Physics Phenomena

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Study Of Momentum and Conservation of Momentum

Momentum and Conservation of Momentum:

Isaac Newton studied a phenomena referred to as momentum.  He defined momentum as the quantity of the motion of a body that is described by the mathematical product of mass and velocity.  In the MKS system (meter, kilogram, second) this would give momentum the units of kg m/s.

Note: That though the units for momentum is similar to the units for force, 1 N = 1 kg m/s^2, it is not the same measurement at all. Momentum is simply kg m/s.

Isaac Newton also explored the conservation of momentum which says that "the total momentum of a closed system before an event occurs is equal to the total momentum after the event occurs".

To express this in another way, though the individual momentum of different objects may change within a closed system, the total momentum of the closed system remains the same.

[Special Note: Conservation laws are found throughout the study of science.  Physicists found that they were more easily able to study certain phenomena such as momentum using conservation laws, where, though individual quantities within a system may change, the overall sum total of a quantity remained constant.  The law of conservation of energy is another example of a conservation law.]

Momentum:  As defined by Newton, momentum is the product of an object’s mass and its velocity.  The symbol for momentum is the lower case p.  Momentum has to do with questions about both starting and stopping objects.  Momentum is not just about the mass of an object or just about the velocity of an object, but it is about the combined effect of both measurements when an object’s motion changes.  Consider this question, “ Which object has more momentum and would be harder to stop, a small sports car traveling with a high velocity or a large truck moving with a low velocity? ”.
 

Calculating Momentum
Example:  Calculate the momentum of a 10 kg object moving at 5 m/s.

G: m = 10 kg, v = 5 m/s
F: the momentum of the truck
E: p = m v
S: p = 10 kg * 5 m/s
A: p = 50 kg m/s

Impulse:  When discussing the prospect of changing the momentum of an object, the term Impulse (symbol J*) is introduced. Impulse is the product of the unbalanced (net) force applied to an object multiplied by the time interval over which the force is applied.  The unit for impulse is the Ns, based on the fact that Impulse (J) equals the product of Force and the time interval over which the force is applied.
(*Some sources use an upper case I, but on this page we use the upper case J, which is a very commonly used symbol for Impulse)
 
 

An impulse causes a change in motion and is equal to the change in momentum that an object experiences because of the impulse.

Calculating Impulse
Example:  Calculate the impulse experienced by a 7.5 kg object when a 20 N force is applied to it for 10 seconds.

G: m = 7.5 kg, F = 20 N, t = 10 s
F: the impulse experienced by the object
E: Impulse = J = F t
S: Impulse = J = 20 N * 10 s
A: Impulse = J = 200 Ns

Impulse and change in momentum:  Impulse is the means by which momentum is changed, so we may say that Impulse equals change in momentum.  [rem: The phrase “change in” is symbolized by /\ (delta) symbol.]
 

Impulse-momentum Theorem
Impulse = Change in momentum, 

J = /\ p

or F /\t = m /\ v

NOTE: Though it is possible to change momentum by changing the mass of an object, most often we study change in momentum as the result of a change in an object’s velocity.
 

Using the Impulse-momentum Theorem
Example:  A force of 20 N is applied to a 50 kg object over 10 seconds.  Determine the object’s change in velocity.

G: F = 20 N, m = 50 kg, t = 10 s
F: the change in velocity of the object
E: J = /\p, which is F t = m /\ v
S: 20 N * 10 s = 50 kg * /\ v
A:  /\ v = 4 m/s

In the above problem, if the v(initial) was 12 m/s, what would the v(final) equal?

G: v(initial) = 12 m/s, /\ v = 4 m/s
F: the final velocity
E: v(final) = v(initial) + /\ v
S: v(final) = 12 m/s + 4 m/s
A: v(final) = 16 m/s

Newton’s Third Law:  During the study of forces we discussed Newton’s three laws of motion, but most of our attention was on his first and second laws.  We are now going to turn our attention towards his third law.

Authors often use the phrase “the law of action and reaction” when describing Newton’s third law.  This statement describes the forces exerted by two different objects when they interact with one another. A classical example described in many physics texts is that of a cannon launching (firing) a cannon ball. When launched the cannon ball flies forward with a specific amount of momentum. At the same time the cannon experiences a recoil velecity, which results in it moving backwards with an equal amount of momentum. The cannon ball and the cannon both experience the same amount of momentum, only it is in opposite directions. The cannon ball's momentum is in the forward direction, and the cannon's momentum is in the reverse direction.

