You are at the controls of a particle accelerator, sending a beam of 1.80×107 m/s protons (mass m) at a gas target of an unknown element. Yo

Question

You are at the controls of a particle accelerator, sending a beam of 1.80×107 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 1.50×107 m/s. Assume that the initial speed of the target nucleus is negligible and the collision is elastic.

in progress 0
Sigridomena 4 years 2021-08-10T13:53:43+00:00 1 Answers 18 views 0

Answers ( )

  1. giabao
    0
    2021-08-10T13:54:54+00:00
    Reply

    Answer:

    Explanation:

    mass of proton = m

    initial velocity, u1 = 1.80 x 10^7 m/s

    final velocity, v1 = – 1.5 x 10^7 m/s

    (a)

    Let m’ is the mass of unknown nucleus.

    By use of conservation of energy

    m'=\frac{u_{1}-v_{1}}{u_{1}+v_{1}}m

    m'=\frac{1.8\times 10^{7}-(-1.5\times 10^{7})}{1.8\times 10^{7}+(-1.5\times 10^{7})}m

    m’ = 11 m

    Thus, the mass of unknown nucleus is 11 times the mass of proton.

    (b)

    By use of conservation of momentum

    m x u1 + m’ x 0 = m x v1 + m’ x v’

    where, v’ is the velocity of unknown nucleus after collision

    m x 1.8 x 10^7 = – m x 1.5 x 10^7 + 11 m x v’

    3.3 m x 10^7 = 11 m x v’

    v’ = 3 x 10^6 m/s  

    Thus, the velocity of unknown nucleus after collision is 3 x 10^6 m/s

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )