Newton formulated three laws of motion. Newton's first law asserts that objects at rest tend to stay at rest, and objects in motion continue in the same direction at the same speed, unless they are acted upon by an outside force. Thus the first law identifies force, and only force, as the agent of change.

Newton's second law quantifies the relationship between force and the changes it produced: F = ma.

Newton's third law states that forces come only in equal and opposite pairs. That is, when object A exerts a force on B, B exerts an equal force on A, but in the opposite direction. This is a subtle idea which requires further clarification:

Case 1-Press your finger against the table. Can you feel the table pressing back against your finger? Imagine a spring between you and the table. As you compress the spring, it pushes up against you and down against the table. Which receives the greater force...your finger, or the table? (Assume, here, that the spring has no weight.)

Case 2-Now imagine a block sliding onto a cart. The block will skid to a stop, as the cart exerts a backward force on it. Furthermore, the cart will start to roll as the block exerts a reaction force to the right on the cart.

Case 3-Imagine a bullet penetrating a wall of frozen yogurt. The bullet exerts a forward force on the yogurt, spewing it out the far side of the wall, and the yogurt exerts a backward force on the bullet, slowing it down. Though it may seem unreasonable at first, the bullet and yogurt receive equal but opposite forces.

Case 4-A banana is released to fall toward Earth. Earth exerts a downward gravitational force on the banana, but the banana exerts an equal upward force on Earth. Clearly one object responds more dramatically to the force. Which of Newton's laws describes this difference?

The world is made up of action-reaction pairs of forces, and you might well wonder how it is possible to move anything at all if every force you exert is countered by an opposite force. (An educated farm horse tried this argument on its owner for years, but never convinced the farmer it should not pull its cart.) The solution lies in the fact that the forces are exerted on different objects. Thus I may push on a block with a force of 5 pounds, but friction can only supply a force of 3 pounds, so the block accelerates.

the situation:

the force on the finger:

the forces on the block:

the force on the floor:

A philosophical note: If you are ever punched in the nose, take comfort in the knowledge that your assailant has been punched in the fist just as hard.

We may now use our knowledge of Newton's second and third laws to formulate a new expression, or law, which is extremely powerful in solving certain kinds of problems.

Consider two masses forced apart by a compressed spring:

Mass A exerts a force on B, while B exerts an equal and opposite force on A. Each force is exerted for the exact same period of time (again from Newton's third law, when you think about it).

Now we invoke Newton's amazing Law #2:	F = ma
Since	FA = -FB
we can conclude that	mAaA = -mBaB
Now multiply each side by the time interval, t so that	mAaAÇt = -mBaBÇt
But we know from kinematics that Çv = aÇt, so	mAÇvA = -mBÇvB
Under normal circumstances the masses remain constant, so	Ç(mv)A = -Ç(mv)B

This is a result so important that the product, mv, is given its own name, momentum. We see by the relation that when two masses interact, and no forces are exerted from the outside, there will be no change in total momentum. The momentum lost by one will be gained by the other.

The principle that within a system the total amount of momentum remains constant is called the principle of conservation of momentum. In the absence of an outside force, we say that momentum is conserved. For the linear collision of two objects, we express it as m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'.

This conclusion is of enormous physical importance. It can be argued that conservation of momentum is a more fundamental expression of the rules of nature than Newton's laws themselves. One of the beauties of the principle is that it applies easily to a number of problems, frequently involving collisions or ";explosions";. Try some:

1. Two masses start at rest and are pushed apart by a spring as shown. If the 3 kg mass is flung away at 10 m/sec as shown, what velocity does the 7 kg mass have?

Momentum Before = Momentum After m1v1 + m2v2 = m1v1` + m2v2` (3)(0) + (7)(0) = (3)(-10) + (7)(v2`) v2` ~= 4.29 m/sec

2. Again, the same set-up:

Momentum Before = Momentum After m1v1 + m2v2 = m1v1` + m2v2` (8)(0) + (20)(v) = (8)(v) + (20)(2.6) v ~= -6.50 m/sec

3. Once more:

Momentum Before = Momentum After m1v1 + m2v2 = m1v1` + m2v2` 0 = (6)(-1.4) + (m)(44) m = .191 slug

4. A 30 gm rubber ball traveling at 148 cm/sec strikes a motionless 200 gm cucumber. If the ball bounces off backward at 80 cm/sec, how fast will the cucumber be moving?

Momentum Before = Momentum After m1v1 + m2v2 = m1v1` + m2v2` (30)(148) + (200)(0) = (30)(-80) + (200)(v) v ~= 34.2 cm/sec

5. A 76 kg astronaut floats near her capsule, unable to reach it as she has run out of fuel. In desperation she throws a 0.40 kg wrench at -18 m/sec oppositely to the direction she wishes to go. What is her newly acquired speed?

