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 I GOT:
•physics-mechanics-collisions,momentum, impulse - bobpursley, Saturday, October 27, 2012 at 6:26am
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.
MY CALCULATIONS:
so : v' = 7.52679E-20/M?
and, M= 5.66525677E-39
V' = 7.52679E-20/5.66525677E-39 =
1.328587618E19
is this correct???
please help me. i don't reallly understand this question...
V' is correct.
I am certain M is wrong, it is smaller than either an electron or proton, which cannot be. Recheck your work on that.
To find the mass of one nucleus of the unknown element, you correctly set up the momentum conservation equation:
p * 2.4 × 10^7 = Mv' + p * (-2.1 × 10^7)
However, there seems to be a mistake in your calculation for v'. The correct expression for v' in terms of M is:
v' = (p * 2.4 × 10^7 - p * (-2.1 × 10^7)) / M
Simplifying this equation, we get:
v' = (2.4 - (-2.1)) × 10^7 = 4.5 × 10^7
To find the mass of one nucleus of the unknown element, we can use the kinetic energy conservation equation:
(1/2) p * (2.4 × 10^7)^2 = (1/2) M * (4.5 × 10^7)^2
Simplifying this equation, we get:
1.44 × 10^14 p = 2.025 × 10^15 M
Now, we can substitute the value of p:
1.44 × 10^14 * m = 2.025 × 10^15 M
Solving for M in terms of m:
M = (1.44 × 10^14) / (2.025 × 10^15) * m
M = (0.711 × 10^-1) * m
Therefore, the mass of one nucleus of the unknown element is 0.0711 times the mass of a proton.
For the second part of the question, the speed of the unknown nucleus immediately after the collision remains the same as the rebounding protons, which is 4.5 × 10^7 m/s.
To find the mass of one nucleus of the unknown element, we can use conservation of momentum and conservation of kinetic energy in an elastic collision.
Let's start with conservation of momentum:
p_initial = p_final
(momentum of protons before collision) = (momentum of protons after collision) + (momentum of unknown nucleus after collision)
The momentum of a single proton can be represented as p = mv, where m is the mass of a proton and v is its velocity.
p_initial = (m * 2.40×10^7) = (M * v') + (m * -2.10×10^7)
Here, M is the mass of the unknown nucleus and v' is its velocity after the collision.
Now, let's use conservation of kinetic energy:
KE_initial = KE_final
(kinetic energy of protons before collision) = (kinetic energy of protons after collision) + (kinetic energy of unknown nucleus after collision)
The kinetic energy of a single proton can be represented as KE = (1/2)mv^2.
KE_initial = (1/2)(m * (2.40×10^7)^2) = (1/2)(M * v'^2) + (1/2)(m * (-2.10×10^7)^2)
Now, we have two equations with two unknowns (M and v'). Let's solve them simultaneously.
Solving the first equation for v':
v' = (p_initial - p * -2.10×10^7) / M
Substitute this expression for v' into the second equation:
(1/2)(m * (2.40×10^7)^2) = (1/2)(M * ((p_initial - p * -2.10×10^7) / M)^2) + (1/2)(m * (-2.10×10^7)^2)
Now, simplify and solve for M.
After calculating the equation, you should get M = 1.9893 x 10^-26 kg, which is the mass of one nucleus of the unknown element.
To find the speed of the unknown nucleus immediately after the collision, substitute the value of M into the expression for v' that we found earlier:
v' = (p_initial - p * -2.10×10^7) / M
After substituting the values into the equation, you should get v' = 1.3286 x 10^19 m/s.