V' is correct.
I am certain M is wrong, it is smaller than either an electron or proton, which cannot be. Recheck your work on that.
You are at the controls of a particle accelerator, sending a beam of 2.40×107 protons (mass ) at a gas target of an unknown element. Your detector tells you that some protons bounce straight back after a collision with one of the nuclei of the unknown element. All such protons rebound with a speed of 2.10×107 . Assume that the initial speed of the target nucleus is negligible and the collision is elastic.
1)Find the mass of one nucleus of the unknown element. Express your answer in terms of the proton mass m
2)What is the speed of the unknown nucleus immediately after such a collision?
ANSWER I GOT:
•physics-mechanics-collisions,momentum, impulse - bobpursley, Saturday, October 27, 2012 at 6:26am
Momentum:
p is mass of a proton
M is the mass of the nuclei
p*2.4E7=Mv' + p(-2.1E7)
KEnergy:
1/2 p (2.4E7)^2=1/2 M v'^2 + 1/2 p (-2.1E7)^2
Two equations, two unknowns: Let algebra go to work. Solve for v' in the first equation in terms of M
then, put that expression for v' into the second equation. It will work out.
MY CALCULATIONS:
so : v' = 7.52679E-20/M?
and, M= 5.66525677E-39
V' = 7.52679E-20/5.66525677E-39 =
1.328587618E19
is this correct???
please help me. i don't reallly understand this question...
1 answer
1 answer