Introductory Physics: Problems solving, Slides of Physics

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Introductory Physics: Problems solving

D. A. Garanin

21 December 2021

Introduction

Solving problems is an inherent part of the physics course that requires a more active approach than just reading the theory or listening to lectures. Making only the latter, the student can have an illusion of having understood the material but it is not the case until s/he becomes able to apply one’s knowledge to solving problems that is, working actively with the material.

The main purpose of our Introductory Physics course, for the majority of our students, is to acquire a conceptual understanding of physics, to develop a scientific way of thinking. The latter means relying on the scientific definitions, simple logic, and the common sense, opposed to making wild assumptions at every step that leads to wrong results and loss of points.

PHY166 and PHY167 courses are algebra based, while PHY168 and PHY169 are calculus based. Both types of courses require that problems are solved algebraically and an algebraic result, that is, a formula is obtained. Only after that the numbers are plugged in the resulting formula and the numerical result is obtained. One should understand that physics is mainly about formulas, not about numbers, thus the main result of problem solving is the algebraic result, while the numerical result is secondary.

Unfortunately, most of the students taking part in our physics courses reject algebra and try to work out the solution numerically from the very beginning. Probably, bad teachers at the high school taught the students that problem solving consists in finding the “right” formula and plugging the numbers into it. This is fundamentally wrong.

There are several arguments for why the algebraic approach to problem solving is better than the numeric approach.

1. Algebraic manipulations leading to the solution are no more difficult than the corresponding operations with numbers. In fact, they are easier as a single symbol, such as a , stands for a number that usually requires much more efforts to write without mistakes.
2. Numerical calculations are for computers, while algebraic calculations are for humans. Computers do not understand what they are computing, and they are proceeding blindly along prescribed routes. The same does a human trying to operate with numbers. However, the human forgets what do these numbers stand for and loses the clue very soon. If a human operates with algebraic symbols, s/he is not losing the clue as the symbols speak for themselves. For instance, a usually is an acceleration or a distance, m usually is a mass, etc.
3. The value of a formula is much higher than that of the numerical answer because the formula can be used with another set of input values while the numerical result cannot. In all more or less intelligent devices formulas are implemented that work as “black boxed”: one supplies the input values and collects the output values.
4. Formulas allow analysis of their dependence on the input values or parameters. This is important for understanding the formula and for checking its validity on simple

Physics part I

Kinematics

Vectors, coordinates, displacement, distance, velocity, speed, acceleration, projectile motion, etc.

1. Professor’s way to work

A professor going to work first walks 500 m along the campus wall, then enters the campus and goes 100 m perpendicularly to the wall towards his building, after that takes an elevator and mounts 10 m up to his office. The trip takes 10 minutes.

Calculate the displacement, the distance between the initial and final points, the average velocity and the average speed.

Solution : The total trajectory can be represented by three vectors going from 0 to 1, then from 1 to 2, then from 2 to 3. The displacement is the vector sum of the three displacement vectors:

𝐝 = 𝒓 01 + 𝒓 12 + 𝒓23.

It is convenient to choose the coordinate axes xyz that coincide with these three mutually orthogonal vectors, as shown in the figure. Then, using, for any vector

𝐫 = (𝑟𝑥, 𝑟𝑦, 𝑟𝑧),

one writes

𝒓 01 = (0,500,0) m, 𝒓 12 = (100,0,0) m, 𝒓 23 = (0,0,10) m.

The addition of these vectors is performed as follows:

𝐝 = (0 + 100 + 0, 500 + 0 + 0, 0 + 0 + 10) = (100,500,10) m.

The distance 𝑑 between the initial and final points is the magnitude of the displacement 𝐝:

𝑑 = |𝐝| = √𝑑𝑥^2 + 𝑑𝑦^2 + 𝑑𝑧^2 = √100^2 + 500^2 + 10^2

= √10000 + 250000 + 100 = √260100 = 510 m.

The trajectory length (the way length) is given by

x

y

z

500 m

10 m^1 00 m

d ,d

0^0

𝒓 12 = (𝑟12,𝑥, 𝑟12,𝑦) = (−𝑟 12 cos 45°, −𝑟 12 sin 45°) = (−2000 √2 2 , − 2000 √2 2 )

= (−1000√2, − 1000√2, ) m.

