NEWTON'S THIRD LAW states that:

For every action (a force) there is an equal (same magnitude) and opposite reaction. Therefore, if an object exerts a force on a second object, the second exerts an equal an oppositely directed force on the first one.

 1) In the movie "Matrix" (or any other action movie) the good guy punches the bad guy who goes flying away. Of course the good guy stays steady. What it wrong with that?

2) A) If you stand on a skateboard and push on the wall, the wall pushes back on you and you move back. Why the wall is not moving if action = reaction ? (remember ma = F = ma ).
B) If you stand on the ground (not on a skateboard) and you push the wall, why don't you move back ? What is the other pair of forces involved ?

3) A) How Newton's third law can explain the walk ? If you are standing on a small boat (not tied to the dock), what will happen when you walk from the boat to the dock ? 
B) How Newton's third law can explain the motion of a rocket ? Can a rocket still move in a vacuum (space) ? Is it easier for a ship to move in space than  in the Earth's atmosphere ? Why?
C) How Newton's third law can explain your rebound on a trampoline ? What happen to your shoulder if you shoot using a riffle ? To spare your shoulder, is it better to use a heavy riffle or a light one? REmember:  ma = F = ma
How could you avoid being hurt by a riffle ?
D) You are traveling in a bus at highway speed on a nice summer day and an unlucky bug splatters on the front window. According to Newton's third law, Compared to the force that acts of the bug , how much force acts of the bus ? Which undergoes the greater acceleration ? Which therefore suffer the greater damage ? (ma = F = ma)

watch that very short movie (cartoon on Newton's third law, U-tube)

4) AN astronaut in space (no friction acting on his feet) pushes the space ship. WHat happens to him ? to the space ship ?

astronauts should never forget Newton's law. In space there is no friction. So, according to Newton's first law, if something moves, it will keep moving for ever, in the same direction and the same speed. If you want to push something, it will push back on you and it is hard to stop has no gravity brings you back to solid surface and no friction stops you. If you want to grabs something and pulls , it will pull back to you and it will be very chaotic.

See this movie to understand. 

WORKSHEET: REVIEW                    EXTRA CREDITS !!

Impulse and Momentum

Newton wrote his 2nd law as: ∆V = F t / m   If you apply a force F, during a time t, on an object of mass m, the velocity will increase by ∆V. We can rewrite the equation as: F t = m  ∆V
This equation tells you that if you want to change the velocity of an object (stop it, get it to move, speed it up or slow it down),
you can apply a small constant force during a long time, or you can apply a large force during a short time. It has many consequences in every day phenomena.


The product F t is called the impulse.   m  ∆V is the change in momentum ∆p.  
p is the momentum of the object: p = mV or ∆p = m∆V.
The momentum p = m V is a measure of how hard it is to stop an object of mass m.

4) p is a vector. p has the same direction as the  _________________ (p = m V and m is a scalar)

so F t = ∆p    Applying a force on an object, during the time t, change the momentum by ∆p = m∆V.
OR If , during a collision, an object, undergoes a change of momentum ∆p =  m∆V it will apply an impulse = Ft on the second object.

5) So if a car of mass m, with an initial velocity V,  crashes into a wall and comes to a stop (final V =0)
you have ∆V = ____  ,the car  will apply a force F on the wall during the time t such as:  F t = ______ . If the time t is really small (does not takes long for the wall to stop the car), the force F is really large and so are the damages (the wall pushes back on the car with the same force and during the same time, that is with the same impulse). We have: F t = m V. THink. How could you stop a car without damaging  it ? (hint: you want to minimize F but the change in momentum is the same   F t = m V)
Why an air bag is an application of Newton's third law ? Why an egg will crash on concrete but not on a spread cloth ?

 Write the equations :∆p = F t  or  m ∆V = F t on index cards and try to solve:

6) Why a small mass launched by a sling shot can go really fast (why ∆V is large) ?
 hint: the force applied by the elastic is small but ___________________

7) Boxers know that, if he can not avoid being hit, he needs to "ride or roll back with the punch ". This strategy lessens the force of impact because it increases _________________.

8) You can just do the opposite. How do you think a karate expert can break a stack of brick? For a given momentum m ∆V the karate expert wants to get the greater force possible. (the force will break the stack)  So, that time, she/he needs to lessen _____________. If her hand is made to bound upon impact, the force is ever greater. why ?


9) When you jump, you don't keep the legs straight and stiff (ouch !). Why ?


10) A force of 20N acts on a 2.0 kg mass for 10s, compute  A) the impulse B) the change in velocity of the mass.

