physics 12 - rcc-jlo. · pdf fileclose to speed of light. we have discussed this briefly in...

Physics 12Unit 4 – Vector Dynamics in Two Dimensions,

Momentum and Energy

1. Review of Newton’s laws

• In physics 11 we have studied the relationship between the motion of an object and the forces acting on it. The science which investigates such a relationship is called dynamics. Newton has formulated three laws for dynamics based on a vast number of experimental observations, and we have explored them for one-dimensional motions in details in the past.

• In this unit, we will revisit this topic, yet our attention will be placed on the motions of objects in two dimensions.

• Let’s review the three famous laws proposed by Newton.

Unit 4 - Vector Dynamics in 2D, Momentum and Energy 2

(1) Newton’s first law of motion

• The first law states that every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it.

• This law implies the existence of a property called inertia., which describes the tendency for an object with mass to maintain its original motion.

• An change on the inertia of an object requires the application of force.


(2) Newton’s second law

• Using the first law, we can understand force as the stress that we have to apply to change one’s inertia.

• The second law describes how force changes the motion of an object. From experiments, people know that:

• (1) Acceleration is directly proportional to the net force on the object

• (2) Acceleration is inversely proportional to the mass of the object


𝑎 ∝ 𝐹𝑛𝑒𝑡

𝑎 ∝1

𝑚

• Combining these, we can build the mathematical expression for the second law:

or more commonly:

• This formula works nicely in classical mechanics in which the object moves at the speed much smaller than the speed of light. (Special relativity will be required if dealing with an object moving at a speed close to speed of light. We have discussed this briefly in Physics 11 already.)


𝑎 =𝐹𝑛𝑒𝑡𝑚

𝐹𝑛𝑒𝑡 = 𝑚𝑎

• We know that the second law is presented more accurately in the following form

• It allows us to deal with a system with varying mass. (We will see this more in next unit.)

• If mass is constant, this expression is reduced to the typical form of the second law.


𝐹𝑛𝑒𝑡 =∆𝑝

∆𝑡=∆(𝑚𝑣)

∆𝑡

𝐹𝑛𝑒𝑡 =∆𝑚

∆𝑡∙ 𝑣 + 𝑚 ∙

∆𝑣

∆𝑡

𝐹𝑛𝑒𝑡 = 𝑚 ∙∆𝑣

∆𝑡= 𝑚𝑎

(3) Newton’s third law

• When we do push-ups, we do not push ourselves directly; indeed we exert the push on the floor. In the meantime, the floor exerts a force that raises us up.


• We know that force must be applied on an object by another object. This leads to the formation of the third law, which states that when one body exerts a force on a second body, the second body exerts an equal force on the first body, but in the opposite direction. The forces are exerted during the same interval of time.

• Symbolically,

• The force 𝐹𝐴𝐵 is the action on B by A, while 𝐹𝐵𝐴 is the reaction of B on A.


𝐹𝐴𝐵 = −𝐹𝐵𝐴

• The Newton’s laws tell us how a system responds to the forces applied to it, and they can be used for various types of forces. In Physics 11 and 12, we have encountered and will encounter the following forces:


• Example: A 110 kg motorbike carrying a 50-kg rider coasts to a stop in a distance of 51 m. It was originally travelling 15 m/s. What was the stopping force exerted by the road on the motorbike and rider?

[-350 N]


• Example: A truck of mass 2.00 × 103 kg is towing a large boulder of mass 5.00 × 102 kg using a chain of negligible mass. The tension on the chain is 3.00 × 103 N, and the force of friction on the boulder is 1.20 × 103 N.

(a) At what rate will the boulder accelerate? [3.60 m/s2]

(b) How far will the boulder move in 3.0 s, starting from rest? [16 m]


• Example: In the diagram below, a cord of negligible mass connects a 0.500 kg mass to a 1.00 kg mass. Friction in the system is negligible.

• (a) At what rate will the masses accelerate?

• (b) What is the tension in the cord while the masses are accelerating?


