An alpha-particle collides with an oxygen nucleus which is initially at rest. The alpha-particle is scattered at an angle of 67.0 degrees from its initial direction of motion, and the oxygen nucleus recoils at an angle of 52.0 degrees on the other side of this initial direction. What is the ratio, for alpha-particle to oxygen nucleus, of the final speeds of these particles? The mass of the oxygen nucleus is four times that of the alpha particle.
Well, I guess you could say that the alpha-particle and the oxygen nucleus had a "bumpy" encounter. They were just going about their business, when suddenly, BAM! Collision time!
Now, let's break down the situation. The alpha-particle is scattered at an angle of 67.0 degrees from its initial direction, while the oxygen nucleus recoils at an angle of 52.0 degrees on the other side. It's like they were playing a game of "dodgeball" and ended up going in opposite directions.
But wait, there's more! The mass of the oxygen nucleus is four times that of the alpha particle. So, it's like a classic case of "heavyweight vs. featherweight." The oxygen nucleus is the big guy in this scenario.
Now, let's talk about speed. We want to find the ratio of the final speeds of these particles. Well, since the oxygen nucleus is much heavier, it will probably have a slower speed compared to the alpha-particle. It's like watching a race between a tortoise and a hare, with the oxygen nucleus as the tortoise.
So, in conclusion, the ratio of the final speeds will be in favor of the alpha-particle. The alpha-particle will be zooming away with its light-weightedness, while the oxygen nucleus will be crawling along. It's like a comedic version of "Fast & Furious," where the alpha-particle is Vin Diesel and the oxygen nucleus is a slow-moving snail.
Hope that brings a smile to your face! If you have any more physics questions or need a good laugh, feel free to ask!
To solve this problem, we can use the conservation of linear momentum and the conservation of kinetic energy.
Let's say the initial speeds of the alpha particle and the oxygen nucleus are v_alpha and v_oxygen, respectively.
According to the conservation of linear momentum, the total initial momentum before the collision is equal to the total final momentum after the collision:
m_alpha * v_alpha_initial = m_alpha * v_alpha_final + m_oxygen * v_oxygen_final
We know that the mass of the oxygen nucleus (m_oxygen) is four times that of the alpha-particle (m_alpha).
So, the equation becomes:
m_alpha * v_alpha_initial = m_alpha * v_alpha_final + 4m_alpha * v_oxygen_final
Simplifying the equation:
v_alpha_initial = v_alpha_final + 4v_oxygen_final
Now, let's consider the conservation of kinetic energy.
The initial kinetic energy before the collision is equal to the final kinetic energy after the collision:
(1/2) * m_alpha * v_alpha_initial^2 = (1/2) * m_alpha * v_alpha_final^2 + (1/2) * 4m_alpha * v_oxygen_final^2
The masses (m_alpha and 4m_alpha) cancel out, so the equation becomes:
v_alpha_initial^2 = v_alpha_final^2 + 4v_oxygen_final^2
Now, let's solve these two equations simultaneously:
v_alpha_initial = v_alpha_final + 4v_oxygen_final (Equation 1)
v_alpha_initial^2 = v_alpha_final^2 + 4v_oxygen_final^2 (Equation 2)
From Equation 1, we can isolate v_alpha_final:
v_alpha_final = v_alpha_initial - 4v_oxygen_final
Substituting this value in Equation 2:
(v_alpha_initial - 4v_oxygen_final)^2 = v_alpha_final^2 + 4v_oxygen_final^2
Expanding and simplifying the equation:
v_alpha_initial^2 - 8v_alpha_initial * v_oxygen_final + 16v_oxygen_final^2 = v_alpha_final^2 + 4v_oxygen_final^2
v_alpha_initial^2 - v_alpha_final^2 = 12v_oxygen_final^2 + 8v_alpha_initial * v_oxygen_final
(v_alpha_initial + v_alpha_final) * (v_alpha_initial - v_alpha_final) = 12v_oxygen_final^2 + 8v_alpha_initial * v_oxygen_final
Now, recall that the alpha-particle is scattered at an angle of 67.0 degrees and the oxygen nucleus recoils at an angle of 52.0 degrees.
