Study Questions/Problems Week 2

I have selected these as worthy problems. Read them all and do those that seem interesting

Chapter 3: Conceptual Questions 2, 3, 5, 7 Conceptual Exercises 2, 3, 7, 12 Problems 5, 16, 20, 21, 28, 34, 36, 43, 48

Chapter 4: Conceptual Questions 6, 8, 9 Conceptual Exercises 3, 5, 6, 10 Problems 4, 11, 25, 31, 35, 39, 45, 61

Answers/solutions for even numbered problems are on following pages—don’t peek until you have done your best to solve a problem.

Chapter 3: Vectors in Physics

Answers to Even-Numbered Conceptual Questions

2. Vectors A , G , and

J are all equal to one another. In addition, vector

I is the same as vector L .

Answers to Even-Numbered Conceptual Exercises

2. The vectors, in order of increasing magnitude (length), are B , C , A ,

and D .

12. (a) Vectors 1 and 5 are position vectors. (b) Vectors 2, 3, 7, and 8 are velocity vectors. (c) Vectors 4 and 6 are acceleration vectors.

Chapter 4: Two-Dimensional Kinematics

Answers to Even-Numbered Conceptual Questions

6. Just before it lands, this projectile is moving downward with the same speed it had when it was launched. In addition, if it was launched upward at an angle above the x axis, it is moving in a direction that is an angle below the x axis just before it lands.

Therefore, its velocity just before landing is v = 2 m/s( ) x + – 4 m/s( ) y .

8. Maximum height depends on the initial speed squared. Therefore, to reach twice the height, projectile 1 must have an initial speed that is the square root of 2 times greater than the initial speed of projectile 2. It follows that the ratio of the speeds is the square root of 2.

Answers to Even-Numbered Conceptual Exercises 6. (a) In the order of increasing initial speed, we have projectiles A, B,

and C. To see this, note that projectile C has the smallest launch angle. Therefore, if this projectile is to reach the same height as projectiles A and B—that is, for it to have the same y component of velocity as the other projectiles—it must have a larger initial speed to make up for its smaller launch angle. The answer is A > B > C. (b) Because these projectiles all have the same initial y component of velocity, which gives them the same maximum height, they all have the same time of flight. Therefore, the answer is A = B = C.

10. Plot 2 correctly represents the speed of the ball as a function of time. Plot 1 is incorrect because it shows the speed of the ball increasing as the ball rises higher above the ground. Plot 3 is incorrect because it shows the speed going to zero when the ball is at its highest point. This is not the case because the ball was thrown at an angle to the horizontal that is different from 90˚, and hence it has a nonzero horizontal component of velocity at every point of its flight—its speed cannot go to zero.

See following pages for Problem solutions

Solutions to Ch. 3 Problems

5. Picture the Problem: The base runner travels from

C (home plate) to first base, then to A (second base),

then to B (third base), and finally back to C (home

plate).

Strategy: The displacement vector r is the same

as the position vector r if we take home plate to be

the origin of our coordinate system (as it is drawn).

The displacement vector for a runner who has just

hit a double is drawn.

Solution: 1. (a) Write the

displacement vector from C to A in

terms of its x and y components: r = 90 ft( ) x + 90 ft( ) y

2. (b) Write the displacement vector

from C to B in terms of its x and y

components: r = 0 ft( ) x + 90 ft( ) y = 90 ft( ) y

3. (c) For a home run the

displacement is zero: r = 0 ft( ) x + 0 ft( ) y

Insight: The displacement is always zero when the object (or person) returns to its original position.

16. Picture the Problem: The vectors involved in the problem are depicted at right.

Strategy: Subtract vector

v

i from

v

f using the vector component method.

Solution: 1. (a) A sketch of the vectors and their difference is shown at right.

2. (b) Subtract the x components:

v

x= v

f ,xv

i,x= 66 km/h( )cos 75°( ) 45 km/h( )cos 0°( ) = 28 km/h

3. Subtract the y components:

v

y= v

f , yv

i,y= 66 km/h( )sin 75°( ) 45 km/h( )sin 0°( ) = 64 km/h

4. Find the magnitude of v :

v = v

x

2+ v

y

2= 28 km/h( )

2

+ 64 km/h( )2

= 70 km/h

5. Find the direction of v :

v= tan

1v

y

vx

= tan1 64 km/h

28 km/h= 66° +180° = 114° where the angle is measured

counterclockwise from the positive x axis.

Insight: Resolving vectors into components takes a little bit of extra effort, but you can get much more

accurate answers using this approach than by adding the vectors graphically. Notice, however, that when

your calculator returns

66° as the angle in step 5, you must have a picture of the vectors in your head (or

on paper) to correctly determine the direction.

