PROBLEM-SOLVING

GENERAL GUIDELINES

GENERAL PROBLEM SOLVING STRATEGY 

Solving problems require three major steps:PrepareSolveAssess

PREPARE The "Prepare" step of a solution is where you identify important elements of the problem and collect information you will need to solve it.

It's tempting to jump right to the "solve" step, but a skilled problem solver will spend the most time on this step, the preparation.

Preparation includes: 1. Identifying the Physics Principle(s):

Read the problem carefully and identify what is the underlying physics principle of the problem, then, write down the principle using an acronym such as the ones in the following list. If the problem has several steps, write down the principle(s) as appropriate.

Newton's First Law (N1L)Newton's Second Law (N2L)Kinematics in One-Dimension (UAM)Kinematics in Two-Dimension (K2D)Conservation of Energy (COE)Conservation of Momentum (COM)

2. Data: Given and Unknown  Make a table of the quantities whose values you can determine from the problem statement or that can be found quickly with simple geometry or unit conversions. Any relevant constants should be written here. All units should be consistent with the SI values (i.e. kg, m, s). All unit conversion should take place in this section.Also, identify the quantity or quantities that will allow you to answer the question.

3. Sketch, graph, FBD

In many cases, this is the most important part of a problem. The picture lets you model the problem and identify the important elements. As you add information to your picture, the outline of the solution will take shape. If appropriate, select a coordinate axis. If the quantities involved are vectors, be sure to draw an arrow with the tip of the arrow clearly indicating the direction of the vector.When drawing a free-body-diagram (FBD), be sure to draw and clearly label only the forces acting on the system. 

SOLVE The "Solve" step of a solution is where you actually do the math or reasoning necessary to arrive to the solution needed.

This is the part of the problem-solving strategy that you likely think of when you think of "solving problems". But don't make the mistake starting here! If you just choose an equation and plug in numbers, you will likely go wrong and will waste time trying to figure out why. The "Prepare" step will help you be certain you understand the problem before you start putting numbers in equations.

Solving the problem includes: 4. Equation (always solve for unknown)

Write the relevant equation or equations that will allow you to solve for the unknown. Be sure to always solve the equation for the unknown instead of just a 'plug and chug' approach.

5. Substitution

Once you have solved the equation algebraically, substitute the appropriate values.

6. Answer with UnitsWrite down the answer with the appropriate units. Remember that 'naked' numbers make no sense in Physics!

ASSESS

The "Assess" step of your solution is very important. When you have an answer, you should check to see if it makes sense.

7. Check the Answer Ask yourself:

- Does my solution answer the question that was asked? Make sure that you have addressed all parts of the question and clearly written down your solutions.

- Does my answer have the correct units and number of significant figures?

- Does the value I computed make physical sense?

- Can I estimate what the answer should be to check my solution?

- Does my final solution make sense in the context of the material I'm learning?

MOTION An object is in motion if its position changes. The mathematical description of motion is called kinematics.

The simplest kind of motion an object can experience is uniform motion in a straight line. The object experiences translational motion if it is moving without rotating.

Describing Motion

The study of one-dimensional kinematics is concerned with the multiple means by which the motion of objects can be represented. Such means include the use of words, graphs, equations, and diagrams.

One-dimensional motion means that objects are only free to move back and forth along a single line. As a coordinate system for one-dimensional motion, choose this line to be an x-axis together with a specified origin and positive and negative directions.

DISTANCE AND DISPLACEMENTDistance is the length between any two points in the path of an object.

Displacement is the length and direction of the change in position measured from the starting point.

DISTANCE AND DISPLACEMENT

The distance an object travels tells you nothing about the direction of travel, while displacement tells you precisely how far, and in what direction, from its initial position an object is located.  Distance is the total length of travel and displacement is the net length of travel accounting for direction.

The displacement is written:

Displacement is positive. Displacement is negative.

