Okay, so I need to figure out the velocity of particle A in this annihilation process. Let me start by understanding the problem step by step.

The reaction given is p + p̄ → 2A⁺ + 2A⁻. The antiproton is moving slowly, but since it's moving slowly, maybe its kinetic energy is negligible compared to the rest mass energy. The mass of particle A is given as 300 MeV/c². The question is asking for the velocity of particle A, and the options are A to D with different beta values (as a fraction of the speed of light c).

First, I remember that in particle physics, when particles annihilate, their rest mass energy is converted into other forms of energy, like kinetic energy of the products. But since the antiproton is moving slowly, I think we can approximate its kinetic energy as being much smaller than its rest mass energy. Wait, but the antiproton's rest mass is the same as the proton's, which is about 938 MeV/c². If the antiproton is moving slowly, maybe its kinetic energy is not negligible compared to the proton's rest mass, but perhaps the problem assumes that the proton's kinetic energy is much smaller than its rest mass? Hmm, the problem says the antiproton is "slowly moving," which might imply that its kinetic energy is small compared to its rest energy. But since the mass of A is 300 MeV/c², which is much smaller than the proton mass, maybe the energy available from the annihilation is significant.

Wait, the process is p + p̄ → 2A⁺ + 2A⁻. The initial particles are a proton and an antiproton. Their rest masses are each about 938 MeV/c². So the total rest mass energy is 2 * 938 = 1876 MeV. The products are two A⁺ and two A⁻, each with 300 MeV/c², so total rest mass energy is 4 * 300 = 1200 MeV. The difference between the initial rest mass and the final rest mass is 1876 - 1200 = 676 MeV. This energy must be the kinetic energy of the products. Since the products are four particles (two A⁺ and two A⁻), their total rest mass is 1200 MeV, so the kinetic energy is 676 MeV, which is distributed among them.

But wait, in the center-of-momentum frame, the total initial momentum is zero because the proton and antiproton are moving towards each other with equal and opposite momenta (since the antiproton is moving slowly, maybe the proton is moving with some velocity, but the overall center of mass frame would have zero momentum). Wait, but the problem states that the antiproton is moving slowly. If the antiproton is moving slowly, then the proton's velocity must be such that the total momentum is zero. Let me think.

Wait, the reaction is p + p̄ → 2A⁺ + 2A⁻. Let's denote the proton as p and the antiproton as p̄. The initial momentum is the sum of the momentum of the proton and the antiproton. If the antiproton is moving slowly, then its momentum is small compared to the proton's. But for the total momentum to be zero (assuming it's a head-on collision), the proton must be moving in the opposite direction with a momentum equal in magnitude but opposite in direction to the antiproton's. But if the antiproton is moving slowly, then the proton's momentum must be small as well. Wait, but the problem states that the antiproton is moving slowly, but doesn't specify the proton's velocity. Hmm, maybe I need to consider that the proton is moving in the opposite direction with a certain velocity, but the antiproton is moving slowly, so its velocity is much smaller.

Alternatively, maybe the problem is considering the antiproton at rest? No, the problem says "the antiproton is slowly moving," which suggests that its velocity is non-zero but small. However, to simplify, perhaps we can assume that the total initial momentum is zero. Let me check that.

If the antiproton is moving slowly, then its momentum p_p̄ is small, and the proton's momentum p_p must be equal in magnitude but opposite in direction to p_p̄, so that the total initial momentum is zero. So, p_p = -p_p̄. If the antiproton's velocity is v ≪ c, then its momentum is p_p̄ ≈ m_p̄ v. Similarly, the proton's momentum is p_p ≈ m_p v_p. Setting them equal in magnitude: m_p v_p = m_p̄ v. Since m_p = m_p̄ (proton and antiproton have the same mass), this implies v_p = v. So both particles have the same speed but opposite directions. Therefore, the total initial momentum is zero. But the problem states that the antiproton is moving slowly, so v is small. However, this leads to the initial total energy being 2γ m_p c², where γ is the Lorentz factor for each particle. Since v is small, γ ≈ 1 + v²/(2c²). But maybe for the sake of the problem, we can approximate the initial energy as 2 m_p c², neglecting the kinetic energy. But wait, if the initial kinetic energy is non-negligible, then the energy available for the products is 2 m_p c² - 2 m_p c² (rest mass) = 2 (γ m_p c² - m_p c²) = 2 m_p c² (γ - 1). But if the antiproton is moving slowly, then γ ≈ 1, so the kinetic energy per proton is small. However, the problem states that the antiproton is moving slowly, but perhaps the proton is moving with some velocity, making the total initial energy higher. Hmm, this is getting a bit confusing.