• Example: You can bold a box of 5 kg against a rough wall, with a static coefficient of friction 0.40, and prevent it from slipping down by pressing hard horizontally. How much horizontal force will keep an object from moving vertically?


2. Pulley systems• Pulley is a simple machine that supports the movement of heavy

objects. It consists of a wheel (or wheels) on an axle and a rope (or cable).


• Pulley systems provide a mechanical advantage in a movement. Mechanical advantage (MA) is a measure of the force amplificationachieved by using a tool or a machine.

• The MA is defined as

• Using static equilibrium:


MA =load

effort

MA =effort arm

resistance arm

• Example: A mover is trying to lift a piano slowly up to a second-story apartment. He is using a rope looped over two pulleys as shown. What force must he exert on the rope to slowly lift the piano whose weight is 1600 N? What is the mechanical advantage of this pulley machine?


• Example: The weight of the 2.0 kg mass exerts a force on the system causing both masses to move. Given the information, find the acceleration of the system and the tension in the string.


• Example: Two hanging masses are attached to one horizontal mass. Note that the two tensions are not the same. What is the net force on the system? What is the acceleration of the system? What is the tension in each rope?


• Example: Given the information, find the unknown mass of the cart.


3. Newton’s laws involving inclines• The calculation involving an object sliding on an inclined plane is

more complicated because, in this scenario, the acceleration due to gravity is no longer vertical to the plane. The strategy of resolving vectors into components is therefore necessary to solve the corresponding dynamic problem.


• Example: The skier in the diagram below has begun descending the 30° slope. If the coefficient of kinetic friction is 0.10, what is her acceleration?


• Solution: The corresponding free body diagram is as shown:

• The normal force is

• The friction is

• Therefore,


𝐹𝑁 = 𝐹𝐺𝑦 = 𝑚𝑔 cos 𝜃

𝐹𝑓𝑟 = 𝜇𝐹𝑁 = 𝜇𝑚𝑔 cos 𝜃

𝑚𝑎𝑥 = 𝐹𝐺𝑥 − 𝐹𝑓𝑟 = 𝑚𝑔 sin 𝜃 − 𝜇𝑚𝑔 cos 𝜃

𝑎𝑥 = (9.8)(sin 30°) − (0.10)(9.8)(cos 30°) = 4.0 m/s2

• Example: In the diagram below, a 4.0 kg mass rests on a 30°frictionless slope and is pulled by a 3.0 kg mass connected to it over a pulley by a cord. What is the acceleration of the system and the tension in the cord?


• Example: Find the acceleration of the system and the tension on the cord.


• Example: (Challenging!) Two masses 𝑚𝐴 = 2.0 kg and 𝑚𝐵 = 5.0 kg are on inclines and are connected together by a string as shown below. The coefficient of kinetic friction between each mass and its incline is 0.30. Determine their acceleration.


4. Momentum and impulse

• Recall that the Newton’s second law was originally postulated in terms of a quantity (mass) × (velocity). This quantity is a measure of motion of an object; it is now commonly known as momentum.

• Momentum is a vector quantity denoted by a symbol 𝒑:


𝒑 = 𝑚𝒗

• The Newton’s second law can be represented in terms of momentum:

• The Newton’s second law can be stated in words, in terms of momentum, as:


𝐹 = 𝑚𝑎 = 𝑚∆𝑣

∆𝑡=𝑚(𝑣𝑓 − 𝑣𝑖)

∆𝑡=𝑚𝑣𝑓 −𝑚𝑣𝑖

∆𝑡=∆(𝑚𝑣)

∆𝑡=∆𝑝

∆𝑡

• The second law formula can be rearranged in the following way:

• The product of the net force and the time interval during which it is applied is called the impulse of the force, 𝑱.

• Mathematically:

• It means that impulse of the force is the change in momentum it causes.


𝐹∆𝑡 = ∆ 𝑚𝑣

𝑱 = 𝑭∆𝑡 = ∆𝒑

• For an isolated system, the total momentum of the two objects colliding one another is a conserved quantity. This relationship is called the conservation of momentum.

