# Newtonian Momentum Ever wondered why some people refer to fast-moving objects (like a wrecking ball!) as having lots of "momentum"? That's because momentum is, you guessed it, another physical construct created by the man himself, Isaac Newton. We define momentum as just the ==mass times the velocity== of the object in question. $ \overrightarrow{p} = m\overrightarrow{v}$ Taking $\overrightarrow{p}$ as the momentum. Note that momentum is also conserved - make sure you save this down! This rule dictates that the initial momentum of a system must equal the final momentum of the system, where the system is a closed loop (meaning that there are no background influences). Generalising this gives us: $\overrightarrow{p_{i}} = \overrightarrow{p_{f}}$ There exists other methods to describe Newton's Second Law - through the original definition of Newton's work, we come to know that Newton had actually used a differentiation of momentum to come to his conclusion. History bomb! > The time rate of change of the momentum of the particle is equal to the net force acting on the particle. \- Isaac Newton Or more simply: $\overrightarrow{F} = \frac{d\overrightarrow{p}}{dt}$ Easy as that! ## Impulse %%diagram%% Impulse is the change in momentum over a set change in time $\Delta t$. Remember that we have said that momentum can only change if there is a net external force acting on it (as can be seen in the revised version of Newton's Second Law!). Therefore, in the event that only one collision occurs between two particles, we can write the change in their momentums as: $\overrightarrow{I} = \Delta \overrightarrow{p} = \overrightarrow{F_{avg}}\Delta t$ This leaves the question - what happens if this force changes over the duration it acts on the other body? In that case, the tried-and-tested method of integrating the force $\overrightarrow{F}(t)$ with respect to time can be used. >[!Abstract]- Integrating for the impulse >We can rearrange Newton's Second Law regarding momentum to get a differential equation: >$dp = F(t)dt$ >Remember that this $d$ represents an instantaneous change in the values of both momentum and time. As a result, we can just integrate on both sides to get the net change in the momentum, taking the time before the collision as the variable $t_i$ and the variable after the collision as $t_f$. >$\int_{t_{i}}^{t_{f}} dp \, dx = \int_{t_{i}}^{t_{f}}F(t)dt$ >The right side of the equation gives us two values for the momentum, the initial momentum $\overrightarrow{p_i}$ minus the final momentum $\overrightarrow{p_{f}}$, therefore giving us the change in momentum. We can therefore rewrite the equation as: >$\overrightarrow{I} = \int_{t_{i}}^{t_{f}}F(t)dt$ >As a result, we have proven that the change in momentum is the impulse. >Typically, if given the force function $F(t)$ we'd be able to evaluate the integral, or by finding the area between the curve and the axis. Sometimes, we'll be given the average magnitude of the force acting on the object, which will enable us to use the first formula above: $\overrightarrow{I} = F\Delta t$ What we've done so far only really concerns discrete collisions between two particles - that is, one collision that we just monitor. However, you may realise that collisions typically aren't limited to being one-shot wonders - instead, many collisions may happen in a short period of time; perhaps a car crash on an icy road's caused a large-scale pileup. So for this specific study, imagine we've got a big roadblock in front of us on the highway and for some reason or another everybody has severe tunnel vision. There is a steady stream of cars now impacting that barrier. How do we find the change in momentum? We can take the forces of each of the cars as an average, $F_{avg}$, where the cars all have a uniform mass and velocity. Therefore, if the change in momentum for one car is $\Delta p$, we can take the total change in momentum of all the cars as: $I = -n\Delta p$ Where $n$ is the number of collisions against the wall, showing us that the change in momentum is in the opposite direction as the collision. Using this knowledge, we can also find the average force as shown below: >[!Abstract]- Finding the average force >Remember that the formula for impulse is $I = F_{avg}\Delta t$. Rearranging to make $F_{avg}$ the subject gives us $F_{avg} = \frac{I}{\Delta t}$. >We can substitute the secondary equation for impulse given above to get: >$F_{avg}=-\frac{n}{\Delta t}\Delta p$ Collisions at this stage are