3.1 You leave your home and drive 4.83 km North on Preston Rd. to go to the grocery store. After shopping, you go back home by traveling South on Preston Rd.a. What distance do you travel during this trip?

x1 = 4.83 km

x2 = 4.83 km

distance = x1+ x2

= 4.83 + 4.83 = 9.66 km

b. What is your displacement?

x1 = 4.83 km, Nx2 = 4.83 km, S

displacement = Δx = x2 - x1

= 4.83 - 4.83 = 0 km

UAM

SPEED If an object takes a time interval t to travel a distance x, then the average speed of the object is given by:  

vx

t Units: m/s

3.2 A ship steams at an average speed of 30 km/h. a. What is the speed in m/s?

v = 30 km/h

= 8.33 m/s

UAM

30 km

hv

1000 m

1 km

30

3.61 h

3600 s

b. How far in km does it travel in a day?

t = 24 hv

x

t x vt = 30(24)

= 720 km

c. How long in hours does it take to travel 500 km?

x = 500 km

tx

v

500

30= 16.67 h

AVERAGE SPEED AND AVERAGE VELOCITY Average velocity is the displacement divided by the amount of time it took to undergo that displacement.

The difference between average speed and average velocity is that average speed relates to the distance traveled while average velocity relates to the displacement.

Speed: how far an object travels in a given time interval

Velocity: includes directional information:

v1 = 100 km/h t1 = 2 h

v2 = 75 km/h t2 = 2 h

v3 = 80 km/h t3 = 1 h

3.3 A car travels north at 100 km/h for 2 h, at 75 km/h for the next 2 h, and finally turns south at 80 km/h for 1 h. a. What is the car’s average speed for the entire journey?

Total distance traveledx = v tx1 = v1t1 = 100 (2) = 200 km

x2 = 75 (2) = 150 km

x3 = 80 (1) = 80 km

xT = 200 + 150 + 80

= 430 km

Total timetT = 2 + 2 + 1 = 5 hours

vx

t

430

5= 86 km/h

UAM

b. What is the car’s average velocity for the entire journey?

x1 = 200 km, N

x2 = 150 km, N

x3 = 80 km, S

Displacement:200 + 150 – 80 = 270 km Total time = 5 h

= 54 km/h, N270 km

5 h

xv

t

x

a.

b.

t

x

t

Object starts at the origin and moves in the positive direction with constant velocity.

Object starts to the right of the origin and moves in the negative direction with constant velocity ending at the origin.

3.4 Give a qualitative description of the motion depicted in the following x-versus-t graphs:

xc.

d.

t

x

t

Object starts to the right of the origin and moves in the positive direction with constant velocity.

Object starts to the left of the origin and moves in the positive direction with constant velocity ending at the origin.

The slope of the graphs yields the average velocity. When the velocity is constant, the average velocity over any time interval is equal to the instantaneous velocity at any time.

2 1

2 1

m

s

y yyslope

x x x

3.5 Give a qualitative description of the motion depicted in the following v-versus-t graphs:

va.

b.v

t

t

Object moving to the right at a slow constant speed.

Object moving to the left at a fast constant speed.

ACCELERATION Acceleration is the rate of change of velocity. The change in velocity Δv is the final velocity vf minus the

initial velocity vo.

Acceleration happens when: An object's velocity increases

An object's velocity decreases

An object changes direction

The acceleration of an object is given by:

av v

tf o

Units: m/s2

EQUATIONS FOR MOTION UNDER CONSTANT ACCELERATION:

x v t ato 1

22

v v atf o

v v axf o2 2 2

2

o fv vx t

av v

tf o

v v atf o

3.6 An object starts from rest with a constant acceleration of 8 m/s2 along a straight line. a. Find the speed at the end of 5 s

vo = 0 m/s

a = 8 m/s2

t = 5 s

= 0 + 8(5)= 40 m/s

b. Find the average speed for the 5 s interval

vv vo f

2

0 40

2= 20 m/s

UAM

c. Find the distance traveled in the 5 s

x v t ato 1

22 = 0 + ½(8)(5)2

= 100 m

x vtor:

= 20(5)

= 100 m

3.7 A truck's speed increases uniformly from 15 km/h to 60 km/h in 20 s. a. Determine the average speed

vo = 15/3.6 = 4.17 m/s

vf = 60/3.6 = 16.7 m/s

t = 20 s

vv vo f

2

417 16 7

2

. .= 10.4 m/s

b. Determine the acceleration

av v

tf o

16 7 4 17

20

. .= 0.63 m/s2

UAM

c. Determine the distance traveled

x vt= 10.4(20)= 208 m

3.8 A skier starts from rest and slides 9.0 m down a slope in 3.0 s. In what time after starting will the skier acquire a speed of 24 m/s? Assume that the acceleration is constant.

