You are at the controls of a particle accelerator, sending beam of 3.90 x 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 3.60 x 10^7 m/s. assume that the initial speed of the target nucleus is negligible and that the collision is elastic.

(a) find the mass of one of the nuclei of the unknown element. express you answer in terms of the proton mass m.
(b) what is the speed of the unknown nucleus immediately after such a collision?

Mass =25.m

Speed = 4.0×10^7m/s

Explanation:

●Nuclei bounce back in the direction where they came from, we may write the conservation of momentum and energy.

●Mass of proton = M

● initial velocity of proton = vi = 3.9 ×10^7m/s

●final velocity of proton = vb = 3.6×10^7m/s

Note that the:

Target nuclei have a mass M, are initially immobile with speed vn after collision

Using conservation of energy:

m.vi²/2 = m.vb²/2 + M.vn²/2

So M.vn² = m.(vi² - vb²)

And conservation of momentum:

m.vi = M.vn - m.vb

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

Therefore:

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

= m.(vi+vb)²/(vi² - vb²)

= m.(3.9×10^7 +3.6×10^7)²

((3.9×10^7)²) - ((3.6×10^7)²)

= 25.m

B.

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

=((3.9×10^7)²) - ((3.6×10^7)²)

(3.9+3.6)

= 4.0×10^7 m/s

this isn't the answer but they're the graphs

wind moving at higher speeds will transfer more energy to the water, resulting in stronger waves.