• Using the Newton’s third law, we know that when two objects, A and B, collide,

• That means

• Therefore


𝐹𝐴𝐵 = −𝐹𝐵𝐴

∆(𝑚𝑣𝐵)

∆𝑡= −

∆(𝑚𝑣𝐴)

∆𝑡

∆𝑝𝐵 = −∆𝑝𝐴

∆𝑝𝐴 + ∆𝑝𝐵 = 0

• Example: A rifle bullet of mass 0.060 kg leaves the muzzle with a velocity of 600 m/s. If the 3.0 kg rifle is held very loosely, with what velocity will it recoil when the bullet is fired?

[-12 m/s]


• Example: A 6000 kg railway car is coasting along the track at 7.0 m/s. Suddenly a 2000 kg load of coal is dumped into the car. What is its new velocity?


• Example: A 1200 kg car traveling 33 m/s collides head-on with a 1800 kg car traveling 22 m/s in the opposite direction. If the cars stick together, what is the velocity of the wreckage immediately after impact?


• We have studied in Physics 11 about the conservation of momentum for motions in 1D. In fact, the same principle can be applied to collisions in even 2D or 3D.

• However, since momentum is a vector quantity, it is the sum of the momentum vectors which is conserved before and after the collision, not just the magnitudes.

• Mathematically,


𝒑1,𝑖 + 𝒑2,𝑖 +⋯ = 𝒑1,𝑓 + 𝒑2,𝑓 +⋯

𝑚1𝒗1,𝑖 +𝑚2𝒗2,𝑖 +⋯ = 𝑚1𝒗1,𝑓 +𝑚2𝒗2,𝑓 +⋯

Initial momentum vector sum Final momentum vector sum

• There are three common scenarios in which the conservation of momentum is considered.

(1) Explosion

• Before the explosion, the momentum of the stationary object is zero. Hence, the vector sum of the momenta after explosion has to be zero.


• Example: An object, at rest, explodes into three pieces, each traveling parallel to the ground. The first piece has a mass of 3.0 kg and travels at 4.0 m/s 30° N of E. The second piece has a mass of 4.0 kg and travels at 3.0 m/s 30° S of E. Find the velocity of the third piece if its mass is 5.0 kg.


(2) Oblique collision, two particles sticking together

• If two masses collide and stick together, the resulting momentum would be equal to the sum of the individual initial momentum vectors of the two masses. That means,


𝒑1 + 𝒑2 = 𝒑𝑓

• Example: A 100 kg football player going 3.0 m/s North, tackles another player of mass 150 kg going 1.5 m/s East. The players entangle. What is their combined speed and direction?


(3) Oblique collision, moving mass strikes a stationary mass

• The moving object before collision gives rise to the final momenta of the two objects. Therefore,


𝒑1 = 𝒑′1 + 𝒑′2

• Example: A 2.0 kg ball going 5.0 m/s strikes a stationary 4.0 kg ball. After the collision, the second ball goes off at 1.11 m/s at 54° from the direction of the original ball. What is the velocity of the first ball?


• The total momentum of an isolated system remains the same in a collision without external force due to the conservation of momentum. In order for the momentum of a system to be changed, an impulse has to be applied.

• In two dimension, it means

• What does this relationship imply in the following situation?


𝑱 = ∆𝒑 = 𝒑𝑓 − 𝒑𝑖 = 𝒑𝑓 + −𝒑𝑖

• Using vector addition, we know that

• That implies there exists a perpendicular force exerted by the surface to the ball. This force is simply the normal force due to the surface! (Recall the Newton’s third law!)


𝑭 =∆𝒑

∆𝑡

• Example: A 2.0 kg ball going 10 m/s bounces off a wall at an angle of 40° to the wall. (Both incoming and outgoing angles are 40°). After the bounce the speed is still 10 m/s.

(a) What is the impulse on the ball?

(b) What is the change in velocity?


5. Work and energy

• When an object is subjected to an applied force and is displaced, we say that work is done on it.