Another example is when you push on a wall the “wall pushes back”.
"Granted that the wall is not “making an effort to push back actively”, but it is by the very design and choice of materials that      were used to construct it that it does resist your effort, thus, in a sense, it pushes back."

In such a case we might say that its resistance represents a force pushing back.  That "resistance force" is due to the very nature of the wall based upon the architects' design and the contractors' efforts in building the wall.  Likewise, the stools, chairs, and the tables or benches used in your science classroom are designed to push back against your weight or the weight of any other object placed on them.  Weight is a downward force and the force they supply as they support that downward force is an equal but opposite force that prevents the stool, chair, table or bench from collapsing.  In the study of forces this “upward force” that opposes gravity and is perpendicular to the surface upon which an object resides is called the normal force.

Example:  Consider two objects such as a book and a table.  The book is resting on the surface of the table.  It is pressing down with a force, the force of gravity, on the top of the table.  By the very nature of the design of the table and the choice of materials it is made from, it resists that downwards force and exerts an upward force equal but opposite in direction, meaning upwards in this case, to that of the weight of the book.

An individual or single force cannot exist. For one object to produce a force on a second object the second object must produce a force on the first object.  When baseballs and tennis balls are stuck, they both experience a force as the result of an impulse during contact.  But these forces are not isolated, single forces.  As they are being struck, the baseball bat and the tennis racquet are also experiencing equal but opposite forces, respectively, as the result of an impulse during contact.  This can be described by the statement, "The momentum gained by one object, when it makes contact with a second object, is lost by the other object."

Law of Conservation of Momentum:  When studying conservation laws, such as the law of conservation of momentum, the term isolated system is used.  An "isolated system" is a system that only experiences internal changes, but experiences no changes as a whole due to external net (unbalanced) forces from without.  A system can be very complex involving a large number of objects in a very large space.  It can also be simply two object interacting with one another without anything else interfering with them.  It is in this context that we will look at the conservation of momentum.

The Law of Conservation of Momentum says, “The total momentum of an isolated system does not change.”  What this means is that the initial total momentum of an isolated system before an internal event like a collision between two objects takes place is equal to the final total momentum of that same isolated system after the event.  The total momentum of an isolated system always remains the same.  In a simple system, having only two objects, the momentum that one object loses in a collision is gained by the other object.

Collisions:  There are two categories of simple two object collisions.  One is the elastic collision and the other is the inelastic collision.  Both are explained below.

Elastic CollisionIn this situation one object collides with a second object after which each object moves off separate from the other.  At the moment of collision momentum is transferred between the two objects in such a way that the total momentum of the two together is conserved.  In such a collision we expect velocity to change for each object, but not the mass.  An example of such a collision is seen in both the game of billiards and the game of pool.  Momentum is transferred form one ball to another ball during a collision.

Though many automobile collisions involve only two objects the circumstances of crumple zones and the likelihood of pieces of both cars breaking off make the situation more complex.
 
 

Elastic Collision
Example:  Object 1 (mass is 6 kg), moving along with a velocity of 10 m/s, collides with Object 2 (mass is 3 kg), moving along with a velocity of 5 m/s.  If object 1’s velocity is decreased to 4 m/s by the collision, what will object 2’s velocity be after the collision?

G: Before collision: m(1) = 6 kg, v(1) = 10 m/s, m(2) = 3 kg, v(2) = 5 m/s

After the collision: the masses are the same, but v(1)! = 4 m/s and v(2)! = ?
[Note: the ! symbol means a new value for the same variable]

F: the second object's velocity after the collision

E: p(1)Before + p(2)Before  =  p(1)After + p(2)After   .
and so
m(1) * v(1) + m(2) * v(2) = m(1) * v(1)! + m(2) * v(2)!

S: Step1 6 kg * 10 m/s + 3 kg * 5 m/s  =  6 kg * 4 m/s + 3 kg * v(2)!

To simplify this long equation’s substitution, look at the following steps:

Step2  60 kg m/s + 15 kg m/s  =  24 kg m/s + 3 kg * v(2)!

Step3  75 kg m/s  =  24 kg m/s + 3 kg * v(2)!

Step4  75 kg m/s – 24 kg m/s  =  3 kg * v(2)!

Step5  51 kg m/s  =  3 kg * v(2)!

Step6  51 kg m/s  /  3 kg  =  v(2)!

