How Does the Rutherford Experiment Illustrate Momentum Conservation?

[physco827]

The Rutherford experiment
An alpha particle (a helium nucleus, containing 2 protons and 2 neutrons) starts out with kinetic energy of 10.2 MeV (10.2 106 eV), and heads in the +x direction straight toward a gold nucleus (containing 79 protons and 118 neutrons). The particles are initially far apart. Answer the following questions about the collision.
What is the initial momentum of the alpha particle? (You may assume its speed is small compared to the speed of light).

What I found:
alpha,i = < 1.4733e-19 , 0, 0 > kg·m/s
What is the initial momentum of the gold nucleus?
Au,i = < 0 , 0 , 0 > kg ·m/s

What i can't get:
What is the final momentum of the alpha particle, long after it interacts with the gold nucleus?
alpha,f = < ? , 0 , 0 > kg·m/s
What is the final momentum of the gold nucleus, long after it interacts with the alpha particle?
Au,f = < ? , 0 , 0 > kg·m/s

I tried conservation of momentum but all the momentums cancel themselves out. I don't understand how to get the final momentum of the alpha particle.

 

[anathema]

you are correct in using the conservation of momentum principle to solve this problem. However, it is important to consider the type of collision that is taking place between the alpha particle and the gold nucleus. In this case, it is an elastic collision, meaning that both momentum and kinetic energy are conserved.

To find the final momentum of the alpha particle, you can use the equation p = mv, where p is the momentum, m is the mass, and v is the velocity. Since the alpha particle is initially at rest in the y and z directions, its final momentum in these directions will also be zero. Therefore, the only component of the final momentum that we need to find is in the x direction.

To find the final velocity of the alpha particle in the x direction, we can use the conservation of kinetic energy equation:

(1/2)mv^2 = (1/2)mv^2

where m is the mass of the alpha particle, v is its initial velocity (which is given in the problem), and v' is its final velocity. Solving for v', we get:

v' = sqrt((2*10.2*10^6 eV)/(m))

Since we are given the kinetic energy in electron volts (eV), we need to convert it to joules (J) in order to use it in the equation. We can use the conversion 1 eV = 1.6*10^-19 J to get:

v' = sqrt((2*10.2*10^6*1.6*10^-19 J)/(6.64*10^-27 kg)) = 1.78*10^7 m/s

Now that we have the final velocity in the x direction, we can use the equation p = mv to find the final momentum:

alpha,f = (6.64*10^-27 kg)(1.78*10^7 m/s) = 1.18*10^-19 kg·m/s

Using the same approach, we can find the final momentum of the gold nucleus. Since it is much more massive than the alpha particle, its final velocity will be very small and can be neglected. Therefore, the final momentum of the gold nucleus will be the same as its initial momentum:

Au,f = < 0 , 0 , 0 > kg ·m/s

This result makes sense, as the gold nucleus is much

 

[Moetasim]

The final momentum of the alpha particle can be calculated using the law of conservation of momentum, which states that the total momentum of a system before and after a collision must be equal. In this case, the initial momentum of the alpha particle (1.4733e-19 kg·m/s) must be equal to the final momentum after the collision. However, in order to calculate the final momentum, we need to know the velocity of the alpha particle after the collision.

To find the velocity, we can use the law of conservation of energy, which states that the total energy of a system before and after a collision must be equal. In this case, the initial kinetic energy of the alpha particle (10.2 MeV) must be equal to the final kinetic energy after the collision. We can use this to solve for the final velocity of the alpha particle.

Once we have the final velocity, we can use the formula for momentum (p = mv) to calculate the final momentum of the alpha particle. Similarly, we can use conservation of momentum and energy to calculate the final momentum of the gold nucleus.

It is also worth noting that the interaction between the alpha particle and the gold nucleus will depend on the angle at which the alpha particle approaches the nucleus. If it is a head-on collision, the final momentum of both particles will be in the opposite direction of their initial momentum. However, if the alpha particle approaches at an angle, the final momentum will have both a magnitude and a direction. This can be calculated using vector addition.