Alternatively, maybe the problem is considering that the antiproton and proton are colliding at rest, but that's not possible because they are moving towards each other. Wait, maybe the problem is in the lab frame where the antiproton is moving slowly, and the proton is moving in the opposite direction with a certain velocity. But without knowing the proton's velocity, it's hard to calculate. Alternatively, maybe the problem is treating the system in the center-of-momentum frame. Let me think.

In the center-of-momentum frame, the total initial momentum is zero. The total energy in that frame is the sum of the energies of the proton and antiproton. The energy of each is γ m_p c². So total energy is 2 γ m_p c². The total rest mass energy is 2 m_p c². The kinetic energy available is 2 γ m_p c² - 2 m_p c² = 2 m_p c² (γ - 1). This energy is converted into the kinetic energy of the four particles A⁺ and A⁻. Since the products are two A⁺ and two A⁻, each with rest mass 300 MeV/c², their rest mass energy is 4 * 300 = 1200 MeV. The initial rest mass energy is 2 * 938 = 1876 MeV. So the available energy for kinetic energy is 1876 - 1200 = 676 MeV. Therefore, the total kinetic energy of the products is 676 MeV. But since the products are moving, their kinetic energy depends on their velocities.

But in the center-of-momentum frame, the products must have equal and opposite momenta to conserve momentum. So the two A⁺ and two A⁻ would have momenta arranged such that the total is zero. Let's assume that the two A⁺ and two A⁻ are moving in opposite directions with equal and opposite momenta. But given that they are four particles, maybe two moving in one direction and two in the opposite. However, in reality, perhaps they are all moving in a symmetric way. But this might complicate things. Alternatively, perhaps the problem assumes that the particles are produced at rest. But that's not possible because the initial rest mass is higher than the final rest mass. Therefore, the products must have kinetic energy.

Wait, but the problem asks for the velocity of particle A. Since there are two A⁺ and two A⁻, perhaps each A⁺ and A⁻ has the same speed but opposite direction. But since the problem is symmetric, maybe each A⁺ and A⁻ has the same speed. Let me proceed with that assumption.

Let me denote the velocity of each A particle as v. Then, each A⁺ has a relativistic energy E_A = γ m_A c², where γ = 1 / sqrt(1 - v²/c²). The total energy of all four particles would be 4 γ m_A c². The initial total energy in the center-of-momentum frame is 2 γ_p m_p c², where γ_p is the Lorentz factor of the proton and antiproton. Wait, but if the antiproton is moving slowly, then its γ is approximately 1 + v_p²/(2c²), but since the proton's velocity is v_p = v (from earlier assumption), then γ_p ≈ 1 + v²/(2c²). Similarly, the proton's energy is γ_p m_p c² ≈ m_p c² (1 + v²/(2c²)). Therefore, the total initial energy is 2 γ_p m_p c² ≈ 2 m_p c² (1 + v²/(2c²)).

The total energy after the collision is 4 γ m_A c². The difference between the initial and final energies is the kinetic energy of the products. So:

4 γ m_A c² = 2 γ_p m_p c²

But wait, this can't be right because the initial energy is 2 γ_p m_p c², and the final energy is 4 γ m_A c². But the initial energy must equal the final energy (conservation of energy). Therefore:

2 γ_p m_p c² = 4 γ m_A c²

Simplify:

γ_p m_p = 2 γ m_A

But m_A = 300 MeV/c², and m_p = 938 MeV/c². So:

γ_p * 938 = 2 γ * 300

=> γ_p = (2 * 300 / 938) γ

=> γ_p ≈ (0.639) γ

But γ_p is the Lorentz factor of the proton and antiproton. Since the antiproton is moving slowly, γ_p ≈ 1 + v_p²/(2c²). Similarly, the proton's velocity is v_p = v (since p_p = m_p v_p ≈ m_p v). Therefore, γ_p ≈ 1 + v²/(2c²). 