From trigonometry, we know that:
v_alpha_initial = v_alpha_final / sin(67.0)
v_oxygen_final = v_alpha_final / sin(52.0)
Substituting these values in the previous equation:
(v_alpha_final / sin(67.0) + v_alpha_final) * (v_alpha_final / sin(67.0) - v_alpha_final) = 12 * (v_alpha_final / sin(52.0))^2 + 8 * (v_alpha_final / sin(67.0))^2
Now, we can simplify the equation further by dividing both sides by v_alpha_final and rearranging:
(1/sin(67.0) + 1) * (1/sin(67.0) - 1) = 12 * (1/sin(52.0))^2 + 8 * (1/sin(67.0))^2
Solving this equation will give us the ratio of the final speeds of the alpha-particle to the oxygen nucleus.
To solve this problem, we need to use conservation of momentum and conservation of kinetic energy.
Let's denote the initial speed of the alpha-particle as v_alpha and the initial speed of the oxygen nucleus as v_oxygen. Since the oxygen nucleus is initially at rest, its initial speed (v_oxygen) is zero.
According to conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision.
The momentum of the alpha-particle before the collision is given by p_alpha = m_alpha * v_alpha, where m_alpha is the mass of the alpha-particle.
The momentum of the oxygen nucleus before the collision is zero, since it is initially at rest.
After the collision, the alpha-particle is scattered at an angle of 67.0 degrees, so let's denote its final speed as v_alpha_final and its final momentum as p_alpha_final. The oxygen nucleus recoils at an angle of 52.0 degrees in the opposite direction, so let's denote its final speed as v_oxygen_final and its final momentum as p_oxygen_final.
Using conservation of momentum, we have:
m_alpha * v_alpha = p_alpha_final
0 = p_oxygen_final
Now, let's consider conservation of kinetic energy.
The kinetic energy of the alpha-particle before the collision is given by KE_alpha = (1/2) * m_alpha * v_alpha^2.
The kinetic energy of the oxygen nucleus before the collision is zero, since it is initially at rest.
After the collision, the alpha-particle still has kinetic energy, so we can express its final kinetic energy as KE_alpha_final = (1/2) * m_alpha * v_alpha_final^2.
The oxygen nucleus also gains some kinetic energy, so we can express its final kinetic energy as KE_oxygen_final = (1/2) * m_oxygen * v_oxygen_final^2, where m_oxygen is the mass of the oxygen nucleus.
Using conservation of kinetic energy, we have:
(1/2) * m_alpha * v_alpha^2 = (1/2) * m_alpha * v_alpha_final^2 + (1/2) * m_oxygen * v_oxygen_final^2
To find the ratio of the final speeds of the particles, we can divide the equation above by the kinetic energy of the oxygen nucleus:
[(1/2) * m_alpha * v_alpha^2] / [(1/2) * m_oxygen * v_oxygen_final^2] = [(1/2) * m_alpha * v_alpha_final^2 / (1/2) * m_oxygen] / (v_oxygen_final^2 / 1)
Canceling out the common terms and substituting the given mass relationship (m_oxygen = 4 * m_alpha), we have:
(v_alpha^2 / v_oxygen_final^2) = (v_alpha_final^2 / 4)
Now, let's substitute the given angles into the equation.
We know that the alpha-particle is scattered at an angle of 67.0 degrees from its initial direction, so the angle between its initial and final direction is 67.0 degrees.
Similarly, the oxygen nucleus recoils at an angle of 52.0 degrees on the other side, so the angle between its initial and final direction is also 52.0 degrees.
Using the relationship between the scattering angle and the initial and final speeds, we can write:
tan(67.0 degrees) = v_alpha_final / v_alpha --(1)
tan(52.0 degrees) = v_oxygen_final / 0 --(2)
Since tan(52.0 degrees) is undefined, we cannot directly solve for v_oxygen_final using equation (2).
However, we can use the fact that for small scattering angles, sin(theta) ~ tan(theta). In this case, we can approximate tan(52.0 degrees) as sin(52.0 degrees).
Using this approximation, we can rewrite equation (2) as:
sin(52.0 degrees) = v_oxygen_final / 0 --(3)
Since sin(52.0 degrees) = 0, we find that v_oxygen_final = 0.
Now, substituting v_oxygen_final = 0 into the equation (v_alpha^2 / v_oxygen_final^2) = (v_alpha_final^2 / 4), we have:
(v_alpha^2 / 0^2) = (v_alpha_final^2 / 4)
0 = (v_alpha_final^2 / 4)
Therefore, we find that v_alpha_final = 0 as well.
Now, let's find the ratio of the final speeds of the particles:
Ratio = v_alpha_final / v_oxygen_final = 0 / 0
Since both v_alpha_final and v_oxygen_final are zero, the ratio is undefined.
Therefore, the ratio of the final speeds of the alpha-particle to the oxygen nucleus is undefined.