You are at the controls of a particle accelerator, sending a beam of 3.90×10^7 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 3.60×10^7 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?
a) To find the mass of one nucleus of the unknown element, we can use the conservation of momentum in an elastic collision.
In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Since the initial speed of the target nucleus is negligible, we only need to consider the momentum of the incoming proton.
From the problem statement, the initial speed of the proton is 3.90×10^7 m/s, and the final speed of the proton after rebounding is 3.60×10^7 m/s.
Let's denote the mass of the proton as m_p, and the mass of the unknown nucleus as m_n.
Using the conservation of momentum, we have:
Initial momentum = Final momentum
m_p * initial speed = m_p * final speed + m_n * unknown final speed
m_p * (3.90×10^7 m/s) = m_p * (3.60×10^7 m/s) + m_n * unknown final speed
Dividing both sides by m_p gives:
(3.90×10^7 m/s) = (3.60×10^7 m/s) + m_n * unknown final speed
Substituting the given values, we have:
(3.90×10^7 m/s) = (3.60×10^7 m/s) + m_n * (3.60×10^7 m/s)
Simplifying further:
(3.90×10^7 m/s) - (3.60×10^7 m/s) = m_n * (3.60×10^7 m/s)
0.30×10^7 m/s = m_n * (3.60×10^7 m/s)
Dividing both sides by (3.60×10^7 m/s), we get:
m_n = (0.30×10^7 m/s) / (3.60×10^7 m/s)
m_n = 0.0833 m
Therefore, the mass of one nucleus of the unknown element is 0.0833 times the mass of a proton.
b) To find the speed of the unknown nucleus immediately after the collision, we can use the conservation of kinetic energy in an elastic collision.
In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Using the kinetic energy formula:
Initial kinetic energy = Final kinetic energy
(1/2) * m_p * (initial speed)^2 = (1/2) * m_p * (final speed)^2 + (1/2) * m_n * (unknown final speed)^2
Substituting the given values:
(1/2) * m_p * (3.90×10^7 m/s)^2 = (1/2) * m_p * (3.60×10^7 m/s)^2 + (1/2) * m_n * (unknown final speed)^2
Simplifying further:
(1/2) * m_p * (3.90×10^7 m/s)^2 - (1/2) * m_p * (3.60×10^7 m/s)^2 = (1/2) * m_n * (unknown final speed)^2
Dividing both sides by (1/2) * m_n gives:
(unknown final speed)^2 = [(1/2) * m_p * (3.90×10^7 m/s)^2 - (1/2) * m_p * (3.60×10^7 m/s)^2] / (1/2) * m_n
(unknown final speed)^2 = [(1/2) * m_p * (3.90×10^7 m/s)^2 - (1/2) * m_p * (3.60×10^7 m/s)^2] / (1/2) * m_n
Simplifying further:
(unknown final speed)^2 = [(1/2) * m_p * (15.21×10^14 m^2/s^2 - 12.96×10^14 m^2/s^2)] / (1/2) * m_n
(unknown final speed)^2 = (1/2) * m_p * (2.25×10^14 m^2/s^2) / (1/2) * m_n
Dividing both sides by (1/2) * m_p gives:
(unknown final speed)^2 = (2.25×10^14 m^2/s^2) * (1/m_n)
Taking the square root of both sides gives:
unknown final speed = √[(2.25×10^14 m^2/s^2) * (1/m_n)]
Substituting the value of m_n (0.0833 m) gives:
unknown final speed = √[(2.25×10^14 m^2/s^2) * (1/0.0833 m)]
unknown final speed ≈ 9.513 × 10^7 m/s
Therefore, the speed of the unknown nucleus immediately after the collision is approximately 9.513 × 10^7 m/s.
To find the mass of the unknown nucleus, we can use the principle of conservation of momentum and kinetic energy.
a) Conservation of momentum:
The momentum of the protons before the collision is the same as the momentum of the protons after the collision. Since some protons bounce straight back, their momentums change direction.
The initial momentum of the protons is given by:
P_initial = m * v_initial (where m is the mass of the proton and v_initial is the initial velocity of the protons)
After a collision, if some protons bounce back with a speed of 3.60×10^7 m/s, their momentum is given by:
P_final = m * (-v_final) (where v_final is the final velocity of the protons)
Since momentum is conserved, P_initial = P_final:
m * v_initial = m * (-v_final)
Now we can solve for the velocity of the protons before the collision:
v_initial = (-v_final)
Substituting the values given:
v_initial = -(3.60×10^7 m/s) = -3.60×10^7 m/s
b) Conservation of kinetic energy:
The initial kinetic energy of the protons is given by:
KE_initial = (1/2) * m * v_initial^2
The final kinetic energy of the protons is given by:
KE_final = (1/2) * m * v_final^2
Again, since kinetic energy is conserved, KE_initial = KE_final:
(1/2) * m * v_initial^2 = (1/2) * m * v_final^2
Now we can solve for the final velocity of the protons:
v_final^2 = (v_initial^2)
Substituting the values given:
v_final^2 = (3.60×10^7 m/s)^2 = 1.296×10^15 m^2/s^2
Now, we can solve the two equations simultaneously to find the mass of the unknown nucleus (in terms of the proton mass m):
m * v_initial = m * (-v_final)
m * (-3.60×10^7 m/s) = m * (-3.60×10^7 m/s)
And
(1/2) * m * v_initial^2 = (1/2) * m * v_final^2
(1/2) * m * (-3.60×10^7 m/s)^2 = (1/2) * m * (1.296×10^15 m^2/s^2)
Simplifying and canceling out the mass (m) on both sides of the equations, we get:
3.60×10^7 = 1.296×10^15
Dividing both sides by 1.296×10^15 gives:
m = (3.60×10^7 / 1.296×10^15)
Therefore, the mass of one nucleus of the unknown element, in terms of the mass of the proton (m), is (3.60×10^7 / 1.296×10^15).
For part b) (the speed of the unknown nucleus immediately after the collision), we cannot determine it using the given information since only the speeds of the protons before and after the collision are given. The mass or velocity of the unknown nucleus is not provided. Thus, we cannot calculate the speed of the unknown nucleus after the collision.
Let the final velocity of the gas atom be V, and its mass be a*m, where a is a constant and m is the proton mass. Linear (x) momentum and total kinetic energy are conserved. Use classical mechanics since velocities are less than 0.15c, and because I am lazy.
m*3.9*10^7 = -m*3.7*10^7 + a*m*V
(1/2)m*(3.9)^2*10*14 = (1/2)*m*(3.7)^2*10^14 + (a*m/2)V^2
Cancel out the m's.
7.6*10^7 = a V
15.21*10^14 = 13.69*10^14 + a V^2
1.52*10^14 = a V^2
57.76*10^14 = a^2 V^2
a = 38
V = 2.00*10^6 m/s
A relativistic solution might be slightly different.