Physics Problem Solving Practice/Notes, Study notes of Physics

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Introductory Physics: Problems solving

D. A. Garanin

7 May 2026

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 225°, 𝑟 12 sin 225°) = (2000 (−

= (−1000√2, − 1000√2, ) m

(180° + 45° = 225°) 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. Average speed and average velocity

An object is moving from point A to point B, which are on the opposite apices of a square. The magnitude of the average velocity is v. What is the average speed s?

Solution. Let us denote the length of the side of the square as a. and the travel time as  t.

The distance between A and B is the length of the diagonal of the square 𝑑 = √2𝑎, while the trajectory or way length is 𝑤 = 𝑎 + 𝑎 = 2𝑎. According to the definitions,

∆𝑡 ,^ 𝑣 =

From the above one obtains 𝑠/𝑣 = √2, thus the result is 𝑠 = √2𝑣. (the condition 𝑠 ≥ 𝑣 is satisfied).

This solution shows the strength of the algebraic approach. The solution is general and works for any value of v. Also, to solve the problem, one does not need the numerical values of a and  t. It is sufficient to know that they exist.

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

𝑣 = 𝑎𝑡, 𝑥 =

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.

First method. 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

Now, one finds the time as

𝑡 =

𝑎 =^

𝑣^2 /(2𝑥) =

Second method. Also, one can relate 𝑥 to 𝑣 as follows

𝑥 =

𝑎𝑡 × 𝑡 =

After that one finds

𝑡 =

and, further,

𝑡 =^

𝑣^2

𝑎 3 𝑡^1.

This is the analytical or symbolic or algebraic answer or formula for the total time. (This result will not be used, however). 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

The preparatory work done, let us now find the total distance covered. Using the first method, we find it as the area under the curve 𝑣(𝑡) that consists of two triangles and one rectangle, see the figure. The parameters of them have been calculated above. So we write

∆𝑥 = ∆𝑥 1 + ∆𝑥 2 + ∆𝑥 3 =

2 𝑡^1 𝑣^1 + ∆𝑡^2 𝑣^1 +

2 ∆𝑡^3 𝑣^1.

Here we must substitute the expressions for the quantities that are not given in the problem formulation: 𝑣 1 and ∆𝑡 3. We prefer not to factor 𝑣 1 to keep the contributions of each interval separately. The result reads

∆𝑥 =

2 𝑎^1 𝑡^1

2 + ∆𝑡 2 𝑎 1 𝑡 1 +^1

𝑎 3 𝑡^1 ) 𝑎^1 𝑡^1

or, finally,

2 𝑎^1 𝑡^1

2 + ∆𝑡 2 𝑎 1 𝑡 1 −^1

𝑎 3 𝑡^1

This is our symbolic answer for the distances covered in the motion.

Substituting the numerical values from the problem’s formulation, one obtains

2 2 × 10

2 + 30 × 2 × 10 −^1

(−3)^10

Now, let us find the total distance covered using the formula for the displacement in the motion with a constant acceleration

∆𝑥 ≡ 𝑥 − 𝑥 0 = 𝑣 0 ∆𝑡 +

2

in the form appropriate to each of the motion intervals. One has

∆𝑥 = ∆𝑥 1 + ∆𝑥 2 + ∆𝑥 3 =

2 𝑎^1 𝑡^1

2 + 𝑣 1 ∆𝑡 2 + [𝑣 1 ∆𝑡 3 +^1

2 𝑎^3

(∆𝑡 3 )^2 ]

𝑎 1 𝑡 12 + 𝑣 1 ∆𝑡 2 + [𝑣 1 +

𝑎 3 ∆𝑡 3 ] ∆𝑡 3.

Substituting here the expressions for 𝑣 1 and ∆𝑡 3 , one obtains

∆𝑥 =

2 𝑎^1 𝑡^1

2 + ∆𝑡 2 𝑎 1 𝑡 1 + [𝑎 1 𝑡 1 +^1

2 𝑎^3 (−

𝑎 3 𝑡^1 )] (−

𝑎 3 𝑡^1 )

2 𝑎^1 𝑡^1

2 + ∆𝑡 2 𝑎 1 𝑡 1 + [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

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 (the red formula).

6. Rocket motion (1D)

A rocket starts vertically up and moves with the acceleration 20 𝑚/𝑠^2 during 20 seconds. Then it continues its motion ballistically. Find the maximal height reached and the corresponding time. Find the time of hitting the ground and the corresponding speed.