• Work can be defined as the product of the force component in the direction of motion and the displacement:

• The unit of work done is Joule (J), or Newton·meter (Nm).


𝑊 = 𝐹⫽𝑑

• In one dimension, the work done definition is simply:

because 𝐹 and 𝑑 are always parallel.

• In two dimensions, however, we have to consider the component of the force parallel to the displacement of the object. If the angle between them is 𝜃, then the formula becomes:


𝑊 = 𝐹𝑑

𝑊 = 𝐹𝑑 cos 𝜃

• Example: How much work does the man do to lift the 150 N of firewood from the ground up to a height of 1.2 m?


• Example: How much work does the hiker do on the 90.0 N backpack she carries to the top of the 850 m high hill?


• Example: A child is pulling a wagon along a driveway, exerting a force along the handle of the wagon. If the force exerted on the handle is 45 N, and the angle 𝜃 is 28°, how much work will be done pulling the wagon 17 m along the driveway?


• A question to think about: A ball on the end of a rope is following a circular path because of a constant force exerted on the ball by the rope in a direction toward the center of the circle. How much work is done on the ball by this force when it completes one rotation?


• When the force applied to an object is not constant over the distance it moves, the work done can be calculated in two ways:

(1) Force-distance diagram

• The area under the force-distance diagram is the total work done on the object.


• Example: For a spring with a force constant of 12 N/m, the total work done by stretching it for 0.5 m is:

𝑊 =1

20.5 × 6 = 1.5 J

(2) Calculus (Optional)

• The work done on an object by a varying force can be determined by means of integration.

• Mathematically, if a force 𝑭(𝑥) is acting on an object so that it moves from 𝑥 = 𝑎 to 𝑥 = 𝑏, then the work done will be given by

• If the form of 𝐹(𝑥) is known, this integral can be evaluated directly.

• If 𝐹(𝑥) is a constant, this formula will reduce to the typical work done formula, 𝑊 = 𝐹𝑑.


𝑊 = න𝑎

𝑏

𝐹 𝑥 𝑑𝑥

• Example: A girl is pushing a laboratory supply cart down the hallway of a school. She pushes in a direction parallel with the floor. On a linoleum floor, she exerts a force of 30.0 N over a distance of 10.0 m. She then has to push the cart over a carpeted floor for a distance of 10.0 m, and this requires a force of 50.0 N. Finally, she must push it another 7.5 m on another stretch of linoleum floor, again with a force of 30.0 N. How much work does she do on the cart during the whole trip?


• Example: How much work would be done moving an object from a distance of 1.0 m to a distance of 5.0 m subjected to a gravitational attraction depicted below?


• There are three common types of work that can be done on objects.

(1) Work done against gravity

• When an object is lifted upward, work is done on a mass against the resisting force of gravity.

• The energy used to do this is converted to the gravitational potential energy, 𝐸𝑝, of the object.

• Since 𝐹𝑎𝑝𝑝 = 𝐹𝑔 = 𝑚𝑔 and 𝑑 = ∆ℎ,

• The term 𝐸𝑝 = 𝑚𝑔ℎ is called potential energy.


𝑊 = 𝐹𝑎𝑝𝑝𝑑

𝑊 = 𝑚𝑔∆ℎ = ∆𝐸𝑝

(2) Work done against inertia

• When a force is applied on a moving object, it is accelerated, and its motion is no longer the same as before.

• Consider an unbalanced force acts on a ball so that its speed changes from 𝑣𝑖 to 𝑣𝑓.

• Since 𝐹𝑛𝑒𝑡 = 𝑚𝑎,


𝑊 = 𝐹𝑑 = 𝐹𝑛𝑒𝑡𝑑 = 𝑚𝑎𝑑

• From kinematics, we know that

• Substituting these back to the previous equation gives

• The expression, 𝐸𝑘 =1

2𝑚𝑣2, is called kinetic energy, and is the

energy stored as the energy of speed.

• This is the net work done that can be used!


