Physics 1: kinematics and dynamics midterm

First-semester lectures started a little over a week ago — at the Politecnico di Milano on 14 September — and Physics 1 is already making the rules of the game clear: derivatives and vectors used for real, not as in high school. The first midterms (prove in itinere), where the course has them, fall between the end of October and the beginning of November (at the Politecnico, 29 October – 2 November): from here that is about five weeks. Every university has its own teaching calendar, so check the exact dates on your course’s website; here I give you the method, which holds whatever the exact day of the exam.

The first midterm covers almost everywhere the first part of the syllabus: kinematics of the particle and dynamics. The rest — work-energy, systems of particles, rigid body, oscillations, fluids, gravitation and thermodynamics — comes later, often in the second midterm or in the final written exam. Concentrate on what they are asking you now.

What changes compared with high school

In high school kinematics was done with direct formulas: $s (t) = s_{0} + v_{0} t + \frac{1}{2} a t^{2}$ , full stop. In Physics 1 that formula is not a starting point, it is a special case: the real definition is

v = \frac{d r}{d t} a = \frac{d v}{d t}

and the typical problem gives you $a (t)$ (not constant) and asks you to work back to $v (t)$ and $x (t)$ by integrating, with the initial conditions fixing the constants of integration. If you have not yet done integrals in Analisi 1 (Calculus 1), it is the weak point to invest in straight away: here you will find the dedicated treatment of integrals for Analisi 1, and if before that you lack confidence on limits there is limits explained well. If you arrive weak in calculus you arrive weak here too: they are not two separate subjects in the first semester, they are the same toolbox used twice.

The second new thing is that motions are almost always treated by components, in 2D. A motion in the plane is split into two independent motions along $x$ and $y$ (or radial and tangential, in the circular case), and they are solved separately before recombining the resulting vector with Pythagoras.

Where marks are lost

Inconsistent units. The dimensional check is the first filter, not the last: if you add metres and centimetres without converting, or if the final result for an acceleration comes out in $m/s$ , you have already gone wrong before looking at the number. Do it before substituting, not after writing the result.

Sign errors. Every dynamics problem begins with a stated frame of reference — an axis, a positive direction — and every force, velocity, acceleration has to be projected with respect to it. Changing convention halfway through the exercise (friction “seems” to oppose the motion so I write it negative everywhere, even when the motion reverses) is the single most frequent mistake in a dynamics test.

Reading graphs. In the $v$ - $t$ graph the area under the curve is the displacement and the slope is the acceleration; swapping the two meanings, or swapping $v$ - $t$ with $x$ - $t$ , produces plausible but wrong answers.

Understanding the text before applying formulas. “The block starts from rest”, “the string is inextensible”, “friction is negligible”: each condition in the text eliminates a term from the equation or fixes an initial condition. Whoever reads the problem as a list of numbers to slot into a formula, skipping the conditions, gets the set-up wrong even when the calculations are right.

Circular motion: the piece that surprises most

In high school uniform circular motion only needed to be known by heart: $a_{c} = v^{2} / r$ , directed towards the centre. In Physics 1 the non-uniform case also arrives, where the magnitude of the velocity changes in time, and the acceleration has two independent components:

a_{t} = \frac{d v}{d t} a_{c} = \frac{v ^{2}}{r}

the tangential one (it changes the magnitude of the velocity) and the centripetal one (it changes the direction). They are perpendicular to each other, so the total acceleration is found with Pythagoras: $a = a_{t}^{2} + a_{c}^{2}$ . Forgetting the tangential component when the text says “it is accelerating along the path” is a mistake I often see in those who learned circular motion only in its uniform version.

On relative motion — changing frame of reference between a stationary observer and a moving one — the part you need for the first midterm is usually just the composition of velocities, $v_{A} = v_{A / B} + v_{B}$ : nothing else is needed at this level, the rest (non-inertial frames, apparent forces) comes later.

Dynamics: the free-body diagram first of all

You already know Newton’s three laws. What changes in Physics 1 is the discipline with which you apply them: first the free-body diagram, then the equations. You isolate the body, draw every force acting on it (never the forces it exerts on others), project along the axes you have chosen, and only then write $\sum F_{x} = m a_{x}$ , $\sum F_{y} = m a_{y}$ .

Friction follows two different regimes, and it is a point where easy marks are lost: as long as the body is at rest, static friction equals whatever is needed to keep it at rest, up to a maximum $f_{s} \leq μ_{s} N$ ; when the body slides, kinetic friction has a fixed value, $f_{d} = μ_{d} N$ . They are not the same formula applied twice: they are two different physical regimes, and the text of the problem usually tells you explicitly which one to use, with “the block is at rest” or “the block slides”.

Four worked examples

1. Equation of motion from a non-constant acceleration

A particle moves along an axis with acceleration $a (t) = (6 - 2 t) m/ s^{2}$ . At time $t = 0$ it has velocity $v_{0} = 2 m/s$ and position $x_{0} = 0$ . Find $v (t)$ , $x (t)$ and their values at $t = 3 s$ .

I integrate the acceleration once for the velocity:

v (t) = v_{0} + \int_{0}^{t} a (t^{'}) d t^{'} = 2 + 6 t - t^{2}

and the velocity a second time for the position:

x (t) = x_{0} + \int_{0}^{t} v (t^{'}) d t^{'} = 2 t + 3 t^{2} - \frac{t ^{3}}{3}

At $t = 3 s$ :

v (3) = 2 + 18 - 9 = 11 m/s x (3) = 6 + 27 - 9 = 24 m

Note the mechanism: each constant of integration (here $v_{0}$ and $x_{0}$ ) is fixed by the initial conditions, not left to chance. It is exactly the same procedure you see in Analisi 1 when you work back from a derivative to the starting function.

