You are at the controls of a particle accelerator, sending a beam of 2.70×10 7 m
ID: 1541879 • Letter: Y
Question
You are at the controls of a particle accelerator, sending a beam of 2.70×107 m/s protons (mass m ) 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 2.40×107 m/s . Assume that the initial speed of the target nucleus is negligible and the collision is elastic.
Part A
Find the mass of one nucleus of the unknown element. Express your answer in terms of the proton mass m .
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Part B
What is the speed of the unknown nucleus immediately after such a collision?
You are at the controls of a particle accelerator, sending a beam of 2.70×107 m/s protons (mass m ) 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 2.40×107 m/s . Assume that the initial speed of the target nucleus is negligible and the collision is elastic.
Part A
Find the mass of one nucleus of the unknown element. Express your answer in terms of the proton mass m .
SubmitHintsMy AnswersGive UpReview Part
Part B
What is the speed of the unknown nucleus immediately after such a collision?
m m/sExplanation / Answer
Initial speed of proton u = 2.7 x10 7 m/s
Initial speed of nucleus U = 0
Mass of proton m = m
Mass of nucleus M = M
final velocity of the proton v = -2.4 x10 7 m/s Since it moves opposite to the initial direction of proton.
Coefficient of restitution = 1
relative velocity after collision / relative velocity before collision = 1
V - v /(u -U) = 1
V - v = u - U = u
V = u + v
= ( 2.7 x10 7) +( -2.4 x10 7)
= 0.3 x10 7 m/s
From law of ocnservation of momentum ,
mu + MU = mv + MV
mu + 0 = mv + MV
MV = mu - mv
= m( u - v )
= m [ ( 2.7 x10 7) -(- 2.4 x10 7)]
= m[( 5.1 x10 7)]
M = m[( 5.1 x10 7)/( 0.3 x10 7)]
= 17 m
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