𝑎 =𝑣𝑓 − 𝑣𝑖

𝑡𝑑 =

𝑣𝑓 + 𝑣𝑖

2𝑡

𝑊 = 𝑚𝑣𝑓 − 𝑣𝑖

𝑡

𝑣𝑓 + 𝑣𝑖

2𝑡 =

1

2𝑚 𝑣𝑓

2 − 𝑣𝑖2 = ∆𝐸𝑘

(3) Work done against friction

• As long as an object moves along a horizontal surface with a constantvelocity, work is done against friction.

• In this example, when the mass is moved from #1 to #2 at a constant velocity, the applied force is used to overcome the friction.


• Therefore, the work done by the applied force is

• On a horizontal surface with a coefficient of kinetic friction of 𝜇𝑘,

• If the surface is not horizontal, then an appropriate expression of 𝐹𝑁should be used.

• Note that this work done is wasted as heat (or sound) and lost to the system. In other words, this work done is not useful.


𝑊 = 𝐹𝑎𝑝𝑝𝑑 = 𝐹𝑓𝑑

𝑊 = 𝜇𝑘𝐹𝑁𝑑 = 𝜇𝑘𝑚𝑔𝑑

• In summary, when work is done on an object, it is transformed according to the following relationship:


𝑊 = ∆𝐸𝑝 + ∆𝐸𝑘 +𝑊𝑓

Change in gravitational potential energy

Change in kinetic energy

Heat lost due to friction

∆𝐸𝑝 = 𝑚𝑔∆ℎ∆𝐸𝑘 =

1

2𝑚𝑣𝑓

2 −1

2𝑚𝑣𝑖

2

𝑊𝑓 = 𝜇𝐹𝑁𝑑

• Example: A force of 100 N is applied on a 50 kg cart that is moving with a speed of 6.0 m/s and has a force of friction of 20. N acting on it. At the end of 10 seconds, the cart is going 22 m/s.

(a) How much work was done against inertia?

(b) How much work was done in total?


6. Power• Power is a quantity in physics that measures the rate of doing work

on an object. Since work done refers to the amount of energy input to the object, power is a measure of how fast energy is transferred.

• Symbolically,

• By definition, the unit of power is J/s. A more popular unit for power is Watts (or W). These two units are related by


𝑃 =𝑊

𝑡=∆𝐸

𝑡

1J

s= 1W

• Recall work can be done against gravity, inertia or friction; therefore, power is required to do work.

• In general,

• More precisely,


𝑃 =∆𝐸𝑝 + ∆𝐸𝑘 + 𝐹𝑓𝑑

𝑡

𝑃 =𝑚𝑔∆ℎ +

12𝑚 𝑣𝑓

2 − 𝑣𝑖2 + 𝜇𝐹𝑁𝑑

𝑡

• Example: For the diagram below, if the cart goes from the bottom to the top in 16 seconds, how much power was developed?


• Example: The car below has an initial speed of 2.0 m/s and accelerates to 5.0 m/s by the time it is at the top of the ramp. How much power is developed?


• Example: Here there is a 22° ramp. The cart starts from rest at the bottom of the slope and accelerates to 4.0 m/s by the time it reaches the top of the ramp. With a coefficient of friction 𝜇 = 0.21, how much power was developed?


• In certain situations, the calculation of power developed can be simplified if the velocity of an object is known.

• Recall that 𝑊 = 𝐹𝑑. Substituting this into the power equation gives

• In kinematics, 𝑣𝑎𝑣𝑔 = 𝑑/𝑡. Hence,


𝑃 =𝑊

𝑡=𝐹𝑑

𝑡= 𝐹

𝑑

𝑡

𝑃 = 𝐹𝑣𝑎𝑣𝑔

• Example: A motor driven sled of mass 10.0 kg moves at a constant speed of 15 m/s over a horizontal surface of coefficient of friction 0.12. What power would the motor have to develop to cause this to happen?


7. Conservation of energy

• We have seen in previous science classes that energy is a substance that can be neither created or destroyed in a process; yet it can be transformed from one form to another.

• It means that the total energy contained in a system for an event is constant regardless of how energy is converted between different forms.