On the other hand, the products have velocity v, so their Lorentz factor is γ = 1 / sqrt(1 - v²/c²).

Substituting into the equation:

1 + v²/(2c²) ≈ 0.639 / γ

But γ = 1 / sqrt(1 - v²/c²), so 1/γ = sqrt(1 - v²/c²). Let's denote x = v²/c². Then, 1/γ = sqrt(1 - x). Substituting:

1 + x/2 ≈ 0.639 * sqrt(1 - x)

This is a transcendental equation in x. Let's square both sides to eliminate the square root. But before that, let's rearrange:

sqrt(1 - x) ≈ (1 + x/2) / 0.639

Let me compute 1/0.639 ≈ 1.565.

So:

sqrt(1 - x) ≈ 1.565 (1 + x/2)

But wait, sqrt(1 - x) must be less than or equal to 1, but the RHS is 1.565 (1 + x/2), which is greater than 1.565. This can't be possible because sqrt(1 - x) ≤ 1. Therefore, this suggests that our initial assumption might be incorrect. Perhaps the problem is considering that the proton is moving at a high velocity, not just matching the antiproton's velocity. Wait, maybe I made a wrong assumption earlier.

Wait, let's reconsider. The problem states that the antiproton is moving slowly. So if the antiproton is moving slowly, its velocity is much less than c, so its Lorentz factor is approximately 1. Then, the proton's velocity must be such that the total initial momentum is zero. Since the antiproton is moving slowly, the proton's velocity must be approximately equal in magnitude and opposite in direction to the antiproton's velocity. But if the antiproton is moving slowly, then the proton's velocity would be approximately equal to the antiproton's velocity. Wait, but if the antiproton is moving slowly, its velocity is v ≪ c, so the proton's velocity would also be v ≪ c. Then, the total initial energy would be approximately 2 m_p c² + (p_p + p_p̄) * c. But since p_p = m_p v and p_p̄ = m_p v (since they're moving in opposite directions), the total momentum is zero. Therefore, the total initial energy would be 2 m_p c² + 2 m_p v c. However, the problem states that the antiproton is moving slowly, so v is small, so the kinetic energy is negligible compared to the rest mass. Therefore, the total initial energy is approximately 2 m_p c². The total rest mass of the products is 4 m_A c² = 4 * 300 MeV = 1200 MeV, which is less than the initial rest mass (1876 MeV). Therefore, the available energy for the products is 1876 - 1200 = 676 MeV. But this energy is the kinetic energy of the products. However, if the products are moving, their kinetic energy is (γ - 1) m_A c² each. Since there are four particles, the total kinetic energy is 4 (γ - 1) m_A c². Therefore:

4 (γ - 1) m_A c² = 676 MeV

Substitute m_A = 300 MeV/c²:

4 (γ - 1) * 300 = 676

=> (γ - 1) * 1200 = 676

=> γ - 1 = 676 / 1200 ≈ 0.5633

=> γ ≈ 1.5633

Then, γ = 1.5633, so:

γ = 1 / sqrt(1 - v²/c²) ≈ 1.5633

=> sqrt(1 - v²/c²) ≈ 1 / 1.5633 ≈ 0.639

=> 1 - v²/c² ≈ 0.639² ≈ 0.408

=> v²/c² ≈ 1 - 0.408 = 0.592

=> v/c ≈ sqrt(0.592) ≈ 0.769

So v ≈ 0.769c, which is approximately 0.77c. Looking at the options, B is 0.77c. But wait, let me check my steps again.