Better is to write

𝒓 12 = (𝑟12,𝑥, 𝑟12,𝑦) = (𝑟 12 cos 125°, 𝑟 12 sin 125°) = (2000 (−

= (−1000√2, − 1000√2, ) m

that gives the same result. Now,

𝐝 = (𝑟01,𝑥 + 𝑟12,𝑥, 𝑟01,𝑦 + 𝑟12,𝑦) = (500√3 − 1000√2, 500 − 1000√2) ≈ (−548.2, −914.2) m

The distance is given by

𝑑 = |𝐝| = √𝑑𝑥^2 + 𝑑𝑦^2 = √(−548.2)^2 + (−914.2)^2 ≈ 1066 m

The length of the trajectory is

𝑤 = 𝑟 01 + 𝑟 12 = 1000 + 2000 = 3000 m.

The velocity:

𝐯 =

30 × 60 = (

30 × 60 ,

30 × 60) = (… , … ) m/s.

The magnitude of the average velocity:

∆𝑡 =^

30 × 60 = 0.59 m/s.

The average speed:

𝑠 =

∆𝑡 =^

30 × 60 = 1.67 m/s > 𝑣.

3. A car trip (1D motion)

A car starts from the place with an acceleration 2 m/𝑠^2 and is accelerating during 10 seconds, then travels with the same speed for 30 seconds, then decelerates at the rate 3 m/𝑠^2 until stopping. Show the graph 𝑣(𝑡). Calculate the total time of the trip and the distance covered in each interval and the total distance covered by two methods: 1) Calculation of the area under the line 𝑣(𝑡); 2) Using the formula for the distance in the motion with constant acceleration.

Solution: First, we introduce missing notations: 𝑎 1 = 2 𝑚/𝑠^2 , 𝑡 1 = 10 𝑠, ∆𝑡 2 ≡ 𝑡 2 − 𝑡 1 = 30 𝑠, 𝑎 3 = −3 𝑚/𝑠^2. The time dependence of the car’s velocity is shown in the figure. In the interval 1 the car accelerates according to the formula

Interval 1: 𝑣 = 𝑣 0 + 𝑎 1 𝑡 = 𝑎 1 𝑡,

where we take into account that the initial velocity is zero: 𝑣 0 = 0. At the end of the first time interval, 𝑡 = 𝑡 1 , the velocity reaches the value

𝑣 1 = 𝑎 1 𝑡 1.

This expression is an instance of the formula above.

The velocity remains the same in the second interval of motion:

Interval 2: 𝑣 = 𝑣 1.

The time at the end of the second interval is

𝑡 2 = 𝑡 1 + ∆𝑡 2 = 10 + 30 = 40 𝑠.

In the third interval, the car decelerates according to

Interval 3: 𝑣 = 𝑣 1 + 𝑎 3 (𝑡 − 𝑡 2 )

(this is the velocity formula with shifted time as the motion starts at 𝑡 = 𝑡 2 rather than at 𝑡 = 0). At the end of the motion the car stops that is described by the instance of the formula above with 𝑣 = 0, that is,

0 = 𝑣 1 + 𝑎 3 (𝑡 3 − 𝑡 2 )

that defines 𝑡 3. One obtains

∆𝑡 3 ≡ 𝑡 3 − 𝑡 2 = −

𝑎 3 𝑡^1

and, further,

𝑡 3 = 𝑡 2 + ∆𝑡 3 = 𝑡 1 + ∆𝑡 2 + ∆𝑡 3 = 𝑡 1 + ∆𝑡 2 −

𝑎 3 𝑡^1.

This is the analytical or symbolic or algebraic answer or formula for the total time. In this formula, the result is expressed through the quantities given in the formulation of the problem (this has to be checked each time before submitting the solution for grading!). Now, substituting given numbers, one obtains

v

0 t 1 t 2 t 3 t

v 1

2 𝑎^1 𝑡^1

2 + ∆𝑡 2 𝑎 1 𝑡 1 +^1

2 𝑎^1 𝑡^1 (−

𝑎 3 𝑡^1 )

2 𝑎^1 𝑡^1

2 + ∆𝑡 2 𝑎 1 𝑡 1 −^1

𝑎 3 𝑡^1

2

that coincides with the result obtained by the first method.

4. Motion with constant acceleration

A car started moving from rest with a constant acceleration. At some moment of time, it covered the distance 𝑥 and reached the speed 𝑣. Find the acceleration and the time.

Solution. The formulas for the motion with constant acceleration read

where we have taken into account that the motion starts from rest (all initial values are zero). If 𝑣 and 𝑥 are given, this is a system of two equations with the unknowns 𝑎 and 𝑡. This system of equations can be solved in different ways.

