only one per post, please, it is impossible to read a question and respond.
On the first, you want a momentum change of .577(tan30), so new momentum perpendicular is .577 *7600*120m/s which is equal to mass*3200
solve for mass
on the second.
work it in proton masses.
1*6^2=1*3.6^2+M*v^2
1*6=-1*3.6+ M*v
Idont have time, but rewrite each with the constants on the left.
then solve for v in the second equation, put it backinto thefirst, and solve for M.
I am having some difficulty with a few problems on my homework and I'm hoping to get some insight.
the first one:
a robotic space probe of mass 7600 KG is traveling through space at 120 m/s. Mission control determines that a change in course of 30.0 degrees is necessary and instructs the probe to fire rockets perpendicular to its direction of motion. If the escaping gas leaves the craft's rockets at an average speed of 3200 m/s, what mass of gas should be expelled.
ok, So far on this one I have a diagram of a before and after of the scenario, and I used the conservation of momentum equation (P_{1}+P_{2}=P_{1}'+P_{2}'), used vector addition to find that the speed of the probe after the rockets give it a boost would be 7600.95 m/s. However I think what I have so far is wrong.
the correct answer is 160 kg and I can not come up with it.
If anyone can describe how to work it, I would really appreciate it.
another that I am having trouble with is this:
A proton (mass = 1.67x10^{-27} kg) moves with a speed of 6.00 Mm/s. Upon colliding elastically with a stationary particle of unknown mass, the proton rebounds on its own path with a speed of 3.6 Mm/s. Find the mass of the unknown particle.
so far on this one I have once again drawn a diagram and started work with the conservation of momentum equation. After plugging the givens into it i have 2 unknowns. So I go to the conservation of Kinetic energy equation (KE_{1} + KE_{2} = KE_{1}' + KE_{2}'), plug in the givens and now I have a system of equations.
if I am correct (which I really do not think I am) they should be:
3.006x10^{-14}= 1.08^{-14} + (1/2)XY^{2} and
1.002x10^{-20}= 6.012x10^{-21}+XY
after solving my means of substitution, I came up with the answer of X= 2.13x10^{-19} (x is the mass) and my answer is wrong. For this problem, the correct answer is 6.67x10^{-27} kg.
If anyone could help me out, I would really appreciate it. Thank you.
2 answers
The specific impulse of therocket motot is Isp = 3200/9.8 = 326.53.
The delta velocity input perpendiculr to the probes initial direction is dv = 120tan30ยบ = 69.28m/s.
Since dv = Isp(g)ln(Wo/Wbo) where Wo = the ignition weght and Wbo = the burnout weight,
dv = 69.28 = (326.53)9.8ln(7600/Wbo)making the Wbo = 7437.2kg and the fuel consumed equal to 162.77kg.
The delta velocity input perpendiculr to the probes initial direction is dv = 120tan30ยบ = 69.28m/s.
Since dv = Isp(g)ln(Wo/Wbo) where Wo = the ignition weght and Wbo = the burnout weight,
dv = 69.28 = (326.53)9.8ln(7600/Wbo)making the Wbo = 7437.2kg and the fuel consumed equal to 162.77kg.