Momentum Before = Momentum After m1v1 + m2v2 = m1v1` + m2v2` 0 = (.4)(-18) + (76)(v) v ~= 9.47 * 10-2 m/sec

6. A 1.6 kg can of creamed corn is struck by a 22 gm bullet traveling at 620 m/sec. If the bullet exits at 540 m/sec, how fast is the can moving?

m1v1 + m2v2 = m1v1` + m2v2` (1.6)(0) + (.022)(620) = (1.6)(v1`) + (.022)(540) v1` ~= 1.10 m/sec

7.

Momentum Before = Momentum After m1v1 + m2v2 = m1v1` + m2v2` (6)(8) + (2)(5) = (6)(3) + (2)(v) v = 20.0 m/sec

Another way to think of these problems is that there is an exchange of momentum. That is, whatever momentum is lost by one mass is gained by the other. If you're on top of what we've done so far, try the previous problem with this new method in mind. Momentum Before = Momentum After Momentum Lost = Momentum Gained 6 * 8 - 6 * 3 = 2 * v - 2 * 5 v = 20.0 m/sec

8. Find the velocity of the 18 kg ball.

m1v1 + m2v2 = m1v1` + m2v2` (5)(11) + (18)(-2) = (5)(-3) + (18)(v) v ~= 1.89 m/sec

9. A 30 gm bullet traveling at 480 m/sec hits a 2.5 kg can of cooked tomatoes at rest on a fence post. If the bullet exits at 370 m/sec,
a) what is the velocity of the can?
b) If the fence post is 1.4 m high, how far from its base will the can land?

a) m1v1 + m2v2 = m1v1` + m2v2` (.03)(480) + (2.5)(0) = (.03)(370) + (2.5)(v) v ~= 1.32 m/sec

b) V) s = (1/2)at2 1.4 = (1/2)(9.8)t2 t ~= 1.32 H) s = vt x = (1.32)(.5345) x = .706 m

10. A 5.7 kg rifle fires a 40 gm bullet at 640 m/sec. What is the recoil speed of the rifle?

m1v1 + m2v2 = m1v1` + m2v2` (5.7)(0) + (.040)(0) = (5.7)(v) + (.040)(640) v ~= -4.49 m/sec

11. A baseball player wishes to determine his mass. He glides frictionless along on ice skates at 2.1 m/sec. After catching a 0.80 kg ball thrown at him at 27 m/sec, he finds he is slowed to 1.2 m/sec. What is his mass?

m1v1 + m2v2 = m1v1` + m2v2` (.8)(27) + (m)(-2.1) = (.8)(-1.2) + (m)(-1.2) (.8)(27) - (.8)(-1.2) = (-m)(-2.1) + (m)(-1.2) 22.56 = m (2.1 - 1.2) m ~= 25.1 kg (The player is a precocious 8-yr old)

12. An 82 kg mass dives off the bow of the Queen Mary with at 4.8 m/sec. If the QM's mass is 2.3 x l07 kg, what is her recoil velocity?

m1v1 + m2v2 = m1v1` + m2v2` (0) = (82)(4.8) + (2.3 * 107) v ~= -1.7 * 10-5 m/sec At this rate the ship would drift 1 cm in about 10 min.

Archimedes once said something about how he could move the earth if he were given a long enough lever and a place to stand. It turns out to be even easier. Just throw a baseball straight up into the air. Earth will move slowly in the opposite direction. The ball and Earth will reach the peak (and depth) of their flights at the same time and fall back in towards each other to meet at the same place they started. Don't believe me, try it yourself.

13. A 50 gm bullet, traveling at 580 m/sec, strikes a 9.17 kg block of wood sitting on a 3.2 m high wall, and is imbedded in it.
How fast does the block move after collision?

m1v1 + m2v2 = m1v1` + m2v2` But after collision, v1` = v2` (Bullet is stuck in block) m1v1 + m2v2 = (m1 + m2)v (.050)(580) + (9.17)(0) = (.050 + 9.17)(v) v ~= 3.15 m/sec

The event described in problem 13 is an example of an ";inelastic collision";. That is, there is no elasticity no bounce to it. Inelastic collisions are common whenever sticky things or pointy things collide.

