B.Crowell - Newtonian Physics, Vol

Chapter 0

8. (a) Let’s do 10.0 g and 1000 g. The arithmetic mean is 505 grams. It comes out to be 0.505 kg, which is consistent. (b) The geometric mean comes out to be 100 g or 0.1 kg, which is consistent. (c) If we multiply meters by meters, we get square meters. Multiplying grams by grams should give square grams! This sounds strange, but it makes sense. Taking the square root of square grams (g2) gives grams again. (d) No. The superduper mean of two quantities with units of grams wouldn’t even be something with units of grams! Related to this shortcoming is the fact that the superduper mean would fail the kind of consistency test carried out in the first two parts of the problem.

Chapter 1

11. The bigger scope has a diameter that’s ten times greater. Area scales as the square of the linear dimensions, so its light-gathering power is a hundred times greater (10x10).

12. Since they differ by two steps on the Richter scale, the energy of the bigger quake is 10000 times greater. The wave forms a hemisphere, and the surface area of the hemisphere over which the energy is spread is proportional to the square of its radius. If the amount of vibration was the same, then the surface areas much be in the ratio of 10000:1, which means that the ratio of the radii is 100:1.

Chapter 2

4. 1 light-year = v t

=3.0x108 m/s1 year

×365 days24 hours3600 s

1 year 1 day 1 hour

= 9.5x1015 m

5. Velocity is relative, so having to lean tells you nothing about the train’s velocity. Fullerton is moving at a huge speed relative to Beijing, but that doesn’t produce any noticeable effect in either city. The fact that you have to lean tells you that the train is accelerating.

7. To the person riding the moving bike, bug A is simply going in circles. The only difference between the motions of the two wheels is that one is traveling through space, but motion is relative, so this doesn’t have any effect on the bugs. It’s equally hard for each of them.

10. In one second, the ship moves v meters to the east, and the person moves v meters north relative to the deck. Relative to the water, he traces the diagonal of a triangle whose length is given by the Pythagorean theorem, (v 2+v 2)1/2=21/2v. Relative to the water, he is moving at a 45-degree angle between north and east.

Chapter 3

14.

15. Taking g to be 10 m/s, the bullet loses 10 m/s of speed every second, so it will take 10 s to come to a stop, and then another 10 s to come back down, for a total of 20 s.

16. x = 12at 2 , so for a fixed value of x, we have

t 1/a . Decreasing a by a factor of 3 means that t

will increase by a factor of 3 .

18. (a) Solving x = 12at 2 for a, we find a=2 x/t2=5.51

m/s2. (b) v=2a x =66.6 m/s. (c) The actual car’s final velocity is less than that of the idealized con- stant-acceleration car. If the real car and the idealized car covered the quarter mile in the same time but the real car was moving more slowly at the end than the idealized one, the real car must have been going faster than the idealized car at the beginning of the race. The real car apparently has a greater acceleration at the beginning, and less acceleration at the end. This make sense, because every car has some maximum speed, which is the speed beyond which it cannot accelerate.

19. Since the lines are at intervals of one m/s and one second, each box represents one meter. From t=0 to t=2 s, the area under the curve represents a positive x of 6 m. (The triangle has half the area of the 2x6 rectangle it fits inside.) After t=2 s, the area above the curve represents negative x. To get –6 m worth of area, we need to go out to t=6 s, at which point the triangle under the axis has a width of 4 s and a height of 3 m/s, for an area of 6 m (half of 3x4).

20. (a) We choose a coordinate system with positive pointing to the right. Some people might expect that the ball would slow down once it was on the more gentle ramp. This may be true if there is significant friction, but Galileo’s experiments with inclined planes showed that when friction is negligible, a ball rolling on a ramp has constant acceleration, not constant speed. The speed stops increasing as quickly once the ball is on the more gentle slope, but it still keeps on increasing. The a-t graph can be drawn by inspecting the slope of the v-t graph.

v

t

a

t

(b) The ball will roll back down, so the second half of the motion is the same as in part a. In the first (rising) half of the motion, the velocity is negative, since the motion is in the opposite direction compared to the positive x axis. The acceleration is again found by inspecting the slope of the v-t graph.