A: v(2)! = 17 m/s
 

Inelastic Collision: In a two object collision of this type, the two objects remain stuck together after the collision and move along as though they were a single object having a mass larger than either object"s individual mass.  An example of this type of collision is seen in railroad switching yards, where one railcar rolls down a hill onto a siding where another railcar is sitting.  When the rolling car strikes the stationary car the couplers lock together and the two cars roll together as one until friction slows them to a stop or someone applies a brake or a retarder to their wheels.  Another example of an inelastic collision is a dart being shot at a small stationary block of wood.  If the mass of the block of wood is sufficiently small the dart and the block of wood will move as one single object until stopped by friction or some other force.
 

Inelastic Collision
Example:  A metal 0.50 kg glider carrying a 0.25 kg load on a frictionless air track is moving at 2.5 m/s.  This glider encounters a second identical glider carrying no load which is stationary.  After the two gliders collide they stick together because of magnets attached to the ends of the gliders.  What will the final velocity of the two equal as they continue moving while attached with one another?

G: m(1) = 0.5 kg, m(1)(extra load) = 0.25 kg, v(1) = 2.5 m/s, m(2) = 0.50 kg, v(2) = 0 m/s

F: final velocity of two gliders attached together

E: p(1)Before + p(2)Before  =  p(1)After + p(2)After
and so
m(1) * v(1) + m(2) * v(2) = m(1) * v(1)! + m(2) * v(2)!
but since
v(1)! = v(2)!, the equation can be rewritten as
m(1) * v(1) + m(2) * v(2)  =  ( (m(1) + m(2) ) * v!
[where v! is the combined velocity]

S:  (0.5 + 0.25) kg * 2.5 m/s + 0.5 kg * 0 m/s = (0.75 kg + 0.5 kg) * v!

The simplification steps are as follows

Step1  0.75 kg * 2.5 m/s + 0.5 kg * 0 m/s = 1.25 kg * v!

Step2  1.875 kg m/s + 0 kg m/s  =  1.25 kg * v!

Step3  1.875 kg m/s  =  1.25 kg * v!

Step4  1.875 kg m/s  /  1.25 kg  =  v!

Step5  1.5 m/s  =  v!

A:  v!  =  1.5 m/s
 

Action – Reaction Problems:  Often in the study of physics we encounter a situation where two objects at rest are suddenly flung in opposite directions because of some force acting that separates them.  The mechanism producing the motion in opposite directions can be something as simple as a compressed spring or something as complex as gunpowder or rocket fuel.  Two people on skates facing each other on very smooth ice could also be a model of this type of situation.  One or both of them can push themselves apart.  Also, a skater can move, if they throw something away from themselves.  There is an action reaction situation.  Likewise, people wonder how rockets get launched and fly through the air into and through space beyond the earth's atmosphere.  They think that the rocket exhaust pushes against something, but it doesn’t.  Most often there is nothing to push against in the first place.  A rocket moves by the process of pushing out hot exhaust gasses in the direction opposite to the one they want to go.  This results in the rocket going in the opposite direction from the exhaust gas.  The conservation of momentum tells us that the product of the mass and velocity of the hot exhaust gas going in one direction is equal to the product of the mass and velocity of the rocket going in the opposite direction.  An untethered astronaut floating in space could move back towards the space craft by throwing a tool in the direction opposite to the space craft.
 

Action - Reaction
Example:  A 500 kg cannon fires a 1.5 kg cannon ball through the air. The cannon ball leaves the muzzle of the barrel of the cannon with a velocity of 25 m/s.  What is the recoil velocity of the cannon?

G: Before: m(1)(cannon) = 500 kg, v(1) = 0 m/s, m(2)(cannon ball) = 1.5 kg, v(2) = 0 m/s.  [rem: before the cannon is fired the objects are stationary]

After: v(1)! = ?, v(2) = 25 m/s

F: the recoil velocity of the cannon

E: p(1)Before + p(2)Before  =  p(1)After + p(2)After
and so
E: m(1) * v(1) + m(2) * v(2) = m(1) * v(1)! + m(2) * v(2)!
but v(1) = 0 m/s and v(2) = 0 m/s
whcih makes the left side of the equation equal to 0 kg m/s
and thus
0 kg m/s = m(1) * v(1)! + m(2) * v(2)!

S:  500 kg * 0 m/s + 1.5 kg * 0 m/s = 500 kg * v(1)! + 1.5 kg * 25 m/s

Simplification steps

Step 1  0 kg m/s + 0 kg m/s = 500 kg * v(1)! + 37.5 kg m/s

Step 2  0 kg m/s – 37.5 kg m/s = 500 kg v(1)!

Step 3  - 37.5 kg m/s  /  500 kg  = v(1)!

Step 4  - 0.075 m/s  =  v(1)!

A:  v(1)!  = - 0.075 m/s
 


 

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