Question

You are at the controls of a particle accelerator, sending a beam of 3.00×10^7 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 2.70×10^7 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). B) What is the speed of the unknown nucleus immediately after such a collision?

Answer 1

In the simple case where nuclei bounce back in the direction where they came from, we may write the conservation of momentum and energy.

Protons have a mass m and an initial velocity vi, then bounce back with a velocity vb

Target nuclei have a mass M, are initially immobile and move at a speed vn after collision.

The conservation of energy gives:

m.vi²/2 = m.vb²/2 + M.vn²/2

So M.vn² = m.(vi² - vb²)

The conservation of momentum gives:

m.vi = M.vn - m.vb

So M.vn = m.(vi+vb)

We now have:

(A) M = (M.vn)²/(M.vn²) = (m.(vi+vb))² / m.(vi² - vb²) = m.(vi+vb)²/(vi² - vb²)

= m((3+2.7)*10⁷)²/((3-2.7)*10⁷)²

= 361m

(B) vn = (M.vn²) / (M.vn) = (vi² - vb²) / (vi+vb)

= 3.0×10^6 m/s