = m∆v/∆t  F∆t = m∆v  Ft = p

The above equation (that is highlighted) says that the larger the force acting on an object (over some time interval), the larger the change in momentum. It also says the longer the time the force acts on the object, the larger the change in momentum. Both “t” and “F” are directly proportional to the change in momentum (p).

The product of force and the time interval the force acts (Ft) is called the “impulse”. Some problems will say “Calculate the impulse….” or “The impulse acting on an object is….”. When they do, they simply mean “Ft”, which is also equal to p (which is also equal to mv…they are all equal to each other).

  • p = the “change in “ momentum; this is always equal to the final momentum minus the initial momentum, or pf - pi
  • m = the object’s mass in Kg
  • v = the “change in” velocity of the object; this is always equal to vf - vi (or v – vo)
  • F = the “net” force acting on the object
  • t = the amount of time the net force acts on the object

p = mv = Ft

  • All three of the above “variables” are equal to each other. If you wanted to find “F”, you would need “t” and either ORmv----not both. And, if a question asks you to find the “impulse” (Ft) and you don’t have “F” or “t”, as long as you have a way calculate the change in momentum (p, or mv), you also have the “impulse”.

3. A snowmobile has a mass of 2.5 E2 Kg. A net force is exerted on it for 60 sec. The snowmobile’s initial velocity is 6m/s and its final velocity is 28m/s. (a) What is its change in momentum? (b) What is the magnitude of the net force exerted on it?

a)m = 2.5 E2 kg

t = 60s

vi = 6m/s

vf = 28m/s

p = ?p = mv p = m (vf – vi)5.5 E3 kgm/s

Remember: If a net force is applied to an object, it will accelerate. Anytime an object accelerates, its velocity AND its momentum will change.

b)F =?

  • You could first find “a” using one of the four funky equations and then use Fnet= ma, OR you could also use “t” and the “Δp” that you just calculated and find the force with the following:

Ft = pF = p/t+91.67 N

4. The brakesexert a 6.4 E2N force on a car weighing 15,680N moving at 20m/s. The car finally stops. (a) What is the car’s mass? (b) What is its initial momentum? (c) What is the change in the car’s momentum? (d) How long does the braking force act on the car to bring it to a halt?

a)Fnet = -6.4EE2 N (Fnet = FA - Ffriction due to braking , and FA = 0N since it’s braking)

mg = 15,680 N, down

Vi = 20m/s

Vf = 0m/s

m = ?weight = mg m = 1600Kg

b)pi = ? (This is the momentum the car had before it began breaking….so you would use the velocity the car had before it began breaking)

pi = mvi3.2 E4 Kgm/s

c)p= ?You can do this one of two ways:

  • p = pf – pi and pf = 0 kgm/s (it stops, so its final velocity is zero and its final momentum would also be zero)
  • OR p = mv, where v = vf – vi

-3.2 E4 Kgm/s

  • The “change in” momentum is negative due to it losing momentum (it was moving and then it stopped)

d)t = ?Ft = p OR Ft = mv sincep = mv

50s

Conservation of momentum

Momentum (like matter and energy) is a quantity that can’t be created or destroyed, simply shifted from one form to another. If two cars collide and one loses momentum, the other car must gain the same amount of momentum. The Law of Conservation of Momentum says “Whatever the total momentum of a system is BEFORE a collision, the total momentum of the system AFTER the collision must be the same”.

To put it into an example form that will be easier to understand:

Let’s say car A has +25 kgm/s of momentum (it’s moving to the right) and car B has +10 kgm/s of momentum (it is also moving to the right) before they collide (they are both moving to the right, Car B is in front of Car A, but Car A catches up and rear-ends it from behind). The TOTAL momentum of the system (of both cars combined) before the collision is +35kgm/s (just add them together). What if they then collide? Neither of their masses will change after they collide, but their velocities will most likely change, so their momentums must change, too. After the collision, let’s say Car A onlyhas +15 kgm/s of momentum (it’s still moving to the right, but it’s slowed down). Since the TOTAL momentum of the system after the collision has to equal +35kgm/s and car A has +15 kgm/s, car B’s momentum MUST now be +20kgm/s. This means that car B was hit and it moved off in a forward direction with a greater speed than it had before.

A B A’ B’

+ +

25 kgm/s 10 15 20

“collision”

The formula for conservation of momentum is:

pA + pB =p’A + p’B

The above equation reads “The initial momentum of object A added to the initial momentum of object B before a collision is equal to the final momentum of object A after the collision (p “prime” A) addedto the final momentum of object B after the collision (p “prime” B). Instead of using “initial and final momentum” with the little “i” and the “f” as subscripts, this formula uses a mark called “prime” which is a little ’ to note the “after” conditions. By doing this it eliminates having too many subscripts under any given variable.