14. Two blobs of clay collide and stick together. What is their velocity after collision?

m1v1 + m2v2 = (m1 + m2)(v) (4)(8) + (3)(5) = (4 + 3)(v) v ~= 6.71 m/sec

15. After inelastic collision, the two masses will move together at 2.3 m/sec to the right.
What is the unknown mass?

m1v1 + m2v2 = (m1 + m2)(v) (m)(12) + (3)(-5) = (m + 3)(2.3) (m)(12 - 2.3) = (3 * 2.3) + (3 * 5) m ~= 2.26 kg

16. A 120 gm Twinkie moving at 270 cm/sec strikes a 480 gm Coke resting on a frictionless counter. What is their velocity after collision?

m1v1 + m2v2 = (m1 + m2)(v) (120)(270) + (0) = (120 + 480)(v) v = 54.0 cm/sec

It is interesting to note that even if the counter has some friction the answer will be the same. The duration of impact is so small friction does not have time to alter the final speed until later.

17. A 230 lb lineman traveling at 13.6 ft/sec collides head-on with his 6.7 slug opponent traveling toward him at 2.8 ft/sec. What is their velocity just after impact?

m1v1 + m2v2 = (m1 + m2)(v) (230/32)(13.6) + (6.7)(-2.8) = (230/32 + 6.7)(v) v ~= 5.69 ft/sec

18. A 45 gm bullet imbeds in a 2.1 kg block of wood, sending it flying at 10.7 m/sec. What was the initial velocity of the bullet?

m1v1 + m2v2 = (m1 + m2)(v) (.045)(v) + (2.1)(0) = (2.1 + .045)(10.7) v ~= 510 m/sec

19. 115 pound Tyrone Shoelaces, standing on frictionless roller skates, tries to catch a 0.80 slug medicine ball coming at him at 4.1 ft/sec. A born klutz, Tyrone misses the ball which rebounds off him at 1.2 ft/sec. How fast is he moving after the event?

m1v1 + m2v2 = m1v1` + m2v2` (.8)(4.1) + (115/32)(0) = (.8)(-1.2) + (115/32)(v) v ~= 1.18 m/sec

20. A 4.2 pound can of beans sits on a 7.0 ft fence post. Struck by a .0017 slug bullet traveling at 1300 ft/sec, the can falls to the ground 2.3 ft from the post. How fast does the bullet exit?

Velocity of the can: V) Sy = (1/2)at2 7 = (1/2)(32)t2 t = .661 sec H) Sx = vt 2.3 = v (.661> v = 3.48 ft/sec Speed of bullet: m1v1 + m2v2 = m1v1` + m2v2` (.0017)(1300) + (0) = (.0017)(v) + (4.2/32)(3.48) v ~= 1031 ft/sec

As the trig-weary student might fear, momentum is a vector quantity. That is, its direction is significant and momentum is added by the usual vector methods. Furthermore, it is the vector momentum that is conserved in interactions, direction and all.

21. What is the final speed and direction of the masses after an inelastic collision?

The momentum vectors: After collision: tanØ = 12/30 Ø = 21.8° (5+3)v = (122+302)1/2 So v = 4.04 m/sec

22. What will be the final speed and direction after the bullet imbeds itself in the block?

The momentum vectors: tanØ = 20/9 Ø = 65.8° (4+.03)v = (202 + 92)1/2 v = 5.44 m/sec

23.
a) What is the initial velocity, v?
b) What is the final velocity, v'?

Note that: vx = v`cos33 vy = v`sin33 H) (2)(6) + 0 = (2 + 1.5) v`cos33 v` ~= 4.09 ft/sec V) (1.5)(v) + 0 = (2 +1.5) v`sin33 (1.5)(v) = 3.5 (4.09sin33) v ~= 5.20 ft/sec

24. A really big billiard ball approaches another at rest. They collide and fly off as shown. What is the angle and velocity of the rebound?

H) (2)(6) + 0 = (2)(vcosØ) + (3)(3cos36) v cosØ = 2.36 V) 0 + 0 = (2)(vsinØ) + (3)(3)(3sin36) vsinØ = 2.65 tanØ = (2.65)/(2.36) Ø ~= 48.3° v ~= 3.55 m/sec

25. Milton Yermouth throws a 1.4 kg can of anchovies upward with a velocity of 5.7 m/sec. It is immediately struck from the side by a 0.55 kg rock traveling at 8.8 m/sec. The rock bounces off as shown, so what is the angle and speed of the can's motion?

H) (.55)(.88) + (1.4)(0) = (.55)(6.8cos65) + (1.4)(vcosØ) V) (.55)(0) + (1.4)(5.7) = (.55)(6.8sin65) + (1.4)(vsinØ) THUS: vcosØ = 2.33 vsinØ = 3.28 tanØ = (3.28)/(2.33) Ø ~= 54.6° v ~= 4.02 m/sec

A perfectly elastic collision is one in which there is no kinetic energy lost to heat, sound, or other form of energy. We shall see in the next chapter how the simple relationship for elastic collisions can be derived, but for now you will just have to accept it on faith.

