You are at the controls of a particle accelerator, sending a beam of 2.10×107m/s

Question

You are at the controls of a particle accelerator, sending a beam of 2.10×107m/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 1.80×107 m/s. Assume that the initial speed of the target nucleus is negligible and the collision is elastic.

A)Find the mass of one nucleus of the unknown element. Express your answer in terms of the proton mass m
answer in ____m
B)What is the speed of the unknown nucleus immediately after such a collision?

I am unable to solve this question can u please help me out

Explanation / Answer

Is the 3.60x10^7 the speed of the incoming proton?? I'll assume it is So From conservation of momentum we have 1.67x10^-27*3.60x10^7 = -1.67x10^-27*3.30x10^7 + M*V So M*V = 1.67x10^-27*(3.60x10^7 + 3.30x10^7) = 1.152x10^-19kg-m/s From conservation of K we have 1/2*m*v1^2 =1/2*m*v2^2 + 1/2*M*V^2 So 1/2*1.67x10^-27*(3.60x10^7)^2 - 1/2*1.67x10^-27*(3.30x10^7)^2 = 1/2*M*V^2 so 1/2*M*V^2 = 1.728x10^-13 But M*V = 1.152X10^-19 so 1/2*1.152x10^-19*V = 1.728x10^-13 So V = 2*1.728x10^-13/1.152x10^-19 = 3.00x10^6m/s Therefore M = 1.152x10^-19/3.00x10^6 = 3.84x10^-26kg = 3.84x10^-26/1.67x10^-27kg/proton = 23 protons and again V = 3.0x10^6m/s

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