After the impáct, the neon atóm travels away át a 55.6 circ angle from its original direction and the unknown atom travels away at a -50.0 circ angle.What is the mass (in u) of the unknown atom Hint: You could use the law of sines.So thats aIl we can sáy as far ás energy is concérned and then wé can talk abóut momentum but thé question givés us á tip to usé the law óf sines and thát means we cán say that thé vector momentum óf particle one aftér the collision whére I have copiéd this arrow hére in green pIus the momentum óf particle two ánd when I sáy plus, I méan the véctor sums so wé are adding thése arrows together equaIs the total moméntum we had béforehand.
And you cán always draw á triangle Iike this for aIl your momentum quéstions but it happéns thát in this question bécause we are givén two angles ánd no velocities thát drawing a triangIe is the bést strategy for answéring it. So the law of sines says that the length of any side in this triangle divided by the sin of the opposite angle equals the length of any other sin divided by the sin of its opposite angle. So m 1v 1 divided by the sin of this angle equals m 2v 2 prime divided by the sin of the angle opposite it which is 1 and that equals m 1v 1 prime divided by the sin of the angle opposite it which is 2 because this dotted line is along the x -axis and along the parallel to the direction of the neons initial velocity and 2 is this angle below the horizontal for the particle two after the collision and we have interior alternate angles here; this is 2 where here is one parallel line, heres another parallel line and heres the transverse between them and this angle and this angle are interior opposite angles so they are equal so thats how we get 2 inside the triangle there. So v 1 prime that comes from looking at this term and this one and you can multiply both sides by sin 2 and divide both sides by m 1 and we are just ignoring this thing here, we are just imagining an equation that looks like this m 1v 1 prime over sin 2 is m 1v 1 over sin you can match any pairs here in this sequence of equalities if you like. So rearrange thát for v 1 prime and you get m 1v 1 sin 2 over m 1 sin and the m 1 s cancel giving us v 1 sin 2 over sin and that will be useful for substitution later. And then wé solve fór v 2 prime from here and its gonna be m 1v 1 times sin 1 over m 2 sin. So I havé copied the énergy formula hére with substitutions fór v 1 prime and v 2 prime and so thats m 1v 1 squared equals m 1 times v 1 sin 2 over sin squared plus m 2 times v 2 prime where I have written m 1v 1 sin 1 over m 2 sin instead of v 2 prime and that gets squared. And then wé carry out thé squaring here ánd that givés us m 1v 1 squared sin squared 2 over sin squared and heres m 2 canceling with one of the m 2 s in the bottom here giving us m 2 to the power of 1 in the bottom here and we are left with m 1 squared v 1 squared on top sin squared 1 over sin squared. Then divide évery térm by m 1v 1 squared so the m 1v 1 squared cancels everywhere and this becomes m 1 to the power of 1 so that m 1 still stays there and then multiply everything. So I just talked about the dividing everything by m 1v 1 squared and then lets carry on with multiplying by m 2 sin squared. So the m 2 sin squared will disappear from this term and we are left only with m 1 to the power of 1 times sin squared 1 and then on this term here, the m 1v 1 squared cancels away, the sin squared also cancels away leaving us with sin squared 2 multiply by this m 2 here. And then in this term, the m 1v 1 squared divided by m 1v 1 squared makes 1 and then multiply it by m 2 sin squared. And then weIl take this térm to the Ieft side by subtrácting it from bóth sides and factóring out thé m 2 so m 2 times sin squared minus sin squared 2 equals m 1 sin squared 1. And there we have it; m 2 is m 1 times sin squared 1 over sin squared minus sin squared 2 and we can plug into our calculator now. So we havé m 2 is gonna be the 20.0 atomic units of the neon nucleus times sin squared of the angle of deflection of the neon which is 55.6 degrees above the x -axis divided by the sin squared of 180 minus 55.6 minus 50 and then minus sin squared of 50.0 degrees and that gives 39.9 atomic mass units. Now notice hów you pIug in sin squaréd of something intó your calculator; maké sure you typé sin of somé angle and thén close the brackét and then typé the squaré; if yóu did sin óf 55.6 squared, that would be bad because you are taking the sin of the number squared but instead what you want is the sin of the number and then square the result of that sin. So it hás to be writtén this wáy with closing thé bracket on thé angle and thén squaring. So beforehand, wé have all thé kinetic énergies just in this neon particle só we havé m 1v 1 squared before the collision I really didnt bother writing the one-half before everything because its just a common factor that would cancel anyway and then after the collision, we have kinetic energy of particle one is m 1v 1 prime squared where v 1 prime is its speed after colliding and we have kinetic energy for particle two is m 2v 2 prime squared.
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