# Newton's Laws ## Newton's First Law >[!Tip] If there is no net force on a body, an object will not experience a change to its motion.] >$ F_{net} = 0$ >For example, if an object is already at rest, it will remain at rest. ## Newton's Second Law >[!Note] The net force on a body can be related with the mass and acceleration of the body through the equation] >$F_{net} = m\overrightarrow{a}$ >Where $\overrightarrow{a}$ is the magnitude of the acceleration vector containing the $x, y, z$ components of the object's acceleration. >This is a direct extension of the proper form of this law, characterised as $F = \frac{dp}{dt}$ >A commonly used variation of this law is with defining weight. >$W = mg$ >Where g is the gravitational acceleration on the surface of a body. ## Newton's Third Law >[!Abstract] If there is an action force there will be a reaction force directly opposing the change in motion.] >We can see this in frictional or drag forces, where a virtual 'normal' force is created to oppose a forward-moving force, such as the gravity of a skydiver accelerating them towards the ground. >$\overrightarrow{F_{ab}} = -\overrightarrow{F_{ba}}$ >Take two points A and B. If there is a force between the two and a point lies on a hard surface we can get a generalisation of this law as a *normal* force! >%%diagram%% # The SUVAT Equations There are a number of equations which help define and relate functions of displacement, velocity and acceleration to each other. This allows for a more streamlined solving on challenging mechanics problems. Equations and derivations listed below. >[!EXAMPLE]- The SUVAT Equations >These are the critical equations of motion: >$v = u + at$ >$v^2 = u^2 + 2as$ >$\Delta s = ut + \frac{1}{2}at^2$ >$a = \frac{u+v}{2}$ >Where a is the acceleration, u is the **initial velocity**, v is the **final velocity**, s is the displacement and t is the time. >[!TIP]- Using these equations >These equations only apply if there is a constant acceleration over a certain period of time. Kinematics must be used to extrapolate required data if acceleration is variable. ## Equation Derivations >[!Abstract]- Equation 1 $(v = u + at)$ >Imagine a velocity time graph. This is a general plot of the velocity ($v$) against the time ($t$) > >The gradient of a resulting line therefore **has** to be $a$, the acceleration. Therefore, the final velocity $v$ must be equal to the initial velocity $u$ plus the **increase in the value of velocity** given by $at$. This gives us our equation! >$v = u+at$ >[!Abstract]- Equation 2 ($v^2 = u^2 + 2as$) >We can now define the time $t$ as a function: >$t = \frac {v-u}{a}$ >Now we can substitute this value into the third SUVAT equation, giving us: >$\Delta s = u(\frac{v-u}{a}) + \frac{1}{2}a{(\frac{v-u}{a})}^2$ >We therefore get: >$2 \Delta s = 2u(\frac{v-u}{a}) + a{(\frac{v-u}{a})}^2 $ >We can let the change in $s$ be a change from a constant origin to create: >$2s = \frac{2uv - 2u^2}{a} + a\frac{v^2 - 2uv - u^2}{a^2}$ >$2as = 2uv - 2u^2 + v^2 - 2uv + u^2$ >We can then cancel the terms... >$2as = v^2 - u^2$ >Giving us the expression: $v^2 = u^2 + 2as$ >[!Abstract]- Equation 3 ($\Delta s = ut + \frac{1}{2}at^2$) >Integrate the velocity curve with respect to $t$. >$\int{v}\,dt = \int{u + at}\,dt = ut + \frac {1}{2}at^2 = s$ This can be done as the area under the curve is the displacement. >[!Abstract]- Equation 4($a = \frac{v-u}{t}$) >Remember that the acceleration is given as the gradient of the speed-time graph we create. As a result, the value of acceleration can just be defined as the change in velocity per unit time. $v-u$ gives us the first part of the equation so all we have to do is divide by time. > >In addition, the acceleration $a$ can be rewritten as >$a = \frac{\Delta v}{\Delta t}$ [^1]: No example questions as these should all be givens by this point. For most of them, you could imagine a velocity-time graph and its' different attributes: ![[Pasted image 20230908162656.png]] See how easy this makes it? Graphs always help! ## Equations of Motion FOR POWER - extended $v=\sqrt[3]{ 3\times \frac{lm_{a}P}{m_{b}} }$ # Projectile Motion %%diagram REQUIRED%% Sometimes, we won't be able to use the SUVAT equations above to model the motion of an object. For example, an object may be travelling in an arc. This creates a secondary