• In Physics 12, we only consider three forms of energy:

(i) Potential energy 𝐸𝑝

(ii) Kinetic energy 𝐸𝑘(iii) Heat loss due to friction 𝑊𝑓 = 𝐹𝑓𝑑


Scenario 1: Friction is negligible

• In this case, since the energy loss is so small compared to potential energy and kinetic energy, we simply ignore it.

• Therefore, the change in the potential energy of the system will be equal to the change in its kinetic energy.

• Symbolically,

• More precisely,


−∆𝐸𝑝 = ∆𝐸𝑘

−𝑚𝑔 ℎ𝑓 − ℎ𝑖 =1

2𝑚 𝑣𝑓

2 − 𝑣𝑖2

• Example: In the diagram below, if the ball has a mass of 5.2 kg and an initial speed of 1.4 m/s at the top of the 2.8-m high ramp, what will its speed be at the bottom of the ramp?


• Example: If a cart of mass 10 kg and with an initial speed of 3.5 m/s rolls down a 50-m high frictionless incline and then proceeds to roll up another similar incline to a height of 20 m, what is the speed of the cart at this point?


• Example: A pendulum bob of mass 5.0 kg falls through a height of 25 cm as it swings from maximum height to lowest position.

(a) How fast is it going at the bottom?

(b) What is the energy of the bob at the bottom of the swing?

(c) What is the speed of the bob as it swings up past the bottom of its arc and rises 10 cm from the bottom position?

(d) What is the total energy at this position?


Scenario 2: Friction is not negligible

• The scenario 1 is an ideal case; in reality, friction exists everywhere; therefore heat and other forms of energy are produced. This energy loss must be considered when we apply the theorem of the conservation of energy.

• Hence,


𝐸𝑝,𝑖 + 𝐸𝑘,𝑖 = 𝐸𝑝,𝑓 + 𝐸𝑘,𝑓 + heat

Total energy before the event

Total energy after the event

• Example: Consider the diagram below showing a 60-kg student on a large slide.

(a) In the absence of friction, what would her speed be at the bottom?

(b) If her actual speed at the bottom is 6.0 m/s, how much heat was generated on the section shown?


• Example: An object of mass 12 kg starts from rest and slides down a ramp that has a vertical drop of 6.0 m. Heat generated as the object moves down the ramp is 310 J.

• (a) How fast will the object be going at the bottom of the ramp?

• (b) If the object then slides along a horizontal surface of 𝜇 = 0.25, how far will it travel before coming to a rest?


8. Efficiency

• Due to the presence of friction, the total energy input of a system is not completely converted to kinetic energy and / or potential energy (called useful energy) in an event.

• The percentage of energy input that is transferred into useful energy is called the efficiency.


Efficiency =useful energy transferred

energy input× 100%

• Example: How much useful work can a 30 W motor due in 1 minute if it is 45% efficient?


• Example: A 150 W electric motor takes 10 seconds to lift a 10 kg mass up 7 m. What is the efficiency of the motor?


• Depending on the efficiency of energy transfer, collisions can be classified into three types:

(1) Elastic collision

• In an elastic collision, both momentum and kinetic energy are conserved.

• Mathematically, it means both the equations below must be fulfilled simultaneously


𝑚𝐴𝑣𝐴 +𝑚𝐵𝑣𝐵 = 𝑚𝐴𝑣′𝐴 +𝑚𝐵𝑣′𝐵

1

2𝑚𝐴𝑣𝐴

2 +1

2𝑚𝐵𝑣𝐵

2 =1

2𝑚𝐴𝑣′𝐴

2 +1

2𝑚𝐵𝑣′𝐵

2

• Example: Billiard ball A of mass 𝑚 moving with speed 𝑣𝐴 collides head-on with ball B of equal mass. What are the speeds of the two balls after the collision, assuming it is elastic? Assume ball B is moving initially at 𝑣𝐵.