Wait, when I set the total energy equal to the initial energy, which is 2 m_p c² ≈ 1876 MeV (since m_p ≈ 938 MeV). Then, the rest mass of the products is 1200 MeV. The kinetic energy is 1876 - 1200 = 676 MeV. Then, the kinetic energy of each product is 676 / 4 = 169 MeV. Each product has rest mass 300 MeV, so the total energy per particle is 300 + 169 = 469 MeV. Therefore, γ = 469 / 300 ≈ 1.5633. Then, as before, v ≈ 0.77c. So the answer would be B. But let me cross-check this with another approach.

Alternatively, using conservation of energy and momentum. Since the initial momentum is zero (proton and antiproton moving in opposite directions with equal and opposite momenta), the total momentum after collision must also be zero. Therefore, the momenta of the products must cancel out. Let's assume that the two A⁺ and two A⁻ are moving in pairs. For example, two A⁺ particles move in one direction with velocity v, and two A⁻ particles move in the opposite direction with velocity v. Then, their momenta would be 2 γ m_A v and -2 γ m_A v, which sum to zero. So the total momentum is zero, which is consistent.

The total energy is then 4 γ m_A c². The initial energy is 2 γ_p m_p c², where γ_p is the Lorentz factor of the proton and antiproton. Since the antiproton is moving slowly, γ_p ≈ 1. Therefore, the initial energy is approximately 2 m_p c². The final energy is 4 γ m_A c². Setting them equal:

4 γ m_A c² ≈ 2 m_p c²

=> γ ≈ (2 m_p) / (4 m_A) = m_p / (2 m_A) = 938 / (2*300) ≈ 938 / 600 ≈ 1.5633

Which is the same as before. So γ ≈ 1.5633, leading to v ≈ 0.769c ≈ 0.77c. So the answer should be B.

But wait, let me check if there's another approach. Suppose we consider the entire system's rest frame. The initial total energy is the sum of the energies of the proton and antiproton. Since they are moving towards each other, their energies are γ_p m_p c² each. The total energy is 2 γ_p m_p c². The total momentum is zero because the proton and antiproton have equal and opposite momenta. The total rest mass is 2 m_p c². The available energy for the products is 2 γ_p m_p c² - 2 m_p c² = 2 m_p c² (γ_p - 1). This energy is converted into the kinetic energy of the four particles. The rest mass of the products is 4 m_A c². The total kinetic energy is then 2 m_p c² (γ_p - 1) = 4 (γ - 1) m_A c². Solving for γ:

2 m_p (γ_p - 1) c² = 4 (γ - 1) m_A c²

Cancel c²:

2 m_p (γ_p - 1) = 4 (γ - 1) m_A

Divide both sides by 2 m_A:

(m_p / m_A) (γ_p - 1) = 2 (γ - 1)

But since the antiproton is moving slowly, γ_p ≈ 1. Therefore, γ_p - 1 ≈ 0. So:

(m_p / m_A) * 0 ≈ 2 (γ - 1)

Which implies that γ - 1 ≈ 0, which is not possible because γ is greater than 1. Therefore, this suggests that the initial assumption that γ_p ≈ 1 is incorrect. Wait, but if the antiproton is moving slowly, γ_p ≈ 1, but then the equation leads to a contradiction. Therefore, perhaps the problem is not considering the antiproton's kinetic energy, but only the rest mass. Let me try this alternative approach.

If we neglect the kinetic energy of the antiproton (since it's moving slowly), then the initial total energy is approximately 2 m_p c². The total rest mass of the products is 4 m_A c². The available energy is 2 m_p c² - 4 m_A c² = 2*938 - 4*300 = 1876 - 1200 = 676 MeV. This energy is the total kinetic energy of the four products. Each product has kinetic energy 676 / 4 = 169 MeV. So the total energy per product is 300 + 169 = 469 MeV. Therefore, γ = 469 / 300 ≈ 1.5633. Then, solving for v:

γ = 1 / sqrt(1 - v²/c²) ≈ 1.5633

=> sqrt(1 - v²/c²) ≈ 1 / 1.5633 ≈ 0.639

=> 1 - v²/c² ≈ 0.639² ≈ 0.408

=> v²/c² ≈ 1 - 0.408 = 0.592

=> v ≈ sqrt(0.592) ≈ 0.769c ≈ 0.77c

Which brings us back to the same answer. Therefore, the correct answer should be B, 0.77c.