branch for us to deal with, **projectile motion**, or the study of how particles travel in linear arcs/hyperbolas. [[Conics - Circles, Parabolas & More! (Maths)]] >[!Note]- Projectile Motion Variables >We can think of an object as moving in a triangle with the magnitude of the hypotenuse given as the base value. The initial speed of the particle can be described with the variable $v_{0}$. As a result, the vertical and horizontal components of the projectile (given as $v_x$ and $v_y$ respectively) can be given as: >$u_{x} = v_{0}\cos{\theta}$ >$u_{y} = v_{0}\sin{\theta}$ Where $\theta$ is the angle from the ground/base of which the projectile is launched and $v_{0}$ is the initial speed of the projectile. Whenever a question tells you that a projectile's been launched a certain angle, you can plot a right triangle to visualise the individual components stated above. The acceleration $a$ always points downwards towards the centre of the Earth or any other object. >[!Abstract]- Projectile Motion Laws >With 1-Dimensional acceleration (that is, the horizontal component stays the same), we can create laws such that an object in projectile motion can be modelled: >1. $x_{t} = x_{0} + u_{x} t$ >2. $v_{y} = u_{y} t + at$ >3. $v_{x} = u_{x} t$ >4. $y_{t} = y_{0} + u_{y} t - \frac{1}{2}gt^2$ >5. $v_{{y_{t}}} = u_{y} - gt$ >6. $t = \frac{x}{u_{x}} = \frac{x}{u\cos \theta}$ >Substituting the expression for $t$ into the third equation gives: >$y = (\tan \theta) x - \frac{1}{2}g\frac{x^2}{u_{x}^2}$ >Which is a parametric generalisation for projectile motion. **Where:** - $g$ is the gravitational field strength, which is the acceleration of the object. (It acts downward, leading to a vertical deceleration) - $y_t$, $x_t$ are the vertical and horizontal positions of the object after time $t$ respectively. - $v_{0_{y}}$, $v_{0_{x}}$ are the starting velocities of the object in the $y$ and $x$ coordinates respectively. - $v_{y_{t}}$, $v_{x_{t}}$ are the velocities of the $y$ and $x$ components at a specific time $t$ - $x$ is the total horizontal distance the parabola encloses. For example, if I were to use equation 4 I'd probably be making an equation! Moving on, a couple things we'd typically want to find from projectile motion would be the apex (the highest point) of the object's motion as well as the linear distance it travels if launched from a certain angle. Sometimes, we may even want to maximise the horizontal distance traveled, such as doing shot puts. >[!Abstract]- Finding when the highest point occurs >The vertical velocity at the maximum point is going to be 0 - the initial vertical impulse we give a particle is going to be nullified by the gravity at the point! We can rewrite equation 4 as: >$0 = v_{0}\sin \theta - gt$ >Giving us: $t_{max} = \frac{v_{0}\sin \theta}{g}$ Now, we need to find the highest point, $h$, at the time $t_{max}$. >[!Abstract]- Finding the highest point >We can simply rewrite the highest point as a function of when $y$ is at the time $t_{max}$. >We can thus rewrite the equation >$ h = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^2 $ treating $y_0$ as 0 and $h$ represents any arbitrary point. >This gives us $h = \frac{(v_{0}\sin \theta)^2}{g} - \frac{1}{2}g \frac{(v_{0}\sin \theta)^2}{g^2}$ >Reducing further gives us: >$h = \frac{(v_{0}\sin \theta)^2}{2g}$ Onto finding the horizontal distance travelled if the ball is launched from the same altitude as the ground, given by the variable $x_t$ >[!Abstract]- Finding the horizontal distance travelled >The time taken for the ball to land can just be given as two times the expression for the time taken to reach the highest point, as projectile motion assumes that there is no air resistance. Therefore, it takes the same amount of time to descend from the highest point back down to the origin. >Therefore, the time taken for the ball to land, $t_{total}$, is: >$t_{total} = \frac{2v_{0}\sin \theta}{g}$ >Now, to find the distance, we can use Equation 1 and substitute our expression for $t$. >$x_{t} = \frac{2v^2_{0}\cos \theta \sin \theta}{g} = \frac{v^2_{0}\sin{2\theta}}{g}$ >^[: Double Angle Identity! Check the Trig Identities Note for more: [[Trig Identities - Derivations DLC Ver. (Maths)]].] >[!Abstract]- Object moving vertically upwards >This means $\theta$ can be given as $90\degree$, so $\sin\theta$ is equal to 1. >This creates the identity for $h$: >$h = \frac{v^2_{0}}{2g}$ ## Extra Projectile Motion - The Safety Parabola ![[Pasted image 20230910203841.png]] Let's double back to our parametric equation for a second: $y = x\tan \theta - \frac{1}{2}g\frac{x^2}{(u\cos \theta)^2}$ This isn't a fun equation to deal with - we can't really get much value out of it as there are so many different variables all at once. Let's try to remedy this by turning the $\cos$ into a $\sec$ term: $y = x\tan \theta - \frac{\frac{1}{2}gx^2\sec^2\theta}{u^2}$ [[Trig Identities - Derivations DLC Ver. (Maths)|There's a trig identity for this]] - it's going to allow us to write everything as a function of $\tan$ - giving us equal trigonometric terms: $y = x\tan \theta - \frac{1}{2}\frac{gx^2}{u^2} - \frac{1}{2}\frac{gx^2}{2u^2}\tan^2\theta$ This is a double quadratic for both the $\tan \theta$ and $x$ terms. Letting the equation equal to 0. we get: $0 = (gx^2)\tan^2\theta - (2xu^2)\tan \theta + ({gx^2 + 2u^2y})$ Let $\tan \theta$ be equal to an arbitrary value $t$. This gives us a range of values where the projectile parametrisation is real [[Polynomials Walkthrough (Maths)#The Discriminant|using the discriminant]], namely: $(-2xu^2)^2 > 4(gx^2)(gx^2 + 2u^2y) $ The value of the parabola has no real roots if the term to the right is greater than the term to the left. Any constraints under this will have real roots - so we set up our inequality as: $4x^2u^4 < 4g^2x^4 + 8u^2ygx^2$ $x^2u^4 < g^2x^4 + 2u^2gx^2y$ $u^4 < g^2x^2 + 2u^2gy$ The case where the parabola is equal gives us the maximum possible value for a parabola - giving us: $u^4 = g^2x^2 + 2u^2gy$ $\frac{u^4 - g^2x^2}{2u^2g} = y$ $y = \frac{u^2}{2g} - \frac{gx^2}{2u^2}$ This is our *safety parabola* - the only possible trajectories for a body with an initial speed $v$ and travel time $t$. ## Example Questions >[!Example]- Example Problem: Module 4-4 Problem 23 >A projectile is fired horizontally from a gun that is 45.0 m above flat ground, emerging from the gun with a speed of 250 m/s. (a) How long does the projectile remain in the air? (b) At what horizontal distance from the firing point does it strike the ground? (c) What is the magnitude of the vertical component of its velocity as it strikes the ground? > >**a) Finding airborne time** >To do this, we can use the modified version of the third SUVAT equation: >$y_{t} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^2$ >Substitute the known values into the equation, where $y_0$ is 45 and $y_t$ is 0 (as the ball hits the ground). You should get: >$5t^2 = 45$ >I'm going to pull the number one "poorly-written textbook" cardinal sin here - do some simple algebra to get: >$t = 3$ >**b) Finding the horizontal displacement** >Now that we have the time, we can just throw the values we already have for time and the initial horizontal speed to get the horizontal displacement using the equation: >$x_{t} = x_{0} + v_{0_{x}}t $ >$x_{t} = 0 + 250(3)$ >$x_{t} = 750 m/s$ >**c) Finding the vertical component** >We can use the variables we have gotten from doing the first two questions and plug them into equation 4: >$v_{y_{t}} = v_{0_{y}}-gt$ >$v_{y_{t}} = -10 (3)$ >However, the question is asking us for the **magnitude** of the downwards force, so we just get a value of $30$. >[!Example]- Example Problem: Module 4-4 Problem 26 >A stone is catapulted at time $t = 0$, with an initial velocity of magnitude 20.0 m/s and at an angle of 40.0° above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at $t = 1.10 s$? Repeat for the (c) horizontal and (d) vertical components at $t = 1.80s$, and for the (e) horizontal and (f) vertical components at $t = 5.00 s$. >**Solution:** >**a) Finding the Horizontal component:** >Just $\cos$ it! >$20\cos 40^\circ = 15.3 m/s$ >Multiply by the time to get the horizontal displacement: >$15.3 \times 1.10 = 16.8m$ >**b) Finding the vertical component:** >Just $\sin$ it! >$20\sin_{40}^\circ = 12.9 m/s$ >Given that the starting position is 0 we can use our revised version of $s = ut + \frac{1}{2}gt^2$ to get the vertical displacement: >$s = 12.9 \times 1.1 - \frac{1}{2}(9.8)(1.1)^2 = 8.3m$ >And we just have to repeat it for other times! >[!Example]- Example Problem: Module 4-4 Problem 31 >In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the