A proton moving at v1 = 7.0 Mm/s collides elastically head-on with a second proton moving in the opposite direction at v2 = 8.3 Mm/s. (a) Find the velocity of the first proton after the collision. (b) Find the velocity of the second proton after the collision

(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.

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explanation:

graph slopes whilst gas cools, then it is flat whilst conensation to liquid happens, the coolong liquid then graphs as a slope when it solidifies graph is flat

latent heat