20. Picture the Problem: The vectors involved in the problem

are depicted at right.

Strategy: Add the vectors using the component method in

order to find the components of the vector sum. Use the

components to find the magnitude and the direction of the

vector sum.

Solution: 1. (a) Make estimates from the drawing:

A + B + C 20 m 1.5°

2. (b) Add the vector components:

A + B + C = 0 + 20.0 m( )cos 45° + 7.0 m( )cos 30°( ) x +

10.0 m( ) + 20.0 m( )sin 45° + 7.0 m( )sin 30°( ) y

A + B + C = 20.2 m( ) x + 0.64 m( ) y

3. Use the components to find the magnitude:

A + B + C = 20.2 m( )

2

+ 0.64 m( )2

= 20.2 m

4. Use the components to find the angle:

= tan1 0.64 m

20.2 m= 1.8°

Insight: Resolving vectors into components takes a little bit of extra effort, but you can get much more

accurate answers using this approach than by adding the vectors graphically. Notice, however, that when

your calculator returns the angle of 1.8° in step 4, you must have a picture of the vectors in your head (or

on paper) to correctly determine the direction.

21. Picture the Problem: The vector involved in the problem is depicted at right.

Strategy: Determine the x and y components of and then express them in

terms of the unit vectors.

Solution: 1. Find the x and y components

of r :

rx= 54 m( )cos 42( ) = 40 m

ry= 54 m( )sin 42( ) = 36 m

2. Now express r in terms of the unit

vectors: r = 40 m( ) x + 36 m( ) y

Insight: In general, an arbitrary two-dimensional vector A can always be written as the sum of a

vector component in the x direction and a vector component in the y direction.

28. Picture the Problem: The vectors involved in the problem are depicted at right.

r

x

y

54 m

42°

Strategy: Use the information given in the figure to determine the components of each vector.

Solution: 1.

A = 1.5 m( )cos 40°( ) x + 1.5 m( )sin 40°( ) y

A = 1.1 m( ) x + 0.96 m( ) y

2.

B = 2.0 m( )cos 19°( ) 2.0 m( )sin 19°( )

B = 1.9 m( ) x + 0.65 m( ) y

3.

C = 1.0 m( )cos 180 25°( ) x + 1.0 m( )sin 180 25°( ) y

C = 0.91 m( ) x + 0.42 m( ) y

4.

D = 0 x + 1.5 m( ) y

Insight: Any vector can be resolved into two components. The ability to convert a vector to and from its

components is an essential skill for solving many physics problems.

34. Picture the Problem: You travel due east 1500 ft then due north 2500 ft.

Strategy: The components of the displacement are given, from which we can determine the magnitude and

direction fairly easily. The direction of the average velocity will be the same as the direction of the

displacement. The magnitude of the average velocity is the magnitude of the displacement divided by the

total time of travel. Let north be the positive y direction and east be the positive x direction.

Solution: 1. Find the direction of the displacement:

= tan1

ry

rx

= tan1 2500 ft

1500 ft= 59° north of east

2. Find the magnitude of the displacement:

r = 1500 ft( )

2

+ 2500 ft( )2

= 2900 ft 0.305 m/ft = 890 m

3. Find the magnitude of the average velocity:

vav

=r

t=

890 m

3.0 min 60 s/min= 4.9 m/s

Insight: The 59° refers to the angle above the positive x axis (east) because the argument of the inverse

tangent function is r

yrx

, or north divided by east.

36. Picture the Problem: The ball rises straight upward, momentarily comes to rest, and then falls straight

downward.

Strategy: After it leaves your hand the only acceleration of the ball is due to gravity, so we expect the

answer to be 9.81 m/s2. To calculate the acceleration we need only consider the initial and final

velocities and the time elapsed. Because of the symmetry of the situation, the final velocity downward

will have the same magnitude as the initial velocity upward. Apply equation 3-5, taking upward to be

the positive direction.

Solution: Apply equation 3-

5: a

av=

vf

vi

t=

4.5 m/s( ) y 4.5 m/s( ) y

0.92 s= 9.8 m/s

2y

Insight: We saw in Chapter 2 how a uniform acceleration will produce a symmetric trajectory, with

the time to rise to the peak of flight equaling the time to fall back down, and with equal initial and final

speeds.

43. Picture the Problem: The vectors involved in this problem are

depicted at right.

Strategy: Let v

yw= your velocity with respect to the walkway,

v

wg= walkway’s velocity with respect to the ground, and add the

vectors according to equation 3-8 to find v

yg= your velocity with

respect to the ground. Then find the time it takes you to travel the 85-

m distance.

