The energy of the system, tension in the string, and speed of each cube are respectively; U = 0.72 J, T = 14 N and (V₂, V₄) = (-22, 14)
<h3>Calculation of energy and tension in a String</h3>
We are given;
Mass of cube 1; m₁ = 2 g
Mass of cube 2; m₂ = 4 g
Distance between cubes; d = 5 cm = 0.05 m
Charges of cubes; Q = q = +2. 0 μc = 2.0 × 10⁻⁶ C
A) Formula for the energy of the system is;
U = kQq/d
U = 8.99 × 10⁹ × (2.0 × 10⁻⁶ C)²/0.05 m
U = 0.03596/0.05
U = 0.72 J
B) Formula for the Tension is;
T = U/d
T = 0.72/0.05
T ≈ 14 N
C) Momentum is conserved, and as such the initial momentum is zero. Thus;
0 = (0.0020 × V₂) + (0.0040 × V₄)
⇒ V₂ = -2V₄
Energy is also conserved and so;
(½ × 0.0020 × (-2V₄)²) + (½ × 0.004 × (V₄)²) = 0.72 J
-0.0040V₄² + 0.002V₄² = 0.72 J
0.0060V₄² = 0.72 J
V₄² = 0.72/0.0060
V₄² = 120
V₄ = √120
V₄ ≈ 11 m/s
Recall that; V₂ = -2V₄
Thus;
V₂ = -2(11) m/s
V₂ = -22 m/s
Thus;
(V₂, V₄) = (-22, 14)
Read more about Conservation of Momentum at; brainly.com/question/7538238
