Question

You are at the controls of a particle accelerator, sending a beam of 3.60 x10^7 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 3.30 x 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?

Answer 1

To find the mass of one nucleus of the unknown element in a particle accelerator, use the principle of conservation of momentum. The speed of the unknown nucleus after a collision can be found using the principle of conservation of kinetic energy.

Step-by-step explanation:

(a) Mass of one nucleus:

To find the mass of one nucleus of the unknown element, we can use the principle of conservation of momentum. Since the collision is elastic, the momentum before the collision must be equal to the momentum after the collision.

We can define the momentum before the collision as:

m x v = (mass of proton) x (initial speed of proton)

And the momentum after the collision as:

m x v' = (mass of nucleus) x (rebound speed of proton)

Let's solve these equations for the mass of the nucleus:

mass of nucleus = (mass of proton x initial speed of proton) / (rebound speed of proton)

(b) Speed of the unknown nucleus:

We can use the principle of conservation of kinetic energy to find the speed of the unknown nucleus after the collision. Since the collision is elastic, the total kinetic energy before the collision must be equal to the total kinetic energy after the collision.

The total kinetic energy before the collision is:

0.5 x (mass of proton) x (initial speed of proton)^2

The total kinetic energy after the collision is:

0.5 x (mass of nucleus) x (rebound speed of proton)^2

Let's solve these equations for the speed of the unknown nucleus:

speed of unknown nucleus = sqrt((2 x (mass of proton) x (initial speed of proton)^2) / (mass of nucleus))

Answer 2

a) mass of unknown nucleus = 0.04245 mp, where mp is the proton mass

b) Speed of the unknown nucleus = (7.067 x 10^7) m/s

Step-by-step explanation:

Considering the initial conditions, the observed collisions are ellastic, i.e, the total kinetic energy are conserved. The proton's mass will refer as
`m_(p)`.

(a)

Total kinetic energy conservation

`(1)/(2)m_(p)v_(p_0)^(2)+(1)/(2)m_(u)v_(u_o)^(2)=(1)/(2)m_(p)v_(p_f)^(2)+(1)/(2)m_(u)v_(u_f)^(2)`

where
`v_(u_o)` represents the initial velocity of the unknown element,
`m_(u)` the mass of the unknown element, and
`v_(u_f)` the final velocity of the unknown element

Linear momentum conservation

`m_(p)v_(p_0)+m_(u)v_(u_o)=m_(p)v_(p_f)+m_(u)v_(u_f)`

Using the initial speed of the target nucleus (unknown) is negligible, i.e, its speed is zero. Thereby, using the relation of linear momentum conservation given above, it is possible to find an expression of the final speed of the unknown nucleus in terms of its mass, which can be inserted in the relation of the kinetic energy conservation to obtain the value of the mass of the unknown elements, as follows;

`m_(u)v_(u_f)=m_(p)v_(p_0)-m_(p)v_(p_f)\\\\v_(u_f)=(m_(p)(v_(p_0)-v_(p_f)))/(m_(u))`

Substituting this expression in the relation of total kinetic energy conservation,

`m_(p)(v^(2)_(p_0)-v^(2)_(p_f))={m_(u)}v^(2)_(u)_(f)`

Then,

`m_(p)(v^(2)_(p_0)-v^(2)_(p_f))={m_(u)}(m^(2)_(p)(v_(p_0)-v_(p_f))^(2))/(m^(2)_(u))\\\\m_(u)= (m_(p)(v_(p_0)-v_(p_f))^(2))/((v^(2)_(p_0)-v^(2)_(p_f)))`

Replacing the given data

`m_(u)= (m_(p)(3.6x10^(7)-3.3x10^(7))^(2))/(((3.6x10^(7))^(2)-(3.3x10^(7))^(2)))`

Then,

`m_(u)=0.04245m_(p)`

(b) Using the relation of the final speed from linear momentum conservation and the above result, the speed of the unknown nucleus is calculated

`v_(u_f)=(m_(p)(v_(p_0)-v_(p_f)))/(m_(u))\\\\v_(u_f)=(m_(p)(3.6x10^(7)-3.3x10^(7)))/(0.04245m_(p))\\\\v_(u_f)=7.067x10^(7) m/s`