For instance, one can express the time from the first equation, 𝑡 = 𝑣/𝑎, and substitute it to the second equation,

2

𝑣^2

From this single equation for 𝑎 one finds

𝑣^2

Also, one can relate 𝑥 to 𝑣 as follows

𝑥 =

2 𝑎𝑡 × 𝑡 =

After that one finds

𝑡 =

and, further,

𝑡 =^

𝑣^2

5. Tennis serve (Giancoli Chapter 3)

Solution. First, we define the coordinate axes and introduce missing notations. The origin of the coordinate system is at the server’s position, 𝑧-axis up and 𝑥-axis to the right. The initial height (serve height) 𝑧 0 = 2.5 𝑚, the height of the net 𝑧 1 = 0.9 𝑚, the height of the ground (the reference height) 0 𝑚, distance server-net 𝑥 1 = 15 𝑚. Find 𝑣0𝑥.

First, use the 𝑥- and 𝑧-formulas to find 𝑣0𝑥:

𝑥 = 𝑣0𝑥𝑡, 𝑧 = 𝑧 0 −

The instance of these general formulas corresponding to the ball passing just above the net reads

𝑥 1 = 𝑣0𝑥𝑡 1 , 𝑧 1 = 𝑧 0 −

2 𝑔𝑡^1

This is a system of two equations with two unknowns: 𝑣0𝑥 and 𝑡 1. The second equation is autonomous (contains only one unknown), so it can be solve to give

Then, from the first equation one finds

𝑡 1 = 𝑥^1 √^

Substituting the numbers into this formula, one obtains

𝑣0𝑥 = 15√^

Now we can find the distance from the server at which the ball lands. We use the instances of the general formulas above corresponding to the ball hitting the ground:

Solution in the laboratory frame. Put the origin of the coordinate system on the ground below the copter. The initial 𝑥-coordinate of the car is 𝑥𝑐,0. If it is found, then the angle 𝜃 can be expressed as

tan 𝜃 =

𝑥𝑐,^.

The formulas for the motion of the package and the car have the form

When the package lands into the car, the following conditions are fulfilled:

𝑧𝑝 = 0, 𝑥𝑝 = 𝑥𝑐.

Substituting these into the general equations, one obtains their instance

0 = ℎ −

This is a system of two equations with two unknowns. The first equation is autonomous and yields the fall time

Substituting this into the second equation, one obtains

Now for the angle one obtains

𝜃 = arctan (

2ℎ) = arctan (^

Substituting the numbers, one obtains

𝜃 = arctan (

√9.8 × 78

2 ) = arctan(0.3258) = 18°.

Solution in the moving frame (frame of the car). The absolute velocity of the copter can be represented as

𝑣 = 𝑣′^ + 𝑢,

where 𝑣′^ is the relative velocity of the copter with respect to the car, 𝑣′^ = 𝑣 − 𝑢. The origin of the coordinate axes in the moving frame, 𝑂′, is moving to the right with the velocity of

the car 𝑢. At 𝑡 = 0 the origins of the laboratory and moving frames coincide, 𝑂′^ = 𝑂. Thus the relation between the 𝑥-coordinate (absolute frame) and 𝑥′-coordinate (moving frame) is

𝑥 = 𝑥′^ + 𝑢𝑡

or, conversely,

𝑥′^ = 𝑥 − 𝑢𝑡

The formulas for the motion of the copter in this frame have the form

As for the car, it is at rest in its own frame:

𝑥′𝑐(𝑡) = 𝑥𝑐,0.

As in the first solution, one finds the fall time,

𝑡𝑓 = √

and substitutes it into the condition:

𝑥′𝑝(𝑡𝑓) = 𝑥′𝑐(𝑡𝑓)

or

The result for 𝑥𝑐,0 coincides with that obtained by the first method.

7. Targeting angle (projectile motion)

A cannon launches missiles with the initial speed 𝑣 0. Find the targeting angles 𝜃 to hit the target at the distance 𝑑 at the same height as the cannon.

Solution. The formulas for the projectile motion have the form

𝑧 = 𝑣0𝑧𝑡 −

The origin of the coordinate system is put at the location of the cannon, thus 𝑥 0 = 𝑧 0 = 0. The distance between the cannon and the landing point is defined by the fall time (or final time or flight time) 𝑡𝑓:

𝑑 = 𝑣0𝑥𝑡𝑓.

