Posted by **HANNE** on Sunday, October 28, 2012 at 5:29am.

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...

## Answer this Question

## Related Questions

- physics-mechanics-collisions,momentum, impulse - You are at the controls of a ...
- physics-mechanics-collisions,momentum, impulse - You are at the controls of a ...
- Physics HELP! - You are at the controls of a particle accelerator, sending a ...
- physics - A beam of protons is moving toward a target in a particle accelerator...
- physics - In a certain particle, a current of 0.50 mA is carried by a 5.0MeV ...
- chemistry - A proton beam is made by accelerating protons in a high voltage ...
- Physics - Each of the protons in a particle beam has a kinetic energy of 3.05 10...
- Science-Periodic Table - whats the connection/relationship between the # of ...
- Physics - Some forms of cancer can be treated using proton therapy in which ...
- Physics - Some forms of cancer can be treated using proton therapy in which ...

More Related Questions