v

a

t

21. This is a case where it’s probably easiest to draw the acceleration graph first. While the ball is in the air, the only force acting on it is gravity, so it must have the same, constant acceleration during each hop. Choosing a coordinate system where the positive x axis points up, this becomes a negative acceleration (force in the opposite direction compared to the axis). During the short times between hops when the ball is in contact with the ground, it experiences a large acceleration, which turns around its velocity very rapidly. These short positive accelerations probably aren’t constant, but it’s hard to know how they’d really look. We just idealize them as constant accelerations. Since our acceleration graph consists of constantacceleration segments, the velocity graph must consist of line segments, and the position graph must consist of parabola.

x

t

v

a

t

22. We have vf2=2a x, so the distance is proportional to the square of the velocity. To get up to half the speed, the ball needs 1/4 the distance, i.e. L/4.

Chapter 4

v t = a

= m v F

= (1000 kg)(50 m/s – 20 m/s) 3000 N

= 10 s

Chapter 5

14. (a)

top spring’s rightward force on connector

...connector’s leftward force on top spring

bottom spring’s rightward force on connector

...connector’s leftward force on bottom spring

hand’s leftward force on connector

...connector’s rightward force on hand

Looking at the three forces on the connector, we see that the hand’s force must be double the force of either spring. The value of x-xo is the same for both

springs and for the arrangement as a whole, so the spring constant must be 2k. This corresponds to a stiffer spring (more force to produce the same extension).

(b) Forces in which the left spring participates:

hand’s leftward force on left spring

...left spring’s rightward force on hand

right spring’s rightward force on left spring

...left spring’s leftward force on right spring

Forces in which the right spring participates:

left spring’s leftward force on right spring

...right spring’s rightward force on left spring

wall’s rightward force on right spring

...right spring’s leftward force on wall

Since the left spring isn’t accelerating, the total force on it must be zero, so the two forces acting on it must be equal in magnitude. The same applies to the two forces acting on the right spring. The forces between the two springs are connected by Newton’s third law, so all eight of these forces must be equal in magnitude. Since the value of x-xo for the whole setup is double what it is for either spring individually, the spring constant of the whole setup must be k/2, which corresponds to a less stiff spring.

16. (a) Spring constants in parallel add, so the spring constant has to be proportional to the cross-sectional area. Two springs in series give half the spring constant, three springs in series give 1/3, and so on, so the spring constant has to be inversely proportional to the length. Summarizing, we have k A/L.

(b) With the Young’s modulus, we have k=(A/L)E.The spring constant has units of N/m, so the units of E would have to be N/m2.

18. (a) The swimmer’s acceleration is caused by the water’s force on the swimmer, and the swimmer makes a backward force on the water, which accelerates the water backward. (b) The club’s normal force on the ball accelerates the ball, and the ball makes a backward normal force on the club, which decelerates the club. (c) The bowstring’s normal force accelerates the arrow, and the arrow also makes a backward normal force on the string. This force on the string causes the string to accelerate less rapidly than it would if the bow’s force was the only one acting on it. (d) The tracks’ backward frictional force slows the locomotive down. The locomotive’s forward frictional force causes the whole planet earth to accelerate by a tiny amount, which is too small to measure because the earth’s mass is so great.

Chapter 6

5. (a) The easiest strategy is to find the time spent aloft, and then find the range. The vertical motion and the horizontal motion are independent. The vertical motion has acceleration —g, and the cannonball spends enough time in the air to reverse its vertical velocity component completely, so we have

vy = vyf—vyi

= —2v sin θ .

The time spent aloft is therefore

t = vy / ay

= 2v sin θ / g .

During this time, the horizontal distance traveled is

R = vx t

= 2 v 2 sin θ cos θ / g .

(b) The range becomes zero at both θ=0 and at θ=90°. The θ=0 case gives zero range because the ball hits the ground as soon as it leaves the mouth of the cannon. A 90 degree angle gives zero range because the cannonball has no horizontal motion.

Chapter 8

8. We want to find out about the velocity vector vBG of the bullet relative to the ground, so we need to add Annie’s velocity relative to the ground vAG to the bullet’s velocity vector vBA relative to her. Letting the positive x axis be east and y north, we have

The bullet’s velocity relative to the ground therefore has components

Its speed on impact with the animal is the magnitude of this vector

(rounded off to 2 significant figures).