Solution: 1. Find

your velocity with

respect to the

walkway:

vyw

=x

tx =

85 m

68 sx = 1.25 m/s( ) x

2. Apply equation 3-

8 to find your

velocity with respect

to the ground:

v

yg= v

yw+ v

wg= 1.25 m/s( ) x + 2.2 m/s( ) x = 3.45 m/s( ) x

3. Now find the time of

travel:

t =x

vyg

=85 m

3.45 m/s= 25 s

Insight: The moving walkway slashed your time of travel from 68 s to 25 s, a factor of 2.7! Note that

we bent the significant figures rules a little bit by not rounding v

yw to 1.3 m/s. This helped us avoid

rounding error.

48. Picture the Problem: The situation is depicted in the figure at

right, except the boat is supposed to be a jet ski.

Strategy: Place the x-axis perpendicular to the flow of the

river, such that the river is flowing in the negative y-

direction. Let v

bw= jet ski’s velocity relative to the water,

v

bg= jet ski’s velocity relative to the ground, and

v

wg= water’s velocity relative to the ground. Set the y

component of v

bw equal to the magnitude of

v

wg so that

they cancel, leaving only an x component of v

bg. Then

determine the angle .

Solution: 1. (a) Set

v

bw, y+ v

wg, y= 0 and

solve for :

vbw, y

= vbw

sin = vwg, y

= sin 1v

wg, y

vbw

= sin 12.8 m/s( )

12 m/s= 13°

2. (b) Increasing the jet ski’s speed relative to the water will increase v

bw and therefore decrease the

angle .

Insight: Airplanes must also make heading adjustments like the jet ski’s in order to fly in a certain

direction when there is a steady wind present.

Solutions to Ch. 4 Problems

4. Picture the Problem: The motion of the particle is depicted at right.

Strategy: Use the given information to independently write the equations of motion in the x and y

directions. There will be a pair of equations for the position of the particle (like equations 4-6) and a pair

for the velocity (like equations 4-6) only the acceleration will not be the same as g .

Solution: 1. (a) Write equations 4-6 for the x position of the particle using

a = 4.4 m/s2( ) x instead of

g :

x = x

0+ v

0xt + 1

2a

xt

2= 0 + 0 + 1

24.4 m/s2( )(5.0)2

= –55 m

2. Now do the same for the y coordinate of the position:

y = y

0+ v

0 yt + 1

2a

yt

2= 0 + 6.2 m/s( )(5.0 s) + 0 = 31 m

3. (b) Write equations 4-6 for the x component of the velocity using

a = 4.4 m/s2( ) x instead of

g :

v

x= v

0x+ a

xt = 0 + 4.4 m/s2( )(5.0 s) = –22 m s

4. Now do the same for the y component of the velocity:

v

y= v

0 y+ a

yt = 6.2

m

s+ 0 = 6.2 m s

5. (c) The x-component of the velocity continually increases in the negative direction, so the speed

increases with time.

Insight: This problem is very much akin to projectile motion, with uniform acceleration in one direction

and constant velocity the perpendicular direction. The only difference is the acceleration is

4.4 m/s2 and

is in the x direction.

11. Picture the Problem: The baseball is launched

horizontally and follows a parabolic trajectory

like that pictured at right.

Strategy: Using equations 4-7 find the time it takes the baseball to travel the 18 m to the plate, and then calculate the vertical distance it will fall during that time.

Solution: 1. (a) Find the time it takes the

ball to

reach the plate knowing that its horizontal

velocity

v

0 remains unchanged throughout the

flight:

t =x

v0

=18 m

32 m/s= 0.563 s

2. Find the vertical drop during the time

of flight: h y = 1

2gt2

= 1

29.81 m/s

2( ) 0.563 s( )2

= 1.6 m

3. (b) If the pitch speed is increased the time the ball travels is less, therefore the drop distance

decreases.

4. (c) Since the moon’s gravity is less the drop distance decreases.

Insight: The 1.6 m drop corresponds to about 5 ft, nearly the height of a player. The 32 m/s (72 mi/h)

speed is a bit slow; a sizzling fastball at 44 m/s (98 mi/h) would drop about 0.82 m, half as much.

25. Picture the Problem: The cork travels along a parabolic arc, maintaining its horizontal velocity but

changing its vertical speed due to the constant downward acceleration of gravity.

Strategy: Find the horizontal component of the initial velocity by dividing the horizontal

distance traveled by the time of flight. Then use the cosine function to find the initial speed of the

cork.

Solution: 1. Find the horizontal speed of the

cork:

vx=

x

t=

1.30 m

1.25 s= 1.04 m/s = v

0x

2. Use the cosine function to find the initial

speed:

v0=

v0x

cos=

1.04 m/s

cos35.0°= 1.27 m/s

Insight: Because gravity acts only in the vertical direction, the horizontal component of the cork’s

velocity remains unchanged throughout the flight.