• Solution: According to the conservation of momentum,

• Since the collision is elastic, energy is conserved. Hence


𝑚𝑣𝐴 +𝑚𝑣𝐵 = 𝑚𝑣𝐴′ +𝑚𝑣𝐵

′

1

2𝑚𝑣𝐴

2 +1

2𝑚𝑣𝐵

2 =1

2𝑚𝑣′𝐴

2 +1

2𝑚𝑣′𝐵

2

• Rewriting these two equations:

• Divide the second equation by the first one to yield the following:

• Adding this equation with the first one we obtain:

• Similarly, we can show


𝑣𝐴 − 𝑣𝐴′ = 𝑣𝐵

′ − 𝑣𝐵

𝑣𝐴2 − 𝑣′𝐴

2= 𝑣′𝐵

2− 𝑣𝐵

2

𝑣𝐴 + 𝑣′𝐴 = 𝑣′𝐵 + 𝑣𝐵

𝑣𝐴 = 𝑣𝐵′

𝑣𝐵 = 𝑣𝐴′

• It means that the balls exchange their velocities as a result of the collision.

• What does it implies?

• If ball B is initially at rest, then during the collision, ball B will acquire the original velocity of ball A, while ball A will come to rest.


• Example: A proton of mass 1.01 u traveling with a speed of 3.60 × 104

m/s has an elastic head-on collision with a helium nucleus with a mass of 4.00 u initially at rest. What are the velocities of the proton and helium nucleus after the collision?


(2) Inelastic collision

• If the momentum of a system is conserved while its kinetic energy isnot conserved in a collision, the collision is said to be an inelastic collision.

• Although the kinetic energy is not conserved, the total energy of the system is always conserved.

• In this scenario, we only consider


𝑚𝐴𝑣𝐴 +𝑚𝐵𝑣𝐵 = 𝑚𝐴𝑣𝐴′ +𝑚𝐵𝑣𝐵

′

• Example: A 2.0 kg ball collides at 10. m/s with a much larger stationary 8.5 kg ball as shown. After the collision, the 2.0 kg ball changes its speed to 7.6 m/s at 40° N of E.

(a) At what speed does the 8.5 kg ball move after the collision?

(b) What is the efficiency of this system?


(3) Completely inelastic collision

• If two objects stick together as a result of a collision, the collision is said to be completely inelastic.

• In this type of collision, the kinetic energy is maximally transformedinto other forms of energy consistent with the momentum conservation.

• Symbolically it means that


𝑚𝐴𝑣𝐴 +𝑚𝐵𝑣𝐵 = 𝑚𝐴 +𝑚𝐵 𝑣𝑓

• An illustrative example of completely inelastic collision is a ballistic pendulum, which is a device used to measure the velocity of a projectile.


• The motion of a ballistic pendulum can be divided into two parts.

• In the first part, a bullet collides with the wooden block and getembedded in it. It is assumed that the contact time is so short that the block is essentially stationary before the bullet is fully embedded. Therefore, there exists no external force, and the momentum of the system is conserved.

• Hence

• This part is a typical inelastic collision.


𝑚𝑣 + 0 = 𝑚 +𝑀 𝑣′

𝑣 =𝑚 +𝑀

𝑚∙ 𝑣′

• In the second part, the block with the bullet starts to swing up from the bottom. In this case, since gravity is acting on the system, the momentum of the system is not conserved.

• However, the mechanical energy of the system is still conserved. Thus,

• It gives

• Using this the following working equation for a ballistic pendulum can be obtained:


1

2𝑚 +𝑀 𝑣′2 = 𝑚 +𝑀 𝑔ℎ

𝑣′ = 2𝑔ℎ

𝑣 =𝑚 +𝑀

𝑚2𝑔ℎ

• Example: A 0.015 kg bullet is fired horizontally into a 3.0 kg block of wood suspended by a long cord. The bullet sticks in the block. Compute the original velocity of the bullet if the impact causes the block to swing 10 cm above its initial level. Also, calculate the efficiency of this system.


physics 12 - rcc-jlo. · pdf fileclose to speed of light. we have discussed this briefly in...

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