11) A car that weight 7840N is accelerated from rest to a velocity of 25.0m/s eastward by a force of 1,000N. A) what was the car's change in momentum ? B) how long the force act to change the car's momentum?


12) What force is needed to bring a 1.10 10³  kg car moving at 22.0 m/s to a halt in 20.0 s?

13) A net force of 2.00 10³ N acts on a rocket of mass 1.00 10³ kg. How long does it take this force to increase the rocket's velocity from 0.0m/s to 2.00 10² m/s?


14) A car weighting 15 680N and moving at 20.0 m/s is acted upon by a 6.40 10² N force until it is brought to a halt.
A) What is the car mass ?
B) What is its initial momentum ?
C) what is the change in momentum ?
D) How long does the braking force act on the car to bring it to a halt?

15) THe velocity of a 6.00 10² mass is changed from 10.0m/s to 44.0 m/s in 68.0s by an applied, constant force.
A) what change in momentum does the force produce ?
B) what is the magnitude of the force ?

16) What is the final velocity of a rocket of mass 2.0 10⁴ kg, starting from rest, if a net force of 1.5 10⁵ N acts upon it for 15.0s ?




LAW OF CONSERVATION OF MOMENTUM

This law is a consequence of Newton's third law. Let's say 2 objects interact with each other: 2 cars colliding, a car (object A) crashing in to a wall (object B), a tennis ball (object A) bouncing off a racket (object B) , a trampoline (object A) bouncing you (object B), a riffle (object A) firing a bullet (object B), a rocket (object A) pushing out the gas (object B), you best friend pushing you on an ice rink  ....

Force_{(object A on object B)} = - Force_{(object B on object A)}         or F_{A on B} = - F_{B on A}

 the negative sign indicates that the action and the reaction have same magnitude but opposite direction. REmember a force is a vector/

F_{A on B} = - F_{B on A    or}   F_{A on B} t = - F_{B on A} t              (same time t for the interaction)

so ∆pA = - ∆pB  

This is the law of conservation of momentum.
It only holds when a system of 2 bodies interacting is isolated. No external force are acting on them.

lab: watch the movie (U-tube)

another movie to watch (advanced)


The law of conservation of momentum can be stated as:

THE FINAL MOMENTUM OF THE SYSTEM IS EQUAL TO THE INITIAL MOMENTUM OF THE SYSTEM.
or MOMENTUM before collision = MOMENTUM after the collision

(PA + PB ) before = (PA' + PB') after

or mA VA + mB VB = mA VA'+ mB VB'

STOP ! did you see the extra credits work ? (+5%on your last test)

17)



2 freight cars A and B of equal masses (mÁ=mB = 3.0 10⁵ kg) are on a track. The car A is moving slowly at VA= 2.2 m/s. Car B is at rest.(VB= 0 ) The 2 cars collide and are coupled together. Use the conservation of momentum to predict the resulting velocity of the 2 cars (locked together) after the interaction.  Follow the steps:
A) Fill the table:

before	after
VA =	VA'= VB'= V' = ?
VB =	VB'= VA'= V' = ?
PA = mAVA =	PA'= mA V' = 300, 000 V'
PB = mB VB =	:PB'= mB V' = 300,000 V'
PA + PB =	PA'+ PB'= (mA +mB) V'= 600,000 V'

B) write: PA+ PB = PA'+ PB'  and solve for V'.


18) A steel glider of mass 0.50 kg (mA) moves along an air track with a velocity of 0.75m/s (VA) . It collides with a second steel glider of mass 1.0 kg (mB) moving in the same direction at a speed of 0.38 m/s (VB). After the collision, the first glider continues with a velocity of 0.35 m/s (VA') what is the velocity of the second glider after the collision. Follow the same strategy as before.

Fill the table:

before	after
VA =	VA'= 0.35 m/s
VB =	VB'= ?
PA = mAVA =	PA'= mA VA' =
PB = mB VB =	:PB'= mB VB' =
PA + PB =	PA'+ PB'=

PA + PB = PA'+ PB'  and solve for VB' .


19) A)
Moving at 20.0 m/s (VA) a car of mass 7.00 10² kg (mA) collides with a stationary (VB=0)  truck of mass 1.4 10³ kg (mB) . If the two vehicles interlock as a result of the collision, what is the velocity of the car-truck system (V')?

i) Fill the table:

before	after
VA =20m/s	VA'= V'= ?
VB = 0m/s	VB'= V'=?
PA = mAVA =	PA'= mA V' =
PB = mB VB =	:PB'= mB V' =
PA + PB =	PA'+ PB'= (mA + mB) V'=

ii) PA + PB = PA'+ PB'  solve for V'

B) A 2275 kg (mA)  car going 28m/s (VA = 28m/s) rear-ends an 875kg (mB) compact car going 16m/s (VB) on ice in the same direction. The 2 cars stick together (VA'= VB'= V'). How fast does the wreckage move immediatly after the collision?


