WEBVTT physics/ap-physics-c-mechanics/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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In our final lesson of Mechanics, we are going to go over the free response portion of the AP Physics C Mechanics practice exam.
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Take a minute, print that out, give it a try, and come back here and we will see how things look.
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Taking a look at question number 1 A, it says determine the average speed of glider A for those following time intervals, we want to know first from 0.1 to 0.3 s.
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In that interval, our average velocity is change in position divided by time.
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Our change in position from 0.1 to 0.3 s is 0.2 m and the time interval is 0.2 s.
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The velocity there is 1 m/s.
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For the next part, we are asked to find the average velocity between 0.9 and 1.1 s.
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We used the same formula but now the displacement from 0.9 to 1.1 s, it looks like that is 0.12 m in the same time interval 0.2 s which is just going to be 0.6 m/s.
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And A3, find the average speed from 1.7 to 1.9 s.
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Same formula again, Δ x / Δ t, but now our displacement is 0.04 m in 0.2 s which is just going to be 0.2 m/s.
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And that covers part A.
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The let us move on to part B, we are asked to make a graph.
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Here in part B, we are asked to sketch the speed of the glider as a function of time.
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Our graph will look roughly like this, take your time, make sure you have everything plotted very neatly.
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We will go from 1 to 2 s, so there is 1.5 to 0.5, 0.
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Here is our velocity in meters per second from 0.5, 1, 1.5.
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Our graph should look something like this.
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It should go straight over for a spell then we are going to have a drop down to our final 0.2, make sure you move that over a little bit.
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That looks like that is a little bit light, over to somewhere around there and off like that should give it a shape your graph if you plot it very carefully.
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And for C, use the data to calculate the speed of glider B immediately after it separates from the spring.
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That to me looks like a conservation of momentum sort of question.
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Let us do this one with a momentum table.
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We will have the momentum before, momentum after for A, B, and their total.
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The momentum of A before hand is 1 × 0.9 so that will be 1 × 0.9 = 0.9.
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B is at rest so that is 0.
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The total before is 0.9, and after the collision we have the same mass 0.9 × 0.2 is going to be our .18.
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We also have for B, its mass 0.6 × unknown velocity VB, so 0.9 is going to be equal to 0.18 + 0.6 VB.
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We can solve that for VB, 0.9 -0.18 is going to be 0.72 = 0.6 VB or VB = 0.72/0.6 is going to be 1.2 m/s.
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There is part C, C1.
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C2, asks us to plot the speed of the glider B as a function of time.
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We can even, for our purposes, I'm just going to plot that in blue on our same graph because we have the same axis.
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It is going to come down like that and it is going to then come up here, into roughly the same rate and same points, up to 1.2 m/s at its highest point.
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There is our graph for glider B compared to A.
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Let us give ourselves a little bit more room for part D.
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It shows us the total kinetic energy as a function of time as the collision is elastic.
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The kinetic energy before the collision = the kinetic energy after, therefore it is elastic, yes.
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Justify your answer.
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The kinetic energy before = the kinetic energy after, state that in word somehow.
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For part 2, why is their minimum and kinetic energy at 1 s?
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The kinetic energy is being stored as elastic potential energy at that point while the spring is compressed.
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I would explain that in words too.
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During that deep end kinetic energy, some of the kinetic energy is being transferred to stored potential elastic energy,
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then the spring decompresses again and will be back kinetic energy.
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That covers number 1, moving on to Mechanics question 2.
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Let us see here, a space shuttle astronaut playing around.
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We have 2 masses connected by rigid rod of length L and negligible mass and the device is a small lump of clay of mass M at some speed V knot.
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Determine the total kinetic energy of the system after the collision.
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For M2 part A, we have MV initial must be equal to the total mass × the final velocity which will be 3 M, let us call that V final.
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Complies that V final is just going to be equal to MV knot / 3 M which is V0 / 3.
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If we want the final kinetic energy after the collision, the final kinetic energy is going to be ½ × 3 M the total mass × our final velocity V knot /3²,
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which will be 3 M /2 × V knot² /9, which is MV knot² /6.