2. Projectile motion from a height, with an angle

A projectile starts from a height $h = 20 m$ with initial speed $v_{0} = 25 m/s$ , inclined at $30°$ above the horizontal. Calculate the time of flight, the horizontal range and the speed on reaching the ground (use $g = 9.8 m/ s^{2}$ ).

Axes: $x$ horizontal in the direction of motion, $y$ vertical upwards, origin at ground level below the launch point. The initial components:

v_{x} = v_{0} cos 30° \approx 21.65 m/s v_{y 0} = v_{0} sin 30° = 12.5 m/s

The horizontal motion is uniform, the vertical one uniformly accelerated — they are independent. The projectile hits the ground when $y (t) = 0$ :

h + v_{y 0} t - \frac{1}{2} g t^{2} = 0 ⟹ 4.9 t^{2} - 12.5 t - 20 = 0

Solving (I discard the negative root):

t = \frac{12.5 + 12. 5 ^{2} + 4 \cdot 4.9 \cdot 20}{2 \cdot 4.9} \approx 3.66 s

Range: $x = v_{x} t \approx 21.65 \cdot 3.66 \approx 79.3 m$ .

Vertical velocity on arrival: $v_{y} = v_{y 0} - g t \approx 12.5 - 9.8 \cdot 3.66 \approx - 23.4 m/s$ (negative: it is going down). The arrival speed, by Pythagoras on the two perpendicular components:

v = v_{x}^{2} + v_{y}^{2} \approx 21.6 5^{2} + 23. 4^{2} \approx 31.9 m/s

3. Two connected blocks on an inclined plane with friction

A block of mass $m_{1} = 4 kg$ rests on an inclined plane of $θ = 30°$ , with coefficient of kinetic friction $μ = 0.2$ between block and plane. An inextensible string passes over an ideal pulley at the top of the plane and connects $m_{1}$ to a mass $m_{2} = 6 kg$ hanging vertically. Find the acceleration of the system and the tension in the string, knowing that $m_{2}$ descends and drags $m_{1}$ up the plane.

A separate free-body diagram for each mass. For $m_{2}$ (positive axis downwards, in the direction of motion):

m_{2} g - T = m_{2} a

For $m_{1}$ (positive axis along the plane, in the direction in which $m_{1}$ is pulled, that is upwards): the tension pulls up, the component of the weight along the plane and the friction — which opposes the motion, so points down the plane — pull down:

T - m_{1} g sin θ - μ m_{1} g cos θ = m_{1} a

Adding the two equations (the inextensible string imposes the same $a$ on both masses, and $T$ cancels):

a = \frac{m _{2} g - m _{1} g sin θ - μ m _{1} g cos θ}{m _{1} + m _{2}}

Numerically: $m_{2} g = 58.8 N$ , $m_{1} g sin θ = 19.6 N$ , $μ m_{1} g cos θ \approx 6.79 N$ :

a = \frac{58.8 - 19.6 - 6.79}{10} \approx 3.24 m/ s^{2}

And from the first equation, $T = m_{2} (g - a) = 6 \cdot (9.8 - 3.24) \approx 39.4 N$ .

Sense check: $a$ is positive and less than $g$ , as it must be in a system with two connected masses; $T$ is less than the weight of $m_{2}$ (58.8 N), consistent with the fact that $m_{2}$ is accelerating downwards and is not in free fall.

4. Non-uniform circular motion: total acceleration

A point moves on a circle of radius $r = 0.5 m$ . At a certain instant it has speed $v = 4 m/s$ and tangential acceleration $a_{t} = 2 m/ s^{2}$ (it is accelerating along the path). Find the total acceleration.

Centripetal component:

a_{c} = \frac{v ^{2}}{r} = \frac{16}{0.5} = 32 m/ s^{2}

The two components are perpendicular, so the total magnitude is:

a = a_{c}^{2} + a_{t}^{2} = 3 2^{2} + 2^{2} \approx 32.1 m/ s^{2}

Note the ratio: here $a_{t}$ weighs very little compared with $a_{c}$ (the angle between the total vector and the radial direction is only $3.6°$ ). It is a good reminder that in circular motion, even when the speed is changing in magnitude, the centripetal acceleration dominates as soon as the speed is not very small.

How to tell whether you are ready

The most honest way to know is not to reread your notes, it is to set a stopwatch on problems that resemble those in the midterm. You will find a Physics 1 exam simulation — the correct answer is free, the guided step-by-step solution unlocks with the first lesson. If an exercise blocks you halfway, try to identify where: is it the free-body diagram (you cannot isolate the forces), is it the sign (you changed convention halfway), or is it purely the calculation (integrals, systems)? They are three different problems and have to be treated differently.

If your course goes on to the dynamics of systems and Lagrangian mechanics, that is second-year material: I wrote about it separately in rational mechanics and the exam with Lagrange. For now, concentrate on kinematics and dynamics of the particle: they are the basis on which the whole rest of the course is built.

Circular motion, if you want a text with more exercises, I covered in centripetal force and fictitious forces. The exam that usually comes next, on electricity, is in Physics 2: electric field and Gauss’s law.

If you want a hand

I work online on Physics 1 always starting from the hand-drawn picture of the problem — axes, forces, constraints — because that is where most mistakes are born, long before getting to the calculations. You will find the general picture on the physics page and the details of the university path on the page about university tutoring. If you want to start with a diagnosis, write to me via contact: send me the exercises from your tutorials and the date of the test, and I will tell you honestly what needs doing between now and then.

Physics 1: kinematics and dynamics for the midterm