But let me check if there's another factor I might have missed. For example, the mass of the proton and antiproton is 938 MeV, but the problem states that the antiproton is moving slowly. If the antiproton is moving slowly, its kinetic energy is negligible, so the total initial energy is approximately 2*938 = 1876 MeV. The rest mass of the products is 4*300 = 1200 MeV. The difference is 676 MeV, which is the total kinetic energy of the products. Dividing by four gives 169 MeV per particle. Then, each particle's total energy is 300 + 169 = 469 MeV, leading to γ ≈ 1.5633. This gives v ≈ 0.77c. So the answer is B.

But wait, another approach: using the invariant quantity. The invariant mass squared is (E_total)^2 - (p_total c)^2. Before the collision, the total energy is the sum of the energies of the proton and antiproton. The total momentum is the vector sum of their momenta. Since the antiproton is moving slowly, its momentum is small, but the proton's momentum is equal in magnitude and opposite in direction. Therefore, the total momentum is approximately zero. Therefore, the invariant mass squared is (E_p + E_p̄)^2 - (p_p c - p_p̄ c)^2. Wait, if the antiproton is moving in the opposite direction with momentum p_p̄, and the proton is moving with momentum p_p, then the total momentum is p_p - p_p̄. But if they are moving in opposite directions, then the total momentum is p_p + p_p̄ (since p_p̄ is negative). Wait, no, if the antiproton is moving in the opposite direction, then its momentum is negative. So total momentum is p_p - p_p̄. But if the antiproton is moving slowly, then p_p̄ is small, so total momentum is approximately p_p. However, the problem states that the antiproton is moving slowly, but not necessarily that the proton is moving with a specific velocity. This is getting complicated. Maybe the invariant mass squared is (E_p + E_p̄)^2 - (p_p c - p_p̄ c)^2. Since the antiproton is moving slowly, p_p̄ ≈ m_p v_p̄ ≈ m_p * 0 = 0. Then, E_p̄ ≈ m_p c². E_p ≈ γ m_p c², and p_p ≈ γ m_p v_p. If the proton is moving with velocity v_p, then the invariant mass squared is (E_p + m_p c²)^2 - (γ m_p v_p c)^2. But the invariant mass squared should equal the sum of the rest masses squared, but since the products are created, the invariant mass of the system is the total energy in the center-of-momentum frame. Wait, I'm getting confused here.

Alternatively, the invariant mass of the system before collision is equal to the invariant mass after collision. The invariant mass squared is (E_total)^2 - (p_total c)^2. Before collision, E_total = E_p + E_p̄, and p_total = p_p - p_p̄ (since antiproton's momentum is opposite). If the antiproton is moving slowly, p_p̄ ≈ m_p v_p̄ ≈ 0, so E_total ≈ E_p + m_p c² ≈ γ m_p c² + m_p c². And p_total ≈ p_p ≈ γ m_p v_p. The invariant mass squared is (γ m_p c² + m_p c²)^2 - (γ m_p v_p c)^2. After collision, the invariant mass squared is (4 γ m_A c²)^2 - (0)^2 = (4 γ m_A c²)^2. Therefore:

(γ m_p c² + m_p c²)^2 - (γ m_p v_p c)^2 = (4 γ m_A c²)^2

Divide both sides by c^4:

(γ m_p + m_p)^2 - (γ m_p v_p)^2 = (4 γ m_A)^2

But since the antiproton is moving slowly, v_p ≈ 0, so the second term on the left is negligible. Therefore:

(γ m_p + m_p)^2 ≈ (4 γ m_A)^2

Take square roots:

γ m_p + m_p ≈ 4 γ m_A

Divide both sides by m_p:

γ + 1 ≈ 4 γ (m_A / m_p)

But m_A = 300 MeV/c², m_p = 938 MeV/c². So:

γ + 1 ≈ 4 γ (300 / 938)

Simplify:

γ + 1 ≈ 4 γ * 0.3198

=> γ + 1 ≈ 1.2792 γ

Subtract γ from both sides:

1 ≈ 0.2792 γ

=> γ ≈ 1 / 0.2792 ≈ 3.58

Then, γ ≈ 3.58. Then, v ≈ sqrt(1 - 1/γ²) c ≈ sqrt(1 - 1/12.83) ≈ sqrt(1 - 0.0779) ≈ sqrt(0.9221) ≈ 0.96c. Which would correspond to option D, 0.96c. But this contradicts the previous calculation. Hmm, this suggests that the answer might be D. But why the discrepancy?