20) A Bullet of mass 50.0g (mA, conversion !) strikes a wooden block of mass 5.0 kg (mB = 5kg, VB = 0 m/s) and becomes embedded in the block.
THe block and bullet then flies off at 10m/s (VA' = VB' = 10m/s). What is the original velocity of the bullet ?
                                        

21) A plastic ball of mass 0.200 kg (mA) moves with a velocity of 0.30 m/s (VA) . THis plastic ball collides  with a second plastic ball of mass 0.100 kg (mB)  that is moving along the same line at a velocity of 0.10m/s (VB) . After the collision, the velocity of the 0.100 kg ball is 0.26m/s (VB'). What is the velocity of the second ball (solve for VA').


22) Granny (object A) is skate boarding in a ring at a speed of VA= 3m/s when suddenly she is confronted to Alfred (object B) , at rest (VB = 0), directly in her path. Rather then knocking him over, she picks him up and continue in motion without braking. (they now have the same velocity VA'= VB'= V   like for problem 17). The mass of granny is 80kg (mA) , the mass of Alfred is 40 kg (mB).
A) using the law of conservation of momentum find the speed of GRanny and Alfred together after the collision. Solve for V.
B) After the collision, does GRanny's speed decreases ? increases ? (no friction)
C) After the collision, does Alfred's speed decreases ? increases ? (no friction)

23)  2 skaters are standing still on ice. Skater A has a mass of 60.0kg and skater B has a mass of 30.0kg. We neglect frictions so there is no external force, all the forces are internal and we can use the law of conservation of momentum.
A) Because the are standing still VA = VB= 0,  initial momentum of the system =  _________.
B) Skater A pushes skater B. as a result, skater A moves back (left)  at a velocity of VA'= - 0.2m/s. The negative sign indicates that the skater moves backward, toward the left. Can you find the velocity VB'of skater B ? Follow the steps:

Fill the table:

before	after
VA = 0	VA'= - 0.2 m/s
VB =0	VB'= ?
PA = mAVA =	PA'= mA VA' =
PB = mB VB =	:PB'= mB VB' =
PA + PB =	PA'+ PB'=

PA + PB = PA'+ PB'  and solve for VB' .                                                      

24) What is the recoil velocity of a 1.20 10³ kg (mA)  launcher if it projects a 20.0kg (mB)  mass at a velocity of VB'= 6.00 10² m/s Fill the table:

before	after
VA =0	VA'= ?
VB =0	VB'=  6.00 102 m/s
PA = mAVA =	PA'= mA VA' =
PB = mB VB =	:PB'= mB VB' =
PA + PB =	PA'+ PB'=

PA + PB = PA'+ PB'  and solve for VA' . Expect a negative sign.

25) Upon launching, a 4.0 kg (mA)  model rocket expels 50.0g (mB, conversion) of oxidized fuel from its exhaust at an average velocity if 6.00 10²m/s (negative because the gas moves down) . What is the vertical velocity of the model rocket after the launch ? (disregard gravity). !!! convert g to kg.

Fill the table:

before	after
VA =0	VA'= ?
VB =0	VB'=  - 6.00 102 m/s
PA = mAVA =	PA'= mA VA' =
PB = mB VB =	:PB'= mB VB' =
PA + PB =	PA'+ PB'=

PA + PB = PA'+ PB'  and solve for VA' .

26) 2 campers dock a canoe. One camper steps onto the dock. THis camper has a mass of 80.0 kg (mA)  and moves forward at 4.0m/s VA' positive) . With what speed and direction will the canoe and the other camper move if their combined mass is 110 kg (mB) ? Make the table as before. You are solving for VB'. 

RECOIL PROBLEM

27) A thread holds 2 carts together on a frictionless surface. A compressed spring act on both carts. After the thread is burned , the 1.5kg mass (mA) cart moves back with a velocity of 27cm/s to the left. What is the velocity of the 4.5kg cart ?

hint: total mometum before = total momentum after. The total momentum before = 0 and a car moving to the left has a negative velocity

28) An astronaut at rest in space fires a thruster pistol 35 g of hot gas at 875m/s. The combined mass of the astronaut and the pistol
is 84kg. How fast and in what direction is the astronant moving after firing the pistol ?  
Draw the situation

29) two campers dock a canoe. One camper has a mass of 80kg and moves forward at 4m/s as she leaves the boat to step onto the dock.
With what speed and direction do the canoe and the other camper move if their combined mass is 115kg ?