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?
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.
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...
Don't answer here please, but in the other topic i created, sorry!!!
To answer the question, we can go through the steps provided:
1) Begin with the equation for momentum conservation:
p * 2.4 × 10^7 = M * v' + p * (-2.1 × 10^7)
2) Next, solve for v' in terms of M:
v' = (p * 2.4 × 10^7 + p * (-2.1 × 10^7)) / M
= 2.4 × 10^7 - 2.1 × 10^7
= 0.3 × 10^7
3) Now we have the expression for v'. Next, substitute this into the equation for kinetic energy conservation:
(1/2) * p * (2.4 × 10^7)^2 = (1/2) * M * (0.3 × 10^7)^2 + (1/2) * p * (2.1 × 10^7)^2
4) Simplify the equation:
p * (2.4 × 10^7)^2 = M * (0.3 × 10^7)^2 + p * (2.1 × 10^7)^2
5) Now, substitute the mass of a proton (m) for p:
m * (2.4 × 10^7)^2 = M * (0.3 × 10^7)^2 + m * (2.1 × 10^7)^2
6) Solve for M:
M = (m * (2.4 × 10^7)^2 - m * (2.1 × 10^7)^2) / ((0.3 × 10^7)^2)
= (m * (5.76 × 10^14 - 4.41 × 10^14)) / (0.09 × 10^14)
= (m * 1.35 × 10^14) / (0.09 × 10^14)
= m * (1.35/0.09)
= 15 * m
Therefore, the mass of one nucleus of the unknown element is 15 times the mass of a proton (m).
7) Finally, substitute M back into the equation for v':
v' = 0.3 × 10^7
So, the speed of the unknown nucleus immediately after the collision is 0.3 × 10^7.
To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy.
1) Conservation of momentum: Before the collision, the total momentum of the system is given by the momentum of the protons, which is p * 2.4×10^7. After the collision, the proton bounces back with a speed of -2.1×10^7. The unknown nucleus will have a momentum of Mv' (where M is the mass of the nucleus and v' is its velocity). So we have the equation:
p * 2.4×10^7 = Mv' + p * (-2.1×10^7)
2) Conservation of kinetic energy: Before and after the collision, the total kinetic energy of the system should be the same. The initial kinetic energy is given by (1/2) * p * (2.4×10^7)^2, and after the collision, the kinetic energy is given by (1/2) * M * v'^2. So we have the equation:
(1/2) * p * (2.4×10^7)^2 = (1/2) * M * v'^2 + (1/2) * p * (-2.1×10^7)^2
Now we have two equations with two unknowns (M and v'). We can solve this system of equations to find the answers.
To solve for v' in terms of M, rearrange the first equation:
v' = (p * 2.4×10^7 - p * (-2.1×10^7)) / M
Simplifying this expression:
v' = (p * (2.4×10^7 + 2.1×10^7)) / M
v' = (4.5×10^7 p) / M
Now substitute this expression for v' into the second equation:
(1/2) * p * (2.4×10^7)^2 = (1/2) * M * [(4.5×10^7 p) / M]^2 + (1/2) * p * (-2.1×10^7)^2
Simplifying this equation will give you the value of M.
Once you find the value of M, you can substitute it back into the expression for v' to find the speed of the unknown nucleus immediately after the collision.