Answer:
(a) The final velocity of the first proton after the collision = 8.3 m/s in the opposite to its initial direction
(b) The final velocity of the second proton after the collision = 7 m/s in the opposite to its initial direction
Explanation:
The given parameter of the protons are;
The velocity of the first proton, v₁ = 7.0 Mm/s
The velocity of the second proton, v₂ = -8.3 Mm/s
The type of collision = Elastic collision
In an elastic collision, the kinetic and momentum energies are conserved, therefore, we have, for the initial and final momentums;
(m₁·v₁ + m₂·v₂)₁ = (m₁·v₁ + m₂·v₂)₂
1/2·(m₁·v₁² + m₂·v₂²)₁ = 1/2·(m₁·v₁² + m₂·v₂²)₂
Where, m₁ = m₂ or the protons, we get;
(v₁ + v₂)₁ = (v₁ + v₂)₂
(v₁² + v₂²)₁ = (v₁² + v₂²)₂
Therefore;
7.0 - 8.3 = v₁ + v₂
-1.3 = v₁ + v₂...(1)
7.0² + (-8.3)² = v₁² + v₂²
117.49 = v₁² + v₂²...(2)
From equation (1), we have;
v₁ = -1.3 - v₂
Plugging the value v₁ = -1.3 - v₂ in equation (2) gives;
117.49 = v₁² + v₂² = (-1.3 - v₂)² + v₂² = 2·v₂² + 12·v₂ + 1.69
∴ 2·v₂² + 2.6·v₂ + 1.69 - 117.49 = 0
2·v₂² + 2.6·v₂ - 115.8 = 0
Using the quadratic formula, we have;
v₂ = (-2.6 ± √(2.6² - 4×2×(-115.8)))/(2 × 2)
∴ v₂ ≈ -8.3 m/s or 7 m/s
When v₂ ≈ -8.3, v₁ = -1.3 - v₂ ≈ -1.3 - (-8.3) = 7
When v₂ ≈ 7, v₁ = -1.3 - v₂ ≈ -1.3 - (7) = -8.3
Therefore, the final velocity of the first proton after the collision = 8.3 m/s in the opposite to its initial direction
(b) The final velocity of the second proton after the collision = 7 m/s in the opposite to its initial direction.