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There is A, for part 2, determine the change in kinetic energy as a result of the collision.
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A2, initial kinetic energy is 1/2 M V initial² which is just M V knot² /2.
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Our change in kinetic energy is going to be the final kinetic energy - the initial kinetic energy
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which is MV knot² /6 from up there - MV knot² /2 from right there.
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That is MV knot² / 6 -3 MV knot² / 6 which is just going to be - MV knot² / 3.
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Moving on 2 part B, as we look at B part 1, the assembly is brought to rest, the lump of clay is pulled off,
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we hit again but this time we had it stick to one of the spheres of the assembly over on the side.
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Determine where the center of mass is going to be after the collision.
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Our position vector to the center of mass is just the sum for all of our different masses of MIRI divided by the total mass M so that is going to be,
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if we count the left side as 0, we have got M × 0 + 2 M and the distance L /3 M our total mass which is just going to be 2 L /3.
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Alright and for part 2 B2, on the figure above, indicate the direction of the motion of the center of mass immediately after the collision.
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That is straightforward, if the clay that is coming toward is going up, afterwards we must have the exact same thing so up would be the correct direction there.
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In part 3 B3, determine the speed of the center of mass immediately after the collision.
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The final velocity is just going to be the initial velocity divided by 3, conservation of momentum.
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Moving on to B4, determine the angular speed of the system immediately after the collision.
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B4, the way I would do that is look at the angular momentum, initial = the final angular momentum
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which implies that MVR sin θ = I ω or M, V knot L /3 r sin θ, sin 90 is 1 = Iω.
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We can solve for ω which is just going to be, we will have MV knot L /3 × the moment of inertia.
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To figure out the moment of inertia, if I want to go any further with this, the moment of inertia is just the sum of our Mr²
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which is going to be M × 2 L/3² + 2 M × its distance from the center of mass L /3² which is going to be M × 4 L² /9 + 2 ML² /9, which will give us 6 ML²/9 or 2 ML² /3.
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I can use that in there for my moment of inertia.
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Ω = M V knot L /3 divided by our moment of inertia which is 2 ML² /3
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3 makes a ratio of 1, our masses will cancel out, we will lose one of our L's which implies then that ω is just going to be equal to V knot /2 L.
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B5, determine the change in kinetic energy as a result of the collision.
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B5, our initial kinetic energy was 1/2 M V initial².
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Our final kinetic energy, now we have translational and rotational components.
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We have to add those up, that is going to be ½ × our total new mass 3 M × V knot /3² + our rotational ½ × I
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which we determined up here was 2 ML² /3 × √angular velocity which was V knot /2 L².
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This implies then that our final kinetic energy is going to be equal to, this left hand side we are going to have / 3,
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we are going to have MV knot² /6 + 2 ML² / 6 × V knot² /4 L² is going to be MV knot² / 6 + MV knot² /12, which is going to be MV knot² /4.
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Our change in kinetic energy which is final - initial is going to be MV knot² /4 for right there, - our initial 1/2 MV knot², which is just - MV knot² /4.
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That finishes up question 2, as we move to our third question, this is a pretty complex situation.
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However, they try and make it simpler by walking through the different pieces step by step using there are
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a lot of little pieces we are going to take bit by bit in order to put the whole situation together.
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For M3 A1, that first asks us to draw vector on the block and determine the magnitude of the force.
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We are looking for the normal for on M1.
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There is M1, normal force on it going up and M1 = M1 G.
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For part 2, a frictional force exerted on the block 1 by block 2.
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It is at rest, we have got M1 here, the frictional force is going to be 0.
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A3, find the force T exerted on block 2 by the spring.
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Will here is our M2, by the string there is our tension.
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Tension must be equal to MG.
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For A4, now we are looking at the normal force N2 inserted on block 2 by the tabletop.
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Here is M2, we are looking for the normal force exerted on it.
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There is N2, to figure this out I will probably get to the point where I'm drawing free body diagram just help me out.
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We have the force N2 up, we have M2 G down, and M1 G down.
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Those are all balance so I can write here that N2 = M1 G + N2 G.