vo = 0 m/s, x = 9 m

t = 3 s, vf = 24 m/sx v t ato

1

22

ax

t

22

2 9

32

( )= 2 m/s2

tv v

af o

24 0

2= 12 s

UAM

3.9 A car moving at 30 m/s slows uniformly to a speed of 10 m/s in a time of 5.0 s.a. Determine the acceleration of the car

vo = 30 m/s

vf = 10 m/s

t = 5 sa

v v

tf o

10 30

5= - 4 m/s2

b. Determine the distance it moves in the third second

x = v t +½at v t +½ato 3 32

o 2 22 ( )

vo ( ) )t t ½a(t t3 2 32

22

= 30 (3 - 2) + ½ (- 4) (32 - 22)= 20 m

UAM

3.10 The speed of a train is reduced uniformly from 15 m/s to 7.0 m/s while traveling a distance of 90 m.a. Calculate the acceleration.

vo = 15 m/s

vf = 7 m/s

x = 90 m

v v axf o2 2 2

av v

xf o2 2

2

7 15

2 90

2 2

( )= - 0.98 m/s2

UAM

b. How much farther will the train travel before coming to rest, provided the acceleration remains constant?

xv v

af o2 2

2

20 7

2( 0.98)

= 25 m

3.11 A drag racer starts from rest and accelerates at 7.40 m/s2. How far has it traveled in 1.00 s, 2.00 s, and 3.00 s? Graph the results in a position versus time graph.

21

2x at

a = 7.4 m/s2

t = 1.00 s

= 3.70 m

at t = 2.00 s, x = 14.8 mat t = 3.00 s, x = 33.3 m

21(7.40)(1.00)

2

UAM

This example illustrates one of the key features of accelerated motion; position varies directly with the square of the time.

t

t t

x v a

vo = 0

x = 0

t

t t

x v a

vo = 0

x = 0

t

t t

x v a

vo ≠ 0

x = 0

t

t t

x v a

vo = 0

x = 0

t

t t

x v a

vo = 0

x = 0

Observation:- the sign of the velocity and the acceleration is the same if the object is speeding up and that - the sign of the velocity and the acceleration is the opposite if the object is slowing down.

GRAPHICAL ANALYSIS OF MOTION Graphical interpretations for motion along a straight line (the x-axis) are as follows:

• the slope of the tangent of an x-versus-t graph and define instantaneous velocity,

• the slope of the v-versus-t graph and understand that the value obtained is the average acceleration,

• the area under the v-versus-t graph and understand that it gives the displacement,

• the area under the a-versus-t graph and understand that it gives the change in velocity.

The slope of the tangent of an x-versus-t graph yields the instantaneous velocity,

The slope of the v-versus-t graph gives the average acceleration,

The area under the v-t curve yields the displacement.

A = bh + 1/2 bh = m/s . s = m

3.13. A boat moves slowly inside a marina with a constant speed of 1.50 m/s. As soon as it leaves the marina, it accelerates at 2.40 m/s2. a. How fast is the boat moving after accelerating for 5.0 s?

vo = 1.5 m/s

a = 2.40 m/s2

t = 5.00 s

vf = vo + at

= 1.5 + (2.40)(5) = 13.5 m/s

UAM

5.0

1.5

13.5

v(m/s)

t(s)

b. Plot a graph of velocity versus time.

5.0

1.5

13.5

v(m/s)

t(s)

c. How far has the boat traveled in this time?

c. How far has the boat traveled in this time?

ACCELERATION DUE TO GRAVITYAll bodies in free fall near the Earth's surface have the same downward acceleration of: 

g = 9.8 m/s2

 A body falling from rest in a vacuum thus has a velocity of 9.8 m/s at the end of the first second, 19.6 m/s at the end of the next second, and so forth. The farther the body falls, the faster it moves.

A body in free fall has the same downward acceleration whether it starts from rest or has an initial velocity in some direction.

Galileo Galilei (1564-1642)

Galileo is alleged to have performed free-fall experiments by dropping objects off the Tower of Pisa

The presence of air affects the motion of falling bodies partly through buoyancy and partly through air resistance. Thus two different objects falling in air from the same height will not, in general, reach the ground at exactly the same time.