31. Picture the Problem: The golf ball travels along a parabolic arc, landing at the same level from which

it was launched.

Strategy: The maximum range of a projectile launched from level ground occurs when the launch

angle is 45°. Use equation 4-12 to predict the range of the golf ball when launched at 45°. The

minimum speed of the ball will occur when the ball reaches the peak of its flight. At that point the

vertical component of the velocity is zero and the speed equals the horizontal component of the

velocity, which remains unchanged throughout the flight.

Solution: 1. (a) Find the range of the ball

when it is launched at 45° by using

equation 4-12:

R =v

0

2

gsin 2 =

30.0 m/s( )2

9.81 m/s2

sin 90° = 91.7 m

2. (b) Find the x component of the ball’s

velocity, which corresponds to the

minimum speed during the flight:

v

0x= v

0cos = 30.0 m/s( )cos 45° = 21.2 m/s

Insight: The maximum range will occur at =45° only in the absence of air resistance. In the

presence of the atmosphere you must launch the ball at a lower angle than that in order to maximize the

range of the ball.

35. Picture the Problem: The path of the ball is depicted at right.

Strategy: Use the horizontal component of the ball’s velocity

together with the horizontal distance d to find the time elapsed

between the hit and its collision with the wall. Then use the

time to determine the horizontal and vertical components of the

ball’s velocity when it collides with the wall.

Solution: 1. (a) Use

equations 4-10 to find the

time:

t =d

v0

cos=

3.8 m

14 m/s( )cos34°

=3.8 m

11.6 m/s= 0.33 s

2. Find the vertical

component of the ball’s

velocity when it hits the

wall:

v

y= v

0sin gt = 14 m/s( )sin 34° 9.81 m/s

2( ) 0.33 s( ) = 4.6 m/s

3. Find the magnitude of

the velocity: v = v

x

2+ v

y

2= 11.6 m/s( )

2

+ 4.6 m/s( )2

= 12 m/s

4. Find the direction of the

velocity:

= tan1 4.6 m/s

11.6 m/s= 22° above horizontal

5. (b) No. Since the y component of the ball’s velocity is positive when it strikes the wall, it is still

rising and has not yet reached the peak of its flight.

Insight: The peak of the flight occurs at 0.80 seconds, so the ball is not even halfway to the peak of its

flight (in terms of time). If the wall were not there its peak height would be 3.1 m, so it is more than

halfway to the peak in terms of height.

39. Picture the Problem: The pumpkin travels along a parabolic arc, landing at the same level from which

it was launched.

Strategy: Assume the given range corresponds to a launch angle of 45° and use equation 4-12 to find the

initial speed of the pumpkin.

Solution: Solve equation 4-

12 for v

0:

R =v

0

2

gsin 2

v0=

gR

sin 2=

9.81 m/s2( )(4086 ft 0.305 m/ft)

sin 90.0°= 111 m/s = 248 mi/h

Insight: The maximum range will occur at =45° only in the absence of air resistance. In the presence of

the atmosphere you must launch the pumpkin at a lower angle than that in order to maximize the

downrange distance.

45. Picture the Problem: The lava that reaches maximum altitude is hurled straight upward, comes to rest

momentarily, and falls straight downward again.

Strategy: Use equations 4-6 to determine the launch speed required to achieve the maximum altitude.

Solution: 1. (a) Use equation 4-6 to find v

0:

vy

2= v

0 y

2 2g y = 0 at the peak of flight

v0 y

= 2g y = 2 1.80 m/s2( ) 2.00 105 m( ) = 849 m/s

2. (b) If the launch speed remains the same, the maximum height of the ejected lava on Earth would be

less than it is on Io because the acceleration of gravity on Earth is much greater than 1.80 m/s2.

Insight: On Earth a launch speed of 849 m/s would be 2.5 times the speed of sound—roughly equal to the

muzzle velocity of a rifle bullet! Tidal forces from Jupiter produce these violent volcanic eruptions on Io.

61. Picture the Problem: The rover rebounds from the surface and travels along a parabolic arc before

bouncing off the surface again.

Strategy: Use the given velocity and angle information along with the equation derived in Example 4-

7 to find the maximum height above the surface achieved by the rover on this particular bounce. The

same information together with equation 4-12 can be used to find the downfield range of the rover

between the first and second bounces.

Solution: 1. (a) Find the maximum

height:

ymax

=v

0

2sin

2

2g=

9.92 m/s( )2

sin2

75.0°

2 3.72 m/s2( )

= 12.3 m

2. (b) Find the downfield range:

R =v

0

2

gsin 2 =

9.92 m/s( )2

3.72 m/s2

sin 150.0°( ) = 13.2 m

Insight: The rover spent 5.15 s in the air between the first and second bounces.