Wait, this approach assumes that the invariant mass squared before collision is equal to the invariant mass squared after collision. The invariant mass before collision is sqrt( (E_p + E_p̄)^2 - (p_p c - p_p̄ c)^2 ). If the antiproton is moving slowly, then p_p̄ is small, so E_p̄ ≈ m_p c², and p_p ≈ γ m_p v. Then, the invariant mass squared is approximately (E_p + m_p c²)^2 - (γ m_p v c)^2. After collision, the invariant mass squared is (4 γ m_A c²)^2. Therefore:

(E_p + m_p c²)^2 - (γ m_p v c)^2 = (4 γ m_A c²)^2

But E_p = γ m_p c². Substitute:

(γ m_p c² + m_p c²)^2 - (γ m_p v c)^2 = (4 γ m_A c²)^2

Divide by c^4:

(γ m_p + m_p)^2 - (γ m_p v)^2 = (4 γ m_A)^2

This equation relates γ and v. However, in this case, we have two variables: γ and v. But from the center-of-momentum frame, the total momentum after collision is zero, so the total momentum before collision must also be zero. The total momentum before collision is p_p - p_p̄ ≈ p_p (since p_p̄ is small). Therefore, p_p ≈ 0, which implies that γ m_p v ≈ 0, so v ≈ 0. But this contradicts the previous assumption that the proton is moving at high speed. Therefore, this approach might not be valid.

This is getting very confusing. Let me try to find a different approach.

Alternatively, using the fact that the center-of-momentum energy is the same before and after the collision. The center-of-momentum energy squared is (E_total)^2 - (p_total c)^2. Before collision, E_total = E_p + E_p̄, and p_total = p_p - p_p̄. If the antiproton is moving slowly, then p_p̄ ≈ 0, so E_total ≈ E_p + m_p c², and p_total ≈ p_p. Therefore, the invariant mass squared is (E_p + m_p c²)^2 - (p_p c)^2. After collision, the invariant mass squared is (4 γ m_A c²)^2. Therefore:

(E_p + m_p c²)^2 - (p_p c)^2 = (4 γ m_A c²)^2

But E_p = γ m_p c², p_p = γ m_p v. Substitute:

(γ m_p c² + m_p c²)^2 - (γ m_p v c)^2 = (4 γ m_A c²)^2

Divide by c^4:

(γ m_p + m_p)^2 - (γ m_p v)^2 = (4 γ m_A)^2

This is the same equation as before. Let's substitute m_A = 300 MeV/c² and m_p = 938 MeV/c².

(γ * 938 + 938)^2 - (γ * 938 * v)^2 = (4 γ * 300)^2

Let me factor out 938^2:

[938(γ + 1)]^2 - [938 γ v]^2 = [1200 γ]^2

Divide both sides by 938^2:

(γ + 1)^2 - (γ v)^2 = (1200 / 938)^2 γ²

Compute 1200 / 938 ≈ 1.279. So:

(γ + 1)^2 - γ² v² = (1.279)^2 γ² ≈ 1.635 γ²

Expand the left side:

γ² + 2 γ + 1 - γ² v² = 1.635 γ²

Bring all terms to one side:

γ² + 2 γ + 1 - γ² v² - 1.635 γ² = 0

Combine like terms:

(1 - v² - 1.635) γ² + 2 γ + 1 = 0

=> (-0.635 - v²) γ² + 2 γ + 1 = 0

This is a quadratic equation in γ:

[ - (0.635 + v²) ] γ² + 2 γ + 1 = 0

But this seems complicated. However, we can make an assumption that the proton's velocity is such that the equation can be solved. Alternatively, perhaps there's a relationship between γ and v that can be used here. For example, γ = 1 / sqrt(1 - v²/c²). Let's substitute this into the equation.

