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.

(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?

a) mg = 11m

b) Vg2 = 0.3×10^7m/s

Explanation:

Taking initial directions of the proton as the positive

Vpq= 1.80×10^7m/s

Vp2 = -1.5×10^7m/s

In elastic collision,there is no external net force on the system, the total momentum before the collision is equal to the total momentum after the collision

P1 = P2

In the collision between two particles (A and B), we get:

PA1 + PB1 = PA2 + PB2

Where the momentum of a particle is given by:

P = mc

The collision equation will be:

MaVA + MBVB= MAVA2 + MBVB2 ... eq1

In elastic collision between two particles, the relative velocities before and after the collision have the same magnitude but opposite signs.VA1 - VB1 = VB2 - VA2 ...eq2

b) VP1 - 0 = Vg2 - VP2

Vg2 = Vp1 + Vp2

Vg2 = (1.8×10^7) + (-1.5×10^7)

Vg2 = 0.3 ×10^7m/s

a) substituting into eq 1

(1.8×10^7) + mg ×0 = m × (1.8×10^7) + mg × (0

3×10^7)

mg = [(1.8×10^7) + (1.5×10^7) m] / (0.3×10^7)

mg= 11 m

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6

are there options? cause you ain’t. real specific