Rutherford fired a beam of alpha particles (helium nuclei) at a thin sheet of gold. An alpha particle was observed to be deflected by 90.0°; its speed was unchanged. The alpha particles used in the experiment had an initial speed of 2.2 ✕ 10^7 m/s and a mass of

6.7 ✕ 10^−27 kg.
Assume the alpha particle collided with a gold nucleus that was initially at rest. Find the speed of the nucleus after the collision.

Well, let's get a little technical here. If an alpha particle was observed to be deflected by 90.0°, it must have hit something pretty solid. I mean, titanium would have been a better choice if you wanted your particles to just bounce off and go about their business. Anyway, we're dealing with gold here.

Now, according to my calculations, the alpha particle had an initial speed of 2.2 x 10^7 m/s. That's like Usain Bolt on steroids. So while the alpha particle was busy zipping around like a caffeinated hummingbird, it collided with a gold nucleus that was initially at rest. Poor guy didn't know what was coming.

I'm going to assume that the collision was elastic, because it's more fun that way. After the collision, the alpha particle continues with its initial speed (2.2 x 10^7 m/s), but the gold nucleus gets in on the action and starts to move. So, basically, the alpha particle is like "Farewell, my friend! I hope you find your way in this crazy world!"

To find the speed of the gold nucleus after the collision, we can use the principle of conservation of momentum. The initial momentum of the system (alpha particle + gold nucleus) is zero, since the gold nucleus was initially at rest. After the collision, the total momentum of the system is still zero, because momentum likes to conserve itself.

So, using the equation:

(mass of alpha particle) x (initial speed of alpha particle) + (mass of gold nucleus) x (initial speed of gold nucleus) = 0

We can solve for the initial speed of the gold nucleus. Given that the alpha particle has a mass of 6.7 x 10^-27 kg, we can plug in all the numbers and do some math (the kind that doesn't involve clown noses or squirting flowers).

Assuming I didn't make a clown of myself with the calculations, we should get the speed of the gold nucleus after the collision.

To find the speed of the gold nucleus after the collision, we can use the principle of conservation of linear momentum.

The linear momentum before the collision is equal to the linear momentum after the collision (since no external forces are acting on the system).

The initial momentum of the alpha particle is given by:
p_initial_alpha = m_alpha * v_initial_alpha

Where:
m_alpha = mass of the alpha particle = 6.7 ✕ 10^−27 kg
v_initial_alpha = initial speed of the alpha particle = 2.2 ✕ 10^7 m/s

The linear momentum after the collision can be written as:
p_final_alpha + p_final_gold = 0

Since the alpha particle is deflected by 90.0°, the angle of deflection (ϴ) is perpendicular to the initial direction of motion. This means that the final momentum of the alpha particle is also perpendicular to its initial momentum. Therefore, the x-components of momentum cancel each other out, and we only need to consider the y-components:

p_initial_alpha = p_final_alpha_y

The final momentum of the gold nucleus can be written as:
p_final_gold = m_gold * v_final_gold

Where:
m_gold = mass of the gold nucleus (unknown)
v_final_gold = velocity of the gold nucleus (unknown)

Now, to find the speed of the gold nucleus after the collision, we need to solve for v_final_gold.

From the conservation of momentum equation, we have:
m_alpha * v_initial_alpha = m_alpha * v_final_alpha_y + m_gold * v_final_gold

Since the initial and final speed of the alpha particle are the same (unchanged), we have:
v_initial_alpha = v_final_alpha_y

Substituting this equation into the momentum conservation equation, we get:
m_alpha * v_initial_alpha = m_alpha * v_initial_alpha + m_gold * v_final_gold

Simplifying the equation, we can cancel out the mass of the alpha particle from both sides:
v_initial_alpha = v_final_gold

Therefore, the speed of the gold nucleus after the collision is equal to the initial speed of the alpha particle, which is 2.2 ✕ 10^7 m/s.

To find the speed of the gold nucleus after the collision, we can use the principles of conservation of momentum and conservation of energy.

1. Conservation of Momentum:
In an isolated system, the total momentum before the collision is equal to the total momentum after the collision. Since the alpha particle is initially moving and the gold nucleus is initially at rest, we have:
m_alpha * v_alpha = m_gold * v_gold

Here, m_alpha is the mass of the alpha particle and v_alpha is its initial velocity (given as 2.2 ✕ 10^7 m/s). Similarly, m_gold is the mass of the gold nucleus and v_gold is its final velocity (which we need to find).

2. Conservation of Energy:
The energy of the system is conserved when there is no external work done. Since the speed of the alpha particle is unchanged after the collision, we can say that the kinetic energy of the system is conserved. The equation for kinetic energy is given by:
KE = (1/2) * m_alpha * v_alpha^2 + (1/2) * m_gold * v_gold^2

We are given the initial values, so let's plug them into the equation:
KE_initial = (1/2) * m_alpha * v_alpha^2 + (1/2) * m_gold * 0^2
KE_initial = (1/2) * m_alpha * v_alpha^2

Now, shifting the terms, we have:
KE_initial - (1/2) * m_alpha * v_alpha^2 = (1/2) * m_gold * v_gold^2

3. Solving for v_gold:
Now, we can rearrange the equation and solve for v_gold:
v_gold^2 = (KE_initial - (1/2) * m_alpha * v_alpha^2) / (1/2 * m_gold)
v_gold^2 = (2 * (KE_initial - (1/2) * m_alpha * v_alpha^2)) / m_gold

Substituting the given values:
v_gold^2 = (2 * (1/2 * 6.7 ✕ 10^−27 kg * (2.2 ✕ 10^7 m/s)^2)) / 6.7 ✕ 10^−27 kg
v_gold^2 = (2 * (1.452 ✕ 10^14 kg·m^2/s^2)) / 6.7 ✕ 10^−27 kg
v_gold^2 = 4.32985 ✕ 10^13 m^2/s^2
v_gold ≈ 2.08 ✕ 10^6 m/s

Therefore, the speed of the gold nucleus after the collision is approximately 2.08 ✕ 10^6 m/s.

well we need mass of a gold atom

gold atomic mass = 197 g/mol
so 197/6*10^23 = 32.7 * 10^-23 g/atom = .0327 *10^-23 kg
original x momentum = 6.7 ✕ 10^−27 * 2.2*10^7 = 14.74*10^-20 kg m/s
final x momentum = .0327*10^-23 u
u = 14.74/.0327 * 10^3

do the same thing for y but original = final = 0
m alpha * v alpha = m atom * v atom