Because air resistance increases with velocity, eventually a falling body reaches a terminal velocity that depends on its mass, size, shape, and it cannot fall any faster than that.

FREE FALL

When air resistance can be neglected, a falling body has the constant acceleration g, and the equations for uniformly accelerated motion apply. Just substitute a for g.

Sign Convention for direction of motion:

If the object is thrown downward then g = 9.8 m/s2

If the object is thrown upward then g = - 9.8 m/s2

FREE-FALLAn object thrown upward in the absence of air resistance yields the following graphs.

3.14 A ball is dropped from rest at a height of 50 m above the ground.a. What is its speed just before it hits the ground?

vo = 0 m/s

y = 50 mg = 9.8 m/s2 v v gyf o 2 2 2 9 8 50( . )( ) = 31.3 m/s

b. How long does it take to reach the ground?

tv v

gf o

313 0

9 8

.

.= 3.19 s

UAM

y = 20 mg = - 9.8 m/s2

At highest point vf = 0

v v gyo f 2 2

0 2 9 8 20( . )( ) = 19.8 m/s

3.15 A stone is thrown straight upward and it rises to a height of 20 m. With what speed was it thrown? UAM

3.16 A stone is thrown straight upward with a speed of 20 m/s. It is caught on its way down at a point 5.0 m above where it was thrown. a. How fast was it going when it was caught?

vo = 20 m/s

y = 5 mg = - 9.8 m/s2

v v gyf o 2 2

20 2 9 8 52 ( . )( )

= 17.4 m/sSince it is moving down we write it as: = - 17.4 m/s

b. How long did the trip take?

tv v

gf o

17 4 20

9 8

.

.= 3.8 s

UAM

3.17 A baseball is thrown straight upward on the Moon with an initial speed of 35 m/s. (g = 1.60 m/s2) Finda. The maximum height reached by the ball

vo = 35 m/s

g = - 1.6 m/s2

At highest point vf = 0

yv v

gf o2 2

2

0 35

2 16

2

( . )= 382. 8 m

b. The time taken to reach that height

tv v

gf o

0 35

16.= 21.9 s

UAM

c. Its velocity 30 s after it is thrown

t = 30 s

v v gtf o 35 16 30( . )( ) = - 13 m/s

d. Its velocity when the ball's height is 100 m

y = 100 m

gyvv of 22 )100)(6.1(2352 = 30 m/s

3.18 A rock is thrown vertically upward with a velocity of 20 m/s from the edge of a bridge 42 m above a river. How long does the rock stay in the air?

vo = 20 m/s

y = - 42 mg = - 9.8 m/s2

First, find the velocity of the rock at the moment that it hits the river.

gyvv of 22

)42)(8.9(2202 = 35 m/s

UAM

vf = - 35 m/s

Negative velocity because the rock will be moving toward the river.

g

vvt of

8.9

2035

= 5.6 s

THE MONKEY AND THE ZOOKEPERA monkey spends most of its day hanging from a branch of a tree waiting to be fed by the zookeeper. The zookeeper shoots bananas from a banana cannon. Unfortunately, the monkey drops from the tree the moment that the banana leavesthe muzzle of the cannon and the zookeeper is faced with the dilemma of where to aim the banana cannon in order to feed the monkey.

If the monkey lets go of the tree the moment that the banana is fired, where should he aim the banana cannon?

Banana thrown ABOVE the monkey

Wrong move!

Banana thrown AT the monkey

Happy monkey!

PROJECTILE MOTION

An object launched into space without motive power of its own is called a projectile.

If we neglect air resistance, the only force acting on a projectile is its weight, which causes its path to deviate from a straight line.

The projectile has a constant horizontal velocity and a vertical velocity that changes uniformly under the influence of the acceleration due to gravity.

HORIZONTAL PROJECTION If an object is projected horizontally, its motion can best be described by considering its horizontal and vertical motion separately.

In the figure we can see that the vertical velocity and position increase with time as those of a free-falling body. Note that the horizontal distance increases linearly with time, indicating a constant horizontal velocity.