considered **elastic**, where there is no loss in kinetic energy. As a result, the new velocity of the cars may be considered as just the negative value of the original velocities. [^1] Therefore, the change in velocity can be given with just $2v$, where $v$ is the original velocity. --- [^1]: Very improbable, yes, but I feel modelling a realistic crumple-collision would take pints of blood, sweat and tears to do. If I ever get around to this, I'll update this footnote accordingly! This might be a good spot to end it. Check out our [[Newton's Laws and SUVAT#Example Questions|example questions]] for this topic to brush up on knowledge! ## Collisions and Momentum ### Inelastic Collisions %%diagram (perfect vacuum)%% During collisions, you may hear a loud "bang" - this is a sign that an **inelastic collision** has taken place. This is when the net kinetic energy changes across the system, meaning that some energy is lost as sound or heat. As the total momentum is equal on both sides, we can write this as: $\overrightarrow{p_{1i}} + \overrightarrow{p_{2i}} = \overrightarrow{p_{1f}} + \overrightarrow{p_{2f}}$ This can therefore be given as: $m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{1}v_{2f}$ >[!Abstract]- Momentum of Combining Objects >Let's take the case where the first object is stationary. When the two objects join together, they will end up having an equal momentum - giving us: >$m_{1}v_{1i} = (m_{1}+m_{2})V$ >Isolate the $v$ to get: >$V = \frac{m_{1}}{m_{1}+m_{2}}v_{1i}$ >Now let's take a special case where two objects join together even when the second particle is moving. Since the two objects move with an equal velocity $V$ once they join together, we can rewrite the equation above: >$m_{1}v_{1i} + m_{2}v_{2i} = (m_{1}+m_{2})V$ >We can isolate the V, giving us: >$\frac{m_{1}v_{1i} + m_{2}v_{2i}}{m_{1}+m_{2}} = V$ >$V = \frac{m_{1}}{m_{1}+m_{2}}v_{1i}$ >[!Abstract]- Velocity at the Centre of Mass >Since we have taken the particles to be colliding in a closed system, there cannot be a net force to change the velocity of the particle's centre of mass, $V_{com}$. >Remember that the total momentum is conserved after the collision. As a result, we can write the total momentum $\overrightarrow{P}$ as: >$\overrightarrow{P} = \overrightarrow{p_{1i}} + \overrightarrow{p_{2i}}$ >We can also write the total momentum of the two-body system as: >$\overrightarrow{P} = (m_{1i} + m_{2i})\overrightarrow{v_{com}}$ >We can substitute the expression above into the expression below to get: >$\overrightarrow{v_{com}} = \frac{\overrightarrow{P}}{m_{1}+m_{2}}$ ### Elastic Collisions In an elastic collision, kinetic energy is actually preserved - which can't happen in a non-vacuum, so you have questions which ask you to treat the area at which such a collision happens as a perfect vacuum. Even then, most collisions, such as a baseball bat hitting a baseball, can be approximated to an elastic collision, given the difference between the kinetic energies are so small. In other words, the initial kinetic energy is equal to the final kinetic energy. >[!Abstract]- Finding the final velocities for stationary targets >Considering the collision is elastic, the net kinetic energy is going to equal to zero. As a result, we can take two masses with discrete masses $m_1$ and $m_2$, where $m_1$ is the impactor, and rewrite it as: >$\frac{1}{2}m_{1}v^2_{1i} = \frac{1}{2}m_{1}v^2_{1f} + \frac{1}{2}m_{2}v^2_{2f}$ >Conservation of momentum also applies so the equation can also be written as: >$m_{1}v_{1i} = m_{1}v_{1f} + m_{2}v_{2f}$ >We can reorganise this equation to give Equation 1: >$m_{1}(v_{1i}-v_{2f}) = m_{2}v_{2f}$ >Reorganising the kinetic energy equation (by factorising the $m_1$ terms out in case you haven't caught on!) gives us another equation: >$m_{1}(v^2_{1i} - v^2_{1f}) = m_{2}v^2_{2f}$ >Now divide the two equations above with each other. This will yield us expressions for the final velocities of each mass: >$v_{1f}=\frac{m_{1}-m_{2}}{m_{1}+m_{2}}v_{1i}$ >$v_{2f}=\frac{2m_{1}}{m_{1}+m_{2}}v_{1i}$ >However, we do have a couple other collisions that we need to look at! Say, if the **masses of the two objects were equal**, the final velocity of $m_1$ would be 0 while the velocity of $m_2$ would be the initial velocity of $m_1$, $v_{1i}$. >Say we were hitting a **very large target**, like the cars hitting the wall. If we let $m_1$ be negligible to $m_2$, we can get two other equations: >$v_{1f} \approx -v_{1i}$ >$v_{2f} \approx \frac{2m_{1}}{m_{2}}v_{1i}$ Now let's take a look at a situation where the **projectile** is massive. This time, $v_2f$ can be considered negligible, allowing for another two equations to be created. $v_{1f} = v_{1i}$ $v_{2f} \approx 2v_{1i}$ >[!Abstract]- Elastic Collisions with Moving Targets >When we say "moving", we mean both particles have initial momentum values, meaning that each particle has an initial velocity. Thus, we can write the conservation of linear momentum and kinetic energy as: >$m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}$ >$\frac{1}{2}m_{1}v^2_{1i} + \frac{1}{2}m_{2}v^2_{2i} = \frac{1}{2}m_{1}v^2_{1f} + \frac{1}{2}m_{2}v^2_{2f}$ >We can rewrite these equations and divide the kinetic energy equation by the momentum equation using the method with the stationary targets: >$m_{1}(v_{1i}-v_{1f}) = m_{2}(v_{2f}-v_{2i})$ >$m_{1}(v_{1i}-v_{1f})(v_{1i}+v_{1f}) = m_{2}(v_{2i}-v_{2f})(v_{2i}+v_{2f})$ >All we have to do now is to divide both equations by each other to give us: >$(v_{1i}+v_{1f}) = (v_{2i}+v_{2f})$ >$v_{1f} = v_{2i}+v_{2f} - v_{1i}$ >$v_{2f} = v_{1i}+v_{1f} - v_{2i}$ >We can expand our definition of this by substituting the final velocities into one of the equations above - this'll get us other equations: >$v_{1f} = \frac{m_{1}-m_{2}}{m_{1}+m_{2}}v_{1i} + \frac{2m_{2}}{m_{1}+m_{2}}v_{2i} $ >$v_{2f} = \frac{2m_{1}}{m_{1}+m_{2}}v_{1i} + \frac{m_{2}-m_{1}}{m_{1}+m_{2}}v_{2i}$ You actually need to divide the equation by both sides here - you can't just divide it by one side and expect it to cancel out! Made this mistake the first time. ### Explosions - Deadly! Stationary objects sometimes explode - implode, I'm not too sure of the difference but both are violent in their own special ways. When a can of beans in the microwave explodes, it is the duty of the physicist to pause and think: "Was momentum conserved?" To answer that question, yes, yes it still is. It's a universal given, of course it must! >[!Success] What are Explosions? >Explosions are sudden and very quick deformities that happen in objects with runaway reactions - where large amounts of chemical energy are converted into kinetic or thermal (usually both) energy! It's why people tell you not to microwave cans of baked beans - it'll heat up and explode! Let's, however, take this bean can as our example for today. >[!Danger]- An Explosive System (OOS) >![[Pasted image 20230921195818.png]] > >Since this can isn't moving (unless you've actually microwaved it, so maybe it'll have some rotational kinetic energy) we can write the total momentum of the can as: >$\overrightarrow{p}_{tot} = 0$ >We'll ignore the rotational kinetic energy of the can, giving us a kinetic energy of: >$K = 0$ >When the can inevitably explodes taking your microwave out of commission, momentum is conserved. We can imagine that the can separates into four pieces which fly away in random directions. Using conservation of momentum we can quantify this relationship: >$0 = P_{1} + P_{2} + P_{3} + P_{4}$ >This means that added together the momentum vectors of each of the intrinsic vectors will equal to 0, so conservation of momentum is upheld. However, as the previously stationary object suddenly separates into four, fast moving fragments, kinetic energy is not constant. ## Two-Dimensional Collisions ![[Pasted image 20230913232906.png]] Often times, not even physicists want to generalise collisions to just one dimension - so we use two instead. Particles might not collide on the same axis, so we can create another version of conservation of momentum to account for this extra dimension. >[!Danger]- Describing a 2D-Collision %%revision%% >%%diagram PLEASE%% >If a collision happens along an axis (as in the particle moving to collide moves along the x axis), we can generalise the equation for conservation of momentum from a two-body collision as: >$\overrightarrow{P_{1i}} + \overrightarrow{P_{2i}} = \overrightarrow{P_{1f}} \overrightarrow{P_{2f}}$ >It is the **impulses** of the objects that determine the direction that the particles travel. This is as collisions that are not head-on (meaning at angles to each other) will lead to different ending directions for particles. >Particles hitting each other may deflect at angles we can call $\theta_1$ and $\theta_2$. Let's imagine that the second particle in this example is stationary and thus has no initial momentum. The first particle is moving along the x-axis. We can thus create an equation to describe the horizontal momentum of the objects: >$m_{1}v_{1i} = m_{1}{v_{1f}}\cos \theta_{1} + m_{2}v_{2f}\cos \theta_{2}$ >Now all we have to do is replace cos with sin to get the vertical components! As there is no initial vertical component in the example shown, the angles $\theta_1$ and $\theta_2$ should act in opposite directions, creating the relationship: >$m_{1}v_{1f}\sin \theta_{1} = m_{2}v_{2f}\sin \theta_{2}$ >This shows us that both particles in the collision are deflected in equal angles. As a disclaimer, this only really works if one of the particles is stationary. ## Momentum with Varying Mass NASA scientists don't launch rockets on a whim, to say the least. The blood, sweat and tears of technicians, each of them having trained their entire lives to weld the single, yet incomprehensibly important bolt they weld, all mixing into the fuselage of the rocket ship only to be acquired by the likes of Elon Musk and his terrible online reputation. So yes, this is quite literally rocket science. And it's in the "basic" notes - how fitting for a subject so difficult! A final PSA - ignore this if test day is beginning to close in - although interesting studying this won't lead to good GCSE/IB grades! When we think of things with varying masses, a rocket could appear as a potential candidate. From securing investors to cost-cutting to increasing mission efficiency, the amount of fuel carried by a rocket is pivotal for mission success. Besides, the rocket has to burn loads of fuel just to enter orbit and escape the Earth's gravitational pull! >[!Danger]- The First Rocket Equation >We can create a pristine inertial frame where we are rest relative to a rocket accelerating with no gravitational or atmospheric drag forces affecting it. We can let $M$ be the mass of the rocket and $v$ be the velocity of the rocket. >A rocket burns fuel (mass) to increase its velocity, so we can model the speed and mass of the rocket after a certain time $dt$ later: >$v_{t} = v + dv$ $M_{t} = M - dM$ >The fuel will release exhaust products that have masses $-dM$ and a velocity $u$ relative to us - the relative velocity. This is due to conservation of energy - mass is energy using $E = mc^2$, and as a result must be conserved! Here's a diagram - maybe this'll help: > >![[Pasted image 20230921200543.png]] > >Remember that conservation of momentum **has to** apply, so we can rewrite this relationship with the initial and final momentums of the rocket ship: >$Mv = UdM + (M-dM)(v+dv)$ >$Mv = UdM + Mv - dMv - dMdv + Mdv$ >$Mv = (v+dv-u)dM + Mv - dMv - dMdv + Mdv$ >$Mv = vdM+dvdM-udM + Mv - dMv - dMdv + Mdv$ >$Mv = Mv + Mdv - udM$ >Therefore: >$Mdv = -udM$ >We can divide each side by $dt$ to get a function of the change in relation to time: >$-u\frac{dM}{dt} = M \frac{dv}{dt}$ >As $\frac{dv}{dt}$ is just the acceleration of the object, we can also replace $-\frac{dM}{dt}$ with a variable R, which we define as the **mass rate of fuel consumption**, or the rate of mass loss due to fuel consumption. >This gives us our equation: >$Ru = Ma$ >[!Danger]- The Second Rocket Equation >We can continue to standardise the rocket equation to meet our parameters! The second rocket equation revolves around finding the total speed of a rocket after a burn - very useful for rocketry, where booster rockets are used! >Take the equation: >$-\frac{dM}{dt}u = M \frac{dv}{dt}$ >We can find the total change in the velocity by integrating over two , $v_f$ and $v_i$. We can isolate the equation by dividing both sides by $M$: >$-\frac{1}{M}\frac{dM}{dt}u = \frac{dv}{dt}$ >$\int_{M_{i}}^{M_{f}} -\frac{u}{M} \, dM = \int_{v_{i}}^{v_{f}} \, dv $ >Solving this integral gives us our equation: >$-u \ln{\frac{M_{f}}{M_{i}}} = v_{f} - v_{i}$ [^1]: Check out the Special Relativity Notes for more! [[Special Relativity - Foundations # Sample Problems >[!Example]- Module 9-5 Problem 40 >A space vehicle is traveling at 4300 km/h relative to Earth when the exhausted rocket motor (mass $4m$) is disengaged and sent backward with a speed of 82 km/h relative to the command module (mass $m$). What is the speed of the command module relative to Earth just after the separation? > >**Solution:** >These speed values are not in SI units. Make sure you convert these into metres per second values! Note that when the command module and the rocket motor is stuck together the total mass of the module is $5m$, giving us an initial velocity of: >$4300 \times \frac{1000}{3600} \approx 1194 m / s$ >When the command module separates there is a backwards shift in