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That will cover A4, let us go to the next page for A5.
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The frictional force F2 exerted on block 2 by the tabletop.
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Block 2, there it is the frictional force F2 opposing motion.
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To figure our free body diagram, we would have F2 and over here T which we know is MG.
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I can write then that F2 = MG.
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We have got all our free body diagrams.
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For part B, determine the largest value of M for which the blocks can remain at rest.
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When this happens, our force of friction is going to be μ S2 × N2 which has to equal MG.
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But we just previously determined that N2 was equal to M1 G + M2 G so we can rewrite this as μ S2 × M1 G + M2 G = MG.
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Or solving for M, that is going to be μ S2 × (M1 + M2).
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Moving on to C, now suppose M is large enough that the hanging blocked descends when the blocks are released.
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Assume blocks 1 and 2 are moving easy without slipping, find the acceleration.
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A little bit trickier situation, let us draw our free body diagram for our hanging block.
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We have T and MG.
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MG - T = MA calling down our positive y direction here and T, therefore, = MG – MA.
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And if we go look at our block situation, we have the normal force up, we have their combined weight, M1 + M2 down,
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we have the tension pulling them to the right, and we have our force of friction.
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In this case, as we look in the direction of the motion T - F2 is going to be equal to M1 + M2 A.
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But we know a little bit more here as well, F2 is μ K2 × the normal force, which means that F2 = μ K2 × M1 + M2 G.
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We can write that as T - μ K2 × (M1 + M2 G) = M1 + M2 × A.
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We can combine this equation and that equation, to write that our tension MG – MA.
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We still have our – μ K2 × M1 + M2 G = M1 + M2 × A or with a little bit of rearrangement, trying to get all my G on one side and all of my A on the other.
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We have G × M – μ K2 × (M1 + M2) = a × M1 + M2 + M or solving for A by dividing both sides by (M1 + M2 + M),
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I find out that a = M – μ K2 × (M1 + M2) ÷ the sum of our masses M1 + M2 + M all of that multiplied by G.
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And that should cover us for part C, looking at part D, now suppose M is large enough that the hanging block ascends block 1 is slipping on block 2.
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Find the magnitude A1 of the acceleration the block 1.
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D part 1, our M1 block we have normal force 1 up on it, we have the frictional force 1 to the right, and M1 G down.
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The net force in the x direction, in the direction of motion is F1 = M1 a, which implies then that A1 = F1 / M1.
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We know that F1 = μ K1 M1 G so then A1 = μ K1 M1 G / M1 or making a ratio of 1 from M1/M1.
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A1 = μ K1 × G.
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Part D2, find the magnitude A2 of the acceleration the block 2.
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Our free body diagram for block 2, there it is, we have got N2 up, we have tension to the right, we have friction 1 to the left,
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we have friction 2 to the left, we have M1 G down, and we have M2 G down.
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For our hanging mass M, we have tension up, and MG.
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As we look at this, starting with the one on the right MG - T = MA 2, which implies that T = MG - MA 2.
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And if we look at our left most free body diagram over here, we have T - F1 - F2 = M2A2 which implies then that the T we can replace with MG - MA2.
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This becomes MG - MA2 - F1 - F2 = M2A2 which implies then that MG - F1 - F2 = we will have M2A2 + MA2,
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which implies that A2 is going to be equal to MG - F1 - F2 ÷ M2 + M.
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We know a couple more things about friction.
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We know our frictional force 1 = μ K1 M1 G and we know there frictional force 2 = μ K2 × (M1 + M2 G).
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Therefore, we can write this as A2 B2 = MG - F1 which is μ K1 M1 G - F2, which is μ K2 M1 + M2 × G / M2 + M.
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A little bit of simplification, A2 = M – μ KM 1 - μ K2 × (M1 + M2) ÷ M 2 + M, that whole thing × G.
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Pretty involved there but the previous questions help walk you through that.
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Alright, that ends our free response test.
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Hopefully, you get a good feel for where you are strong and areas you need to work a little bit more.
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Thank you so much for joining us at www.educator.com.
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Good luck and make it a great day everyone.