3.19 A cannonball is projected horizontally with an initial velocity of 120 m/s from the top of a cliff 250 m above a lake.a. In what time will it strike the water at the foot of the cliff?

b. What is the x-distance (range) from the foot of the cliff to the point of impact in the lake?

v0x = 120 m/s

y = 250 mv0y = 0 g

yt

2

8.9

)250(2 = 7.14 s

x = vx t

= 120(7.14) = 857 m

UAM

UM

c. What are the horizontal and vertical components of its final velocity?

d. What is the final velocity at the point of impact and its direction?

vx = 120 m/s

vy = voy + gt

= 0 + 9.8 (7.14) = 70 m/s

22yxR vvv

22 )70()120(

120

70tan 1

= 139 m/s

= 30.2 below horizontal

v (139 m/s, 30.2)

UMUAM

UAM

3.20 A person standing on a cliff throws a stone with a horizontal velocity of 15.0 m/s and the stone hits the ground 47 m from the base of the cliff. How high is the cliff?

vx = 15 m/s

x = 47 mvy = 0

xv

xt

15

47 = 3.13 s

y = ½ gt2

= ½ (9.8)(3.13)2

= 48 m

UM

UAM

PROJECTILE MOTION AT AN ANGLEThe more general case of projectile motion occurs when the projectile is fired at an angle.

Problem Solution Strategy:

1. Upward direction is positive. Acceleration due to gravity (g) is downward thus g = - 9.8 m/s2

2. Resolve the initial velocity vo into its x and y components:

vox = vo cos θ voy = vo sin θ

3. The horizontal and vertical components of its position at any instant is given by: x = voxt y = voy t +½gt2

4. The horizontal and vertical components of its velocity at any instant are given by: vx = vox vy = voy + gt

5. The final position and velocity can then be obtained from their components.

3.23 An artillery shell is fired with an initial velocity of 100 m/s at an angle of 30 above the horizontal. Find:a. Its position and velocity after 8 s

vo = 100 m/s, 30t = 8 sg = - 9.8 m/s2

vox = 100 cos 30 = 86.6 m/svoy = 100 sin 30 = 50 m/s

x = vox t

= 86.6(8) = 692.8 my = voy t + ½ gt2

= 50(8) + ½ (-9.8)(8)2

= 86.4 m

vx = vox = 86.6 m/s

vy = voy + gt

= 50 + (-9.8)(8) = - 28.4 m/s

UAMUM

b. The time required to reach its maximum height

At the top of the path: vy = 0

vy = voy + gt g

vt oy

8.9

50

= 5.1 s

c. The horizontal distance (range)

Total time T = 2t = 2(5.1) = 10.2 s

x = vox t

= 86.6(10.2) = 883.7 m

UM

UAM

3.24 A baseball is thrown with an initial velocity of 120 m/s at an angle of 40above the horizontal. How far from the throwing point will the baseball attain its original level?

vo = 120 m/s, 40g = - 9.8 m/s2

vox = 120 cos 40 = 91.9 m/svoy = 120 sin 40 = 77.1 m/sAt top vy = 0

g

vt oy

8.9

1.77

= 7.9 s

x = vox (2t)

= 91.9(2)(7.9) = 1452 m

UAM UM

3.25 A plastic ball that is released with a velocity of 15 m/s stays in the air for 2.0 s.

a. At what angle with respect to the horizontal was it released?

vo = 15 m/s

t = 2 stime to maximum height = 1 sat the top vy = 0

vy = voy + gt

g

vt o sin

ov

tgsin

15

)1(8.9sin 1 = 40.8º

UAM

b. What was the maximum height achieved by the ball?

y = voy t +½ gt2

= (15)(sin 40.8º)(1) + ½ (-9.8)(1)2

= 4.9 m

UAM

3.26 Find the range of a gun which fires a shell with muzzle velocity vo at an angle θ .

At top vy = 0

vy = voy + gt

= vo sin θ - gt

g

vt o sin

Total time = 2t

g

vv o

o

sin2cos

)cos(sin2 2

g

vx o

x = vxt

K2D

sin θ cos θ = ½ sin 2θ

2sin2

g

vx o

)cos(sin2 2

g

vx o

b. What is the angle at which the maximum range is possible?

Maximum range is 45 since 2θ = 90

c. Find the angle of elevation of a gun that fires a shell with muzzle velocity of 120 m/s and hits a target on the same level but 1300 m distant.

vo = 120 m/s

x = 1300 m

22sin

ov

gx

2)120(

)1300(8.9

2sin2

g

vx o

= 0.885

sin-1(2θ)= 62

θ = 31