Let me denote x = v²/c². Then, γ = 1 / sqrt(1 - x). The equation becomes:

[ - (0.635 + x) ] / (1 - x) + 2 / sqrt(1 - x) + 1 = 0

Multiply through by (1 - x):

- (0.635 + x) + 2 sqrt(1 - x) + (1 - x) = 0

Simplify:

-0.635 - x + 2 sqrt(1 - x) + 1 - x = 0

Combine like terms:

( -0.635 + 1 ) + ( -x - x ) + 2 sqrt(1 - x) = 0

=> 0.365 - 2x + 2 sqrt(1 - x) = 0

Rearrange:

2 sqrt(1 - x) = 2x - 0.365

Divide both sides by 2:

sqrt(1 - x) = x - 0.1825

Square both sides:

1 - x = x² - 0.365 x + 0.0331

Bring all terms to left:

0 = x² - 0.365 x + 0.0331 + x - 1

=> x² + 0.635 x - 0.9669 = 0

Solve the quadratic equation:

x = [ -0.635 ± sqrt(0.635² + 4 * 0.9669) ] / 2

Calculate discriminant:

0.635² ≈ 0.403

4 * 0.9669 ≈ 3.8676

Total discriminant ≈ 0.403 + 3.8676 ≈ 4.2706

sqrt(4.2706) ≈ 2.066

So:

x = [ -0.635 ± 2.066 ] / 2

Take the positive root:

x = ( -0.635 + 2.066 ) / 2 ≈ 1.431 / 2 ≈ 0.7155

Therefore, v²/c² ≈ 0.7155 => v ≈ sqrt(0.7155) ≈ 0.846c. Which is approximately 0.85c, which is not among the answer choices. The closest option is C (0.91c), but this suggests that the answer might be B or D. However, this approach gives a different result. There's a contradiction here.

Wait, but this approach assumes that the antiproton's momentum is negligible, which might not be valid. If the antiproton is moving slowly, its momentum is small, but its energy is still significant (m_p c²). Therefore, the initial total energy is E_p + E_p̄ ≈ γ m_p c² + m_p c². The invariant mass squared is (E_p + m_p c²)^2 - (γ m_p v c)^2. After collision, the invariant mass squared is (4 γ m_A c²)^2. Therefore:

(γ m_p c² + m_p c²)^2 - (γ m_p v c)^2 = (4 γ m_A c²)^2

Divide by c^4:

(γ m_p + m_p)^2 - (γ m_p v)^2 = (4 γ m_A)^2

We can factor out m_p^2:

m_p^2 (γ + 1)^2 - m_p^2 γ² v² = 16 m_A² γ²

Divide both sides by m_p^2:

(γ + 1)^2 - γ² v² = (16 m_A² / m_p²) γ²

But m_A = 300 MeV/c², m_p = 938 MeV/c². So:

(γ + 1)^2 - γ² v² = (16 * 300² / 938²) γ²

Calculate 16 * 300² = 16 * 90000 = 1,440,000

938² ≈ 880,000 (since 900² = 810,000, 938 is 38 more, so (900+38)^2 = 900² + 2*900*38 + 38² ≈ 810,000 + 68,400 + 1,444 ≈ 879,844)

Therefore, 1,440,000 / 879,844 ≈ 1.633

So:

(γ + 1)^2 - γ² v² = 1.633 γ²

Expand (γ + 1)^2:

γ² + 2γ + 1 - γ² v² = 1.633 γ²

Rearrange:

γ² (1 - v²) + 2γ + 1 = 1.633 γ²

Bring all terms to left:

γ² (1 - v² - 1.633) + 2γ + 1 = 0

=> γ² (-0.633 - v²) + 2γ + 1 = 0

Again, substitute γ = 1 / sqrt(1 - x) where x = v²/c². Let me denote x = v²/c²:

=> [ -0.633 - x ] / (1 - x) + 2 / sqrt(1 - x) + 1 = 0

Multiply through by (1 - x):

-0.633 - x + 2 sqrt(1 - x) + (1 - x) = 0

Simplify:

(-0.633 - x + 1 - x) + 2 sqrt(1 - x) = 0

=> (0.367 - 2x) + 2 sqrt(1 - x) = 0

Rearrange:

2 sqrt(1 - x) = 2x - 0.367

Divide by 2:

sqrt(1 - x) = x - 0.1835

Square both sides:

1 - x = x² - 0.367 x + 0.0337

Rearrange:

x² - 0.367 x + 0.0337 + x - 1 = 0

=> x² + 0.633 x - 0.9663 = 0

Solve using quadratic formula:

x = [ -0.633 ± sqrt(0.633² + 4 * 0.9663) ] / 2

Calculate discriminant:

0.633² ≈ 0.400

4 * 0.9663 ≈ 3.865

Total ≈ 0.400 + 3.865 ≈ 4.265

sqrt(4.265) ≈ 2.065

Thus:

x = [ -0.633 ± 2.065 ] / 2

Take positive root:

x = ( -0.633 + 2.065 ) / 2 ≈ 1.432 / 2 ≈ 0.716

Therefore, v ≈ sqrt(0.716) ≈ 0.846c, which is ~0.85c. But this is not matching the previous result of 0.77c. There's a discrepancy here. Which approach is correct?

Alternatively, maybe the problem is designed so that the answer is B.0.77c. Let me check the initial approach again.

If we assume that the antiproton's momentum is negligible, then the initial total energy is 2 m_p c² ≈ 1876 MeV, and the rest mass of the products is 1200 MeV. The difference is 676 MeV, which is the total kinetic energy. Dividing by 4 gives 169 MeV per particle. Then, the total energy per particle is 300 + 169 = 469 MeV. The Lorentz factor γ = 469 / 300 ≈ 1.5633. Then, v ≈ sqrt(1 - 1/1.5633²) ≈ 0.77c. This gives B.0.77c.

Alternatively, if we consider the antiproton's momentum, but since it's moving slowly, perhaps we can approximate its energy as m_p c², and its momentum as negligible. Therefore, the initial total energy is approximately 2 m_p c², leading to the same result. Therefore, the answer should be B.

Alternatively, the correct answer is C.0.91c. But according to the calculations, it's around 0.85c. But the problem states that the antiproton is "slowly moving", which might imply that its velocity is small, so the initial energy is mainly from the proton's rest mass. Wait, but if the antiproton is moving slowly, its energy is close to m_p c², and the proton's energy is also m_p c², so total initial energy is 2 m_p c², which is the same as if both were at rest. Therefore, the kinetic energy available is 2 m_p c² - 4 m_A c². That is, 2*938 - 4*300 = 1876 - 1200 = 676 MeV. Divided by four, 169 MeV per particle. Therefore, the calculation for v is the same as before, leading to B.0.77c.

Alternatively, maybe the problem is considering the entire system's center-of-momentum frame. In that case, the total momentum is zero, so the energy in that frame is the sum of the rest masses. But the problem is asking for the velocity of the products, which are moving in the lab frame. Wait, perhaps the products are moving in the lab frame, but the problem states "the velocity of particle A", so the lab frame is probably the same as the center-of-momentum frame? No, the lab frame has the antiproton moving slowly, so the initial momentum is almost zero. Therefore, the center-of-momentum frame is almost the lab frame. Therefore, the velocity of particle A in the lab frame would be approximately the same as in the center-of-momentum frame.

Alternatively, perhaps the problem is considering that the antiproton's velocity is negligible, so the lab frame is the same as the center-of-momentum frame. Therefore, the products' velocities are determined in the lab frame, which is the same as the center-of-momentum frame. Therefore, the calculation leading to B.0.77c is correct.

But why does the other approach give a different answer? Because in that approach, the antiproton's momentum is considered non-negligible, but the problem states it's moving slowly. Therefore, the first approach is more appropriate.

Therefore, I think the correct answer is B.0.77c.
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