The time 𝑡𝑓 can be found from the first equation:

2 = 𝑡𝑓 (𝑣0𝑧 −^1

8. Hitting an elevated target (projectile motion, Giancoli, Chapter 3)

cos

sin

a <

Equation: sin   a

Solutions:

 1  arcsin a

^  2 ^ arcsin a

Solution. First, we introduce missing notations. The horizontal distance cannon-target 𝑑 = 195 𝑚, the height of the target ℎ = 155 𝑚, the missile flight time 𝑡𝑓 = 7.6 𝑠. Find: 𝑣 0 , 𝜃.

The general formulas for the projectile motion have the form

𝑧 = 𝑣0𝑧𝑡 −

The origin of the coordinate system is put at the location of the cannon, thus 𝑥 0 = 𝑧 0 = 0.

The instance of these formulas, corresponding to the problem’s formulation, is

ℎ = 𝑣0𝑧𝑡𝑓 −

From here, one finds the components of the initial velocity:

𝑡𝑓^ ,^ 𝑣0𝑧^ =

ℎ + 12 𝑔𝑡𝑓^2

𝑡𝑓^.

Now

𝑣 0 = √𝑣0𝑥^2 + 𝑣0𝑧^2 =

√𝑑^2 + (ℎ + 12 𝑔𝑡𝑓^2 )^2

and

𝜃 = arctan

= arctan

ℎ + 12 𝑔𝑡𝑓^2

Substituting the numbers, one obtains…

9. Car jumping (Projectile motion, Giancoli, Chapter 3)

Solution. First, we add missing notations. The horizontal distance 𝑑 = 20 𝑚, the initial height ℎ = 1.5 𝑚, the launching angle in (b) 𝜃 = 10°.

cos 𝜃 √^

2(ℎ + 𝑑 tan 𝜃).

For 𝜃 = 0, this formula simplifies to the solution obtained in (a). For small 𝜃, one can use

tan 𝜃 ≅ 𝜃, cos 𝜃 ≅ 1 −

(𝜃 in radians) so that the value of 𝑣 0 decreases with 𝜃 because if the tan 𝜃 term in the denominator. We have the angle, in radians,

𝜃 = 10°

so that the angle is, indeed, small, and one can use the formulas for the small angles above.

Substituting the numbers, one obtains

cos 10° √^

2(1.5 + 20 tan 10°) = 20 𝑚/𝑠.

This is a serious decrease of the minimal speed in comparison to the case 𝜃 = 0. The reason is that the small tan 𝜃 is multiplied by the large 𝑑.

10. Vertical motion with gravity ― full quadratic equation

A person standing on the edge of a cliff throws a rock straight upwards with an initial speed of 9 m/s. (a) Sketch a plot of velocity versus time and position versus time for the motion of the rock. (b) What will be the maximum height the rock reaches? (c) If the cliff stands at a height of 105 meters from the bottom of the ravine, how long will it take to reach the ground? (d) How fast will it be traveling when it reaches the ground?

Solution : (a) Making a sketch is a guarantee of success in problem solving.

v

t

moving up moving down

t max t f

b) Let us introduce missing notations. Initial velocity 𝑣 0 = 9 𝑚/𝑠, the height of the bottom of the ravine ℎ = −105 𝑚. The reference level is the edge of the cliff.

The time dependences of the velocity and displacement in the motion with constant acceleration 𝑎 = −𝑔 are given by the formulas

Finding the maximum of 𝑧(𝑡) directly from the second formula requires using the calculus. However, one can use the physical argument and point out that when the height reaches its maximum, the vertical velocity must vanish. Thus, from the first equation one obtains

0 = 𝑣 0 − 𝑔𝑡𝑚𝑎𝑥 ⟹ 𝑡𝑚𝑎𝑥 =

𝑔 =^

After that, one finds the maximal height from the height formula substituting 𝑡 ⇒ 𝑡𝑚𝑎𝑥, that is,

2 = 𝑣 0 𝑣^0

2

Substituting numbers, one obtains

2 × 9.8 = 4.1 𝑚

c) The time to reach the ground, that is, the fall time 𝑡𝑓, can be found from the height equation (1) substituting 𝑧 ⇒ ℎ:

ℎ = 𝑣 0 𝑡𝑓 −

This is a quadratic equation that can be rewritten into the canonical form

t

z

moving up moving down

t max tf

h < 0

z max