spike is difficult. Suppose a ball is spiked from a height of 2.30 m with an initial speed of 20.0 m/s at a downward angle of 18.00°. How much farther on the opposite floor would it have landed if the downward angle were, instead, 8.00°? > >**Solution:** > >The downward spike means that the ball has been hit from the highest point possible. We can treat the parabola from which the ball is hit as half of a larger one, with the initial displacement from the ground being zero. >We can find the horizontal velocity at both $18^\circ$ and $8^\circ$: >$20\cos{-18} = 19.0m /s$ >$20\cos{-8} = 19.8 m /s $ >We can find the vertical velocity for both parabolas: >$20\sin{-18} = -6.18 m / s$ >$20\sin{-8} = -2.78 m /s$ >Divide the vertical component values by the height to get the total airtimes for each parabola which we can multiply with the horizontal velocity: >$t_{18} = \frac{2.3}{6.18} = 0.372s$ >$t_{8} = \frac{2.3}{2.78} = 0.827s$ >$s_{18} = 19 \times 0.372 = 7.07m$ >$s_{8} = 19 \times 0.827 = 15.7m$ >Subtract to get: >$\Delta s = 8.63m$ >[!Example]- Extra Problem: HRK Module 3 Problem 60 >A water hose is used to fill a large cylindrical storage tank of diameter $D$ and height $2D$. The hose shoots the water at an angle of 45 $^\circ$ from the same level as the base of the tank and is a distance $6D$ away. For what *range* of launch speeds will the water enter the tank? Ignore air resistance and express your answers in terms of $D$ and $g$. > >**Obligatory Diagram:** > >![[Pasted image 20230915220427.png]] > >**Solution (1):** *Answer donated by Chloe* >It's a SUVAT equation! We'll use the third SUVAT equation to work it out... >$h = v_{0}\sin (45)t - \frac{1}{2}gt^2$ >Note that we can get the horizontal component of this motion with the expression $v_{0}\cos 45 t$. Let's find the lower bound of this motion first - with the numerical value $6D$: >$v_{0}\cos 45 t = 6D$ >$\frac{6D}{t} = v_{0}\cos 45$ >Substitute this back into the original equation! Remember that our vertical displacement is going to remain $2D$: >$2D = \frac{6D}{t}t - \frac{1}{2}gt^2$ >$2D = 6D - \frac{1}{2}gt^2$ >$\sqrt{ \frac{8D}{g} } = t$ >Now plug this back into the original equation we've made. We can finally isolate our first value for $v_0$: >$v_{0} = \frac{\sqrt{ 2 }6D}{\sqrt{ \frac{8D}{g} }}$ >$v_{0} = \sqrt{ \frac{72Dg}{8} }$ >$v_{0} = 3\sqrt{ Dg }$ >**Solution (2):** >We're going to have to use the parametric equation in the "Laws" callout to solve this one! Remember that it is: >$y = x\tan \theta - \frac{1}{2} \frac{gx^2}{v_{0}^2\cos \theta}$ >The vertical displacement remains the same throughout the range - so $y$ is 2D. The angle from launch is also 45$^\circ$ - meaning we can get rid of the $\tan \theta$ as it will equal 1: >$2D = x - \frac{1}{2} \frac{g}{v_{0}^2\cos \theta}x^2$ >This actually is a quadratic equation - albeit with a really, really annoying coefficient! > > # Pressure and Density Pressure is the magnitude of force per unit area of impact. So - imagine a hydraulic press, the object being squished by it would be under **pressure.** $P = \frac{F}{A}$ As a result, the unit for pressure is given by $\frac{N}{m^2}$, or the **pascal**. We write that using the letters *Pa*. >[!Success]- Side Note: Atmospheric Pressure >Sometimes, questions will ask you to factor atmospheric pressure into your calculations. Rigidly speaking, this is given by a value of 1013 hectopascals, but most questions would accept a rounded value of $1.0 \times 10^5$ Pascals, or *Pa*. ## Density Commonly denoted by the value $\rho$, and is a representation of an object's mass against its volume. We usually compare this value to that of water, which sits at a nice 1,000 kilograms per metre cubed. $\rho = \frac{M}{V}$ For an object submerged in a liquid, we take the net pressure of the liquid above it as: $P = \rho gh$ Where $h$ is the height from the surface of the fluid and $\rho$ is the density of the fluid. # What's next? Basics done! Remember - feedback appreciated. For the next instalment in the mechanics' chapter, here: [[Power, Work Done and Energy]] To navigate back to the physics menu, click here: [[The (Incomplete) Physics Almanac]] For more mechanics, click here: [[Mechanics - A Contents Page]] To check out some of the prerequisite maths, click here: [[Maths Contents Page]]