Solution. First, we introduce the notations: the duration of the first stage (powered motion) 𝑡 1 = 20 𝑠, the acceleration in the first stage 𝑎 = 20 𝑚/𝑠^2. The initial velocity is zero.

We choose the origin of time 𝑡 = 0 and put the origin of 𝑧-axis (directed up) at zero, so that the initial conditions are 𝑧 0 = 0 and 𝑣 0 = 0. The formulas for the motion with a constant acceleration at the first stage are

At the end of the first stage, 𝑡 = 𝑡 1 , the velocity and the height read

𝑣 1 = 𝑎𝑡 1 , 𝑧 1 =

These are the initial conditions for the motion on the second stage. The formulas for the motion with the constant acceleration −𝑔 on the second (ballistic) stage are

𝑣 = 𝑣 1 − 𝑔(𝑡 − 𝑡 1 ) 𝑧 = 𝑧 1 + 𝑣 1 (𝑡 − 𝑡 1 ) −

2 𝑔(𝑡 − 𝑡^1 )

Note that the second stage begins at 𝑡 = 𝑡 1 , thus we use the formulas with shifted time. The highest point can be found from the condition 𝑣 = 0 that yields the equation for the time at which the maximal height is reached:

0 = 𝑣 1 − 𝑔(𝑡𝑚𝑎𝑥 − 𝑡 1 ).

The solution is

𝑡𝑚𝑎𝑥 − 𝑡 1 =

𝑔 𝑡^1

and, finally,

𝑡𝑚𝑎𝑥 =

Substituting the numbers, one obtains

Now, the final velocity can be found from the velocity formula:

𝑔 𝑡^1 = −√(1 +

𝑔) 𝑎𝑔𝑡^1 = −√(𝑎 + 𝑔)𝑎𝑡^1.

Substituting the numbers, one obtains

𝑣𝑓 = −√(20 + 9.8)20 × 20 = 488 𝑚/𝑠.

7. 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:

𝑥 2 = 𝑣0𝑥𝑡 2 , 0 = z = 𝑧 0 −

2 𝑔𝑡^2

One finds 𝑡 2 from the second equation:

From this formula, one can find the numerical value of 𝑡 2 that is the total time of the motion. Substituting the formula for 𝑡 2 into the first equation, one obtains

𝑔 = 𝑥^1 √^

Substituting the numbers, one obtains

𝑥 2 = 15√^

Now 𝑥 2 − 𝑥 1 = 18.75 − 15 = 3.75 𝑚 that is well below 7 𝑚. Thus, the ball is “good”.

8. Dropping a package from a copter into a moving car (Giancoli Chapter 3)

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 package in this frame have the form

𝑧𝑝 = ℎ −

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:

Then

tan 𝜃 =

𝑥𝑐,0^ =^

2ℎ =^

Wherefrom one finds 𝜃.

9. 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

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

The first solution to this equation, 𝑡𝑓 = 0, corresponds to the beginning of the motion and should be discarded. The landing time nullifies the expression in the brackets,

2 𝑔𝑡𝑓^ = 0,

wherefrom

Now

𝑑 = 𝑣0𝑥𝑡𝑓 =

The components of the initial velocity can be expressed as

𝑣0𝑥 = 𝑣 0 cos 𝜃 , 𝑣0𝑧 = 𝑣 0 sin 𝜃,

so that

2𝑣 02 sin 𝜃 cos 𝜃 𝑔 =

𝑣 02 sin 2𝜃 𝑔 ,

where the trigonometric identity sin 2𝜃 ≡ 2 sin 𝜃 cos 𝜃 was used. As the maximal value of the sine function is 1 and it is reached for the argument equal to 90°, one can see that 𝑑 reaches its maximum for 𝜃 = 45°. One can rewrite

𝑑 = 𝑑𝑚𝑎𝑥 sin 2𝜃 , 𝑑𝑚𝑎𝑥 =

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

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

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 (hitting the target), is

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 the angle 𝜃 is the solution of the equation

tan 𝜃 =

𝑣 0 𝑥^.

This equation has only one solution

𝜃 = arctan

= arctan

ℎ + 12 𝑔𝑡𝑓^2

Substituting the numbers, one obtains…

11. 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°.

We put the origin of the coordinate system at the foot of the “cliff” (below the end of the takeoff ramp at the level of the roofs of the standing cars). The formulas for the motion with a constant acceleration have the form

𝑧 = ℎ + 𝑣0𝑧𝑡 −