the momentum. This means there is a forward momentum we can express as a conservation of angular momentum relationship: >$82 \times \frac{1000}{3600} \times 4 = v_{ctrl}m$ >$91.1 = v_{ctrl}$ >Just add this onto the original velocity value to get: >$1194 + 91.4 = 1285.4 m / s$ >[!Example]- Module 9-5 Problem 47 >A vessel at rest at the origin of an x-y coordinate system explodes into three pieces. Just after the explosion, one piece, of mass m, moves with velocity (-30 m/s)$\hat{i}$ and a second piece, also of mass m, moves with velocity (-30 m/s)$\hat{j}$. The third piece has mass 3m. Just after the explosion, what is the: >(a) magnitude >(b) direction of the velocity of the third piece? > >**Solution a)** >The fact that the first and second blobs of matter are only on the $x$ (the i vector) and $y$ (the j vector) means that we can find the resultant momentum of the two as a square, giving us: >$\sqrt{ 30m^2 + 30m^2} = \sqrt{ 1800m } \approx 42.4m $ >This is the magnitude of the momentum of the third particle - so divide by $3m$! >$\frac{42.4m}{3m} \approx -14.14 m / s$ > >**Solution b)** >As conservation of momentum must apply the resultant momenta of the three particles are going to have to equal 0. We've already found the resultant momentum of the first two particles - since they're equal in length, they'll be facing southwest (225 $^\circ$). Since the magnitude of the third particle has to balance this out, it'll be facing northeast (45 $^\circ$). > >We have our answer - 45$^\circ$ >[!Example]- Module 9-6 Problem 50 >A 5.20 g bullet moving at 672 m/s strikes a 700 g wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to 428 m/s. >(a) What is the resulting speed of the block? >(b) What is the speed of the bullet – block center of mass? >[!Example]- Module 9-6 Problem 54 >A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. Just before the collision, one ball, of mass 3.0 kg, is moving upward at 20 m/s and the other ball, of mass 2.0 kg, is moving downward at 12 m/s. How high do the combined two balls of putty rise above the collision point? (Neglect air drag.) >[!Example]- Module 9-7 Problem 61 >A cart with mass 340 g moving on a frictionless linear air track at an initial speed of 1.2 m/s undergoes an elastic collision with an initially stationary cart of un-known mass. After the collision, the first cart continues in its origi-nal direction at 0.66 m/s. (a) What is the mass of the second cart?(b) What is its speed after impact? (c) What is the speed of the two-cart center of mass? >[!Example]- Module 9-7 Problem 65 >A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed. (a) What is the mass of the other body? (b) What is the speed of the two-body cen-ter of mass if the initial speed of the 2.0 kg body was 4.0 m/s? >[!Example]- Module 9-7 Problem 70 >In Fig. 9-69, puck 1 of mass m1 􀀃 0.20 kg is sent sliding across a frictionless lab bench, to undergo a one-dimensional elas-tic collision with stationary puck 2. Puck 2 then slides off the bench and lands a distance d from the base of the bench. Puck 1 rebounds from the collision and slides off the opposite edge of the bench, landing a distance 2d from the base of the bench. What is the mass of puck 2? (Hint: Be careful with signs.) >[!Example]- Module 9-8 Problem 71 >In Fig. 9-21, projectile particle 1 is an alpha particle and target particle 2 is an oxygen nucleus. The alpha particle is scattered at angle $\theta_{1} = 64.0 ^\circ$ and the oxygen nucleus recoils with speed 1.20 $\times$ 105 m/s and at angle $\theta_{2} = 51^\circ$. In atomic mass units, the mass of the alpha particle is 4.00 u and the mass of the oxygen nucleus is 16.0 u. What are the (a) final and (b) initial speeds of the alpha particle? >[!Example]- Module 9-9 Problem 76 >A 6090 kg space probe moving nose-first toward Jupiter at 105 m/s relative to the Sun fires its rocket engine, ejecting 80.0 kg of exhaust at a speed of 253 m/s relative to the space probe. What is the final velocity of the probe? >[!Example]- Module 9-9 Problem 79 >A rocket that is in deep space and initially at rest relative to an inertial reference frame has a mass of $2.55 \times 10^5$kg, of which $1.81 \times 10^5$ kg is fuel. The rocket engine is then fired for 250 s while fuel is consumed at the rate of 480 kg/s. The speed of the exhaust products relative to the rocket is 3.27 km/s. > >**(a) What is the rocket’s thrust?** >After the 250 s firing, what are: >**(b) the mass** >**(c) the speed of the rocket?** # What's Next?