Know that in any “collision” the mass of the objects will never change for any of the problems that we do in here; if the momentum of an object changes it’s due to the velocity changing. So, the above formula will be expanded out to:

mAvA + mBvB = mAv’A + mBv’B

In all of the conservation of momentum problems, you will most likely have all of the above variables except one. To avoid getting confused as there are so many variables in the equation, in the “conservation of momentum” problems, label the first object that is mentioned in the problem as the “A” object and the second object the problem mentions as the “B” object.

  1. A .035kgbullet moving at 475 m/s strikes a 2.5Kg wooden block. The bullet passes through the block, leaving at 275m/s. The block was at rest when it was hit. How fast is it (the block) moving when the bullet leaves?

mA = 0.035 kgmB = 2.5 kg

vA = 475 m/svB = 0m/s (at rest initially)

v’A = 275 m/sv’B = ?

mAvA + mBvB = mAv’A + mBv’B

  • Rearrange and solve for v’B.2.8 m/s
  1. A 0.105-Kg hockey puck moving at 48m/s is caught by a 75-Kg goalie at rest. With what speed does the goalie slide on theice?
  • This problem is saying that a moving puck hits a non-moving goalie. The goalie catches the puck and they both move off together (the puck is in the goalie’s glove). If they are “together” after the collision, then both of their final velocities will be the same value. So, v’A will equal v’B.

mA = 0.105 kg mB = 75 kg

vA = 48m/svB = 0m/s (at rest initially)

v’A = ?v’B = ?(v’A = v’B)

mAvA + mBvB = mAv’A + mBv’B

  • Since you can’t have two unknowns in an equation and be able to solve it with just a single equation, you need to replace your two unknowns with a single unknown that will represent both (since they’re both equal to each other you can do this). You can replace them with a big “X” or anything you feel like. I’m going to use V’AB since that will remind me that the number that I end up calculating represents BOTH v’A and v’B.
  • Rewriting the equation with the new symbol for my unknowns gives me:

mAvA + mBvB = mA (v’AB) + mB (v’AB)

  • The two “V’AB’ s” can then be factored out as:

mAvA + mBvB = V’AB (mA + mB)

It should be easy to solve from here.0.067m/s

  1. A .035kg bullet strikes a 5Kg stationary wooden block and embeds itself in the block. The block and bullet fly off together at 8.6 m/s. What was the original velocity of the bullet?
  • This problem is very similar to the previous problem (#6). However, in this problem they give you the final velocities of the objects and ask you to calculate the initial velocity of just the bullet (the block is initially at rest). Watch your units as the bullet’s mass is initially given to you in grams, not kilograms.

1237.17m/s

  1. A 0.50 Kg ball traveling at 6m/s collides head-on with a 1.0-Kg ball moving in the opposite direction at a velocity of -12m/s. The 0.50-kg ball moves away at –14m/s after the collision. What is the velocity of the second ball?
  • This problem is just like the very first problem you did (#5). The only difference is the two objects are now moving in opposite directions before the collision, so one velocity is initially negative and the other velocity is initially positive. The sign of your final answer will tell you the direction the object ends up moving after the collision.

-2m/s

The thing you need to remember when doing conservation of momentum problems is that however much one object LOSES in momentum, the other one must GAIN it, and vice-versa. Some of the “collision” problems might have you calculate “the change in momentum for object A” and you will have enough information to do it (like you’ll have the object’s mass and its initial and final velocities). Then the problem will next ask you to calculate object B’s change in momentum and it will look like you don’t have enough info (like you won’t have the object’s mass or its change in velocity). However, if you just calculated that object A lost 10kgm/s of momentum (pA = -10kgm/s), then object B must have gained that amount if they collided, so its change in momentum must be +10kgm/s. Get it? If not, ask me.

Conservation of Momentum in 2-D

AP only….will only be covered in lecture. However, here’s a hint: It involves finding the horizontal and vertical components of each object’s momentum before the collision and after the collision. The total vertical momentum of both object’s added together before the collision must equal the total vertical momentum (of both object’s added together) after the collision and the total horizontal momentum before the collision must equal the total horizontal momentum after the collision. These problems aren’t difficult, they’re just sooo busy that they are a pain in the rear.

Super Important Note to any of my AP Kids who read these notes….When we get to Energy (the last section in this unit), we will go back and re-visit Conservation of Momentum. The momentum problems you will get to after Energy will be referred to “Elastic Collisions”. In pure elastic collisions, the total momentum AND kinetic energy of the system are BOTH conserved. Up to this point, we haven’t taken into account the Kinetic Energy of the system…we have just been working the problems as if they were purely inelastic collision problems (where KE is not conserved). In Elastic collisions, you won’t have either of the two final velocities of the objects so you will have to solve two equations (conservation of momentum AND conservation of Energy) simultaneously for one variable. The thing to remember when doing “Elastic” collision momentum problems is that the total velocity of the first object (before and after the collision) must equal the total velocity of the second object (before and after the collision).

vA + vA’ = vB + vB’AND mAvA + mBvB = mAv’A + mBv’B

You can solve the equation on the left for one of the variables and plug THAT into the conservation of momentum equation to find the final speeds of the objects.

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Unit 4, Part 1: Momentum Notes