You are at the controls of a particle accelerator, sending a beam of 2.40×107 protons (mass ) 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 2.10×107 . Assume that the initial speed of the target nucleus is negligible and the collision is elastic.

Question

You are at the controls of a particle accelerator, sending a beam of 2.40×107 protons (mass ) 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 2.10×107 . Assume that the initial speed of the target nucleus is negligible and the collision is elastic.

1)Find the mass of one nucleus of the unknown element. Express your answer in terms of the proton mass m

2)What is the speed of the unknown nucleus immediately after such a collision?

Answer 1

Momentum:

p is mass of a proton
M is the mass of the nuclei

p*2.4E7=Mv' + p(-2.1E7)
KEnergy:
1/2 p (2.4E7)^2=1/2 M v'^2 + 1/2 p (-2.1E7)^2

Two equations, two unknowns: Let algebra go to work. Solve for v' in the first equation in terms of M
then, put that expression for v' into the second equation. It will work out.

Answer 2

so : v' = 7.52679E-20/M?

Answer 3

and, M= 5.66525677E-39

V' = 7.52679E-20/5.66525677E-39

is this correct???

Answer 4

and, M= 5.66525677E-39

V' = 7.52679E-20/5.66525677E-39 = 1.328587618E19

is this correct???

Answer 5

1) To find the mass of one nucleus of the unknown element, we can use the principle of conservation of momentum.

Before the collision, the momentum of the protons is given by the equation:

p_initial = (2.40×10^7) * m

After the collision, when the protons bounce straight back, their momentum changes to:

p_final = (-2.40×10^7) * m

According to the conservation of momentum, the initial momentum should equal the final momentum:

p_initial = p_final

(2.40×10^7) * m = (-2.40×10^7) * m

We can divide both sides of the equation by (-2.40×10^7) to solve for the mass:

m = m

Therefore, the mass of one nucleus of the unknown element is equal to the mass of a proton, m.

2) To find the speed of the unknown nucleus immediately after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.

Since the collision is elastic, both momentum and kinetic energy are conserved.

Before the collision, the total initial kinetic energy is given by:

KE_initial = (1/2) * (2.40×10^7)^2 * m

After the collision, the protons bounce back and their final kinetic energy is:

KE_final = (1/2) * (2.10×10^7)^2 * m

According to the conservation of kinetic energy, the initial kinetic energy should equal the final kinetic energy:

KE_initial = KE_final

(1/2) * (2.40×10^7)^2 * m = (1/2) * (2.10×10^7)^2 * m

We can simplify the equation:

(2.40×10^7)^2 = (2.10×10^7)^2

Solving for the unknown speed:

(2.10×10^7)^2 * m = (1/2) * (2.40×10^7)^2 * m

Dividing both sides of the equation by m * (2.10×10^7)^2:

1 = (1/2) * (2.40×10^7)^2 / (2.10×10^7)^2

Simplifying the equation:

1 = (1/2) * (2.40×10^7)^2 / (2.10×10^7)^2

1 = (1/2) * (24/21)^2

Simplifying further:

1 = (1/2) * (8/7)^2

1 = (1/2) * (64/49)

1 = 32/49

Therefore, the speed of the unknown nucleus immediately after the collision is 32/49 times the speed at which the protons were initially moving, or approximately 0.6531 times the initial speed.