9. Since its velocity vector is constant, it has zero acceleration, and the sum of the force vectors acting on it must be zero. There are three forces acting on the plane: thrust, lift, and gravity. We are given the first two, and if we can find the third we can infer its mass. The sum of the y components of the forces is zero, so

0 = Fthrust,y +Flift,y +FW,y

= |Fthrust| sin θ + |Flift| cos θ — mg . The mass is

m= (|Fthrust| sin θ + |Flift| cos θ) / g = 6.9x104 kg

10. (a) Since the wagon has no acceleration, the total forces in both the x and y directions must be zero. There are three forces acting on the wagon: FT, FW, and the normal force from the ground, FN. If we pick a coordinate system with x being horizontal and y vertical, then the angles of these forces measured counterclockwise from the x axis are 90°-ϕ, 270°, and 90°+θ, respectively. We have

Fx,total

= FTcos(90°-ϕ) + FWcos(270°) + FNcos(90°+θ)

Fy,total

= FTsin(90°-ϕ) + FWsin(270°) + FNsin(90°+θ) , which simplifies to

0 = FT sin ϕ – FN sin θ

0 = FT cos ϕ – FW + FN cos θ .

The normal force is a quantity that we are not given and do not with to find, so we should choose it to

Multiplying both the top and the bottom of the fraction by sin θ, and using the trig identity for sin(θ+ϕ) gives the desired result,

(b) The case of ϕ=0, i.e. pulling straight up on the wagon, results in FT=FW: we simply support the wagon and it glides up the slope like a chair-lift on a ski slope. In the case of ϕ=180°-θ, FT becomes infinite. Physically this is because we are pulling directly into the ground, so no amount of force will suffice.

11. (a) If there was no friction, the angle of repose

would be zero, so the coefficient of static friction, μs, will definitely matter. We also make up symbols θ, m and g for the angle of the slope, the mass of the object, and the acceleration of gravity. The forces form a triangle just like the one in section 8.3, but instead of a force applied by an external object, we have static friction, which is less than μsFN. As in that example, Fs=mg sin θ, and Fs<μsFN, so

mg sin θ<μsFN .

From the same triangle, we have FN=mg cos θ, so mg sin θ < μsmg cos θ.

(b) Both m and g canceled out, so the angle of repose would be the same on an asteroid.

Chapter 9

5. Each cyclist has a radial acceleration of v2/r=5 m/ s2. The tangential accelerations of cyclists A and B are 375 N/75 kg=5 m/s2.

C

scale: 5 m/s2

B

A

6. (a) The inward normal force must be sufficient to produce circular motion, so

We are searching for the minimum speed, which is the speed at which the static friction force is just barely able to cancel out the downward gravitational force. The maximum force of static friction is

|Fs| = μsFN ,

and this cancels the gravitational force, so |Fs| = mg .

Solving these three equations for v gives

gr

v = μs .

(b) Greater by a factor of 3 .

7. The inward force must be supplied by the inward component of the normal force,

The upward component of the normal force must cancel the downward force of gravity,

FN cos θ = mg .

Eliminating FN and solving for θ, we find

Chapter 10

10. Newton’s law of gravity tells us that her weight will be 6000 times smaller because of the asteroid’s smaller mass, but 132=169 times greater because of its smaller radius. Putting these two factors together gives a reduction in weight by a factor of 6000/169, so her weight will be (400 N)(169)/(6000)=11 N.

11. Newton’s law of gravity says F=Gm1m2/r2, and Newton’s second law says F=m2a, so Gm1m2/r2=m2a. Since m2 cancels, a is independent of m2.

12. Newton’s second law gives

F = mD aD ,

where F is Ida’s force on Dactyl. Using Newton’s

Dactyl’s mass cancels out, giving

Dactyl’s velocity equals the circumference of its orbit divided by the time for one orbit: v=2πr/T. Inserting this in the above equation and solving for mI, we find

so Ida’s density is

ρ= mI / V

4π2r 3

= GVT 2 .

15. Newton’s law of gravity depends on the inverse square of the distance, so if the two planets’ masses had been equal, then the factor of 0.83/0.059=14 in distance would have caused the force on planet c to be 142=2.0x102 times weaker. However, planet c’s mass is 3.0 times greater, so the force on it is only smaller by a factor of 2.0x102/3.0=65.

Acceleration. The rate of change of velocity; the slope of the tangent line on a v-t graph.

Attractive. Describes a force that tends to pull the two participating objects together. Cf. repulsive, oblique.

Center of mass. The balance point of an object.

Component. The part of a velocity, acceleration, or force that is along one particular coordinate axis.

Displacement. (avoided in this book) A name for the symbol x .

Fluid. A gas or a liquid.

Fluid friction. A friction force in which at least one of the object is is a fluid (i.e. either a gas or a liquid).

Gravity. A general term for the phenomenon of attraction between things having mass. The attraction between our planet and a humansized object causes the object to fall.

Inertial frame. A frame of reference that is not accelerating, one in which Newton’s first law is true

Kinetic friction. A friction force between surfaces that are slipping past each other.

Light. Anything that can travel from one place to another through empty space and can influence matter, but is not affected by gravity.

Magnitude. The “amount” associated with a vector; the vector stripped of any information about its direction.

Mass. A numerical measure of how difficult it is to change an object’s motion.

Matter. Anything that is affected by gravity.

Mks system. The use of metric units based on the meter, kilogram, and second. Example: meters per second is the mks unit of speed, not cm/s or km/hr.

Noninertial frame. An accelerating frame of reference, in which Newton’s first law is violated

Nonuniform circular motion. Circular motion in which the magnitude of the velocity vector

changes

Normal force. The force that keeps two objects from occupying the same space.

Oblique. Describes a force that acts at some other angle, one that is not a direct repulsion or attraction. Cf. attractive, repulsive.

Operational definition. A definition that states what operations should be carried out to measure the thing being defined.

Parabola. The mathematical curve whose graph has y proportional to x2.

Radial. Parallel to the radius of a circle; the in-out direction. Cf. tangential.

Repulsive. Describes a force that tends to push the two participating objects apart. Cf. attractive, oblique.

Scalar. A quantity that has no direction in space, only an amount. Cf. vector.

Significant figures. Digits that contribute to the accuracy of a measurement.

Speed. (avoided in this book) The absolute value of or, in more then one dimension, the magnitude of the velocity, i.e. the velocity stripped of any information about its direction

Spring constant. The constant of proportionality between force and elongation of a spring or other object under strain.

Static friction. A friction force between surfaces that are not slipping past each other.

Système International.. Fancy name for the metric system.

Tangential. Tangent to a curve. In circular motion, used to mean tangent to the circle, perpendicular to the radial direction Cf. radial.

Uniform circular motion. Circular motion in which the magnitude of the velocity vector remains constant

Vector. A quantity that has both an amount (magnitude) and a direction in space. Cf. scalar.

Velocity. The rate of change of position; the slope of the tangent line on an x-t graph.

Weight. The force of gravity on an object, equal to mg.

Algebra

Quadratic equation:

The solutions of ax 2 + bx + c = 0

Logarithms and exponentials:

ln (ab) = ln a + ln b

ea + b = eaeb

ln ex = eln x = x ln ab = b ln a

Geometry, area, and volume

h = hypotenuse

o = opposite side

θ

a = adjacent side

Definitions of the sine, cosine, and tangent: sin θ = oh

cos θ = ah tan θ = oa

Pythagorean theorem: h2 =a2 + o2

Trigonometry with any triangle

Properties of the derivative and integral (for students in calculus-based courses)

Let f and g be functions of x, and let c be a constant. Linearity of the derivative:

The chain rule:

ddx f(g(x)) =f ′(g(x))g ′(x) Derivatives of products and quotients:

ddx sin x = cos x

ddx cos x = –sin x

ddx ex = ex

ddx ln x = x1

The fundamental theorem of calculus:

ddxf dx = f

Linearity of the integral:

cf(x)dx = c f(x)dx

f(x) + g(x) dx =f(x)dx +g(x)dx

Integration by parts:

f dg = fg –g df