In a perfectly elastic collision the relative velocity before impact equals the relative velocity after impact.

26. To get the hang of these, figure out the unknown velocities in each of the perfectly elastic collisions below.

a)The initial relative velocity is 8 - 3 = 5 m/sec So the final will be: v - 6 = 5 v = 11 m/sec b)The initial relative velocity is 12 - -8 = 20 cm/sec SO v -5 = 20 v = 25 cm/sec

27. In each case write the symbols for v_A and v_B after elastic collision.

a)	b)	c)
To maintain the relative velocities
a) Designate vA` = vA` Then vB` = vA` + 5	b) Designate vA` = vA` Then vB` = vA` + 14	c) vA` = vA` vB` = vA` + 25

It may seem to you that truly elastic collisions are rare. Indeed they are, in our range of perception, but at the microscopic scale collisions between atoms in a gas or on a large scale near-misses by stars are perfectly elastic. Many ordinary events are nearly elastic if the time of collision is very small a bat striking a ball, billiard balls colliding, or a bowling ball striking pins are examples. A truly elastic collision would have to be completely silent, as no energy can go into sound.

28. Now try some real elastic collisions:

a) Velocities after collision?	b)	c) Velocities after collision?
a) m1v1 + m2v2 = m1v1` + m2v2` (3)(18) + (20)(0) = (3)(v) + (20)(v + 18) v ~= -13.3 cm/sec v + 18 ~= 4.70 cm/sec	b) m1v1 + m2v2 = m1v1` + m2v2` (7)(6) + (m)(-2) = (7)(2) + (m)(2 + 8) m ~= 23.3 m/sec	c) m1v1 + m2v2 = m1v1` + m2v2` (6)(11) + (10)(4) = (6)(v) + (10)(v + 7) v ~= 2.25 m/sec v + 7 ~= 9.25 m/sec

When a bat hits a ball, a rock goes through a window, or a boxer receives a punch, we are dealing with impulsive forces of short duration.

The effect of a force on an object depends both on the size of the force and its duration a large force acting for a small time can have the same effect as a small force acting for an extended time. We can derive the precise relationship from Newton's second law:

	Çv
F = ma =	m-----
	Çt

So FÇt = mÇv = Ç(mv) when m is constant.

The quantity FÇt is called the impulse, and it equals the change in momentum of the object struck.

29. A 3 kg banana speeds from 4 m/sec to 6 m/sec when it is struck from behind by a rutabaga. What impulse did it receive? Give three examples of force and time combinations that would produce this result.

Since impulse equals change in momentum, we need only find the change in momentum. (NOTE:(kgm)/(sec) = (kgm/sec2) * sec = nt * sec) Çmv = 3 * 6 - 3 *4 = 6 (kgm/sec) = 6 nt * 1 sec = 60 nt * .1 sec = 600 nt * .01 sec

30. A 426 nt force acts for 0.037 sec on a 2.5 kg object. How much does the velocity of the object change?

FÇt = Çmv 426 nt * 0.037 sec = 2.5 kg * Çv Çv = 6.30 m/sec

31. A 20 lb rock strikes a wall at 17 ft/sec. It breaks through, emerging at 11 ft/sec. What impulse did it receive?

Çv = 11 - 17 = -6 ft/sec (always take final - initial value) Impulse = FÇt = Çmv Çmv = (20/32)(-6) Çmv = -3.75 slug ft/sec Çmv = -3.75 lb-sec

32. Liz Onya jumps off a chair with her knees locked straight. She hits the cement floor at 5.5 ft/sec and completely stops in 0.016 sec. If Liz weighs 120 lb, what force in excess of her weight was applied to stop her?

F = (Çmv)/Çt) F = (Çmv)/Çt) = [(120/32)(0 - 5.5)]/(.016) F = (Çmv)/Çt) = 1289 lb (1409 If you count her weight)

If you add Liz's weight to the answer for problem #32, you get 1408 lb of force on her knees. It's for this reason that Liz is now in a cast and less masochistic people bend their knees (or even roll to the ground) when they land to extend the time and reduce the force of their collision. Things like air bags, foam pits, and trampolines do the same.

33. The two masses collide inelastically.
a) What is their velocity after collision?
b) What impulse has the 4 kg mass received?
c) What impulse has the 6 kg mass received?

a) m1v1 + m2v2 = (m1 + m2)(v) (6)(8) + (4)(2) = (6 + 4)(v) v = 5.60 m/sec b) Impulse = Ç(mv) = mÇv Impulse = 4(5.6 - 2.0) Impulse = 14.4 kgm/sec c) Impulse = Ç(mv) = mÇv Impulse = 6(5.6 - 8) Impulse = -14.4 kgm/sec