Respuesta :
The instantaneous velocity of the 15 kg. mass just after collision can be found by the principle of linear momentum.
a) The speed of the 15 kg. just after collision is 2 m/s.
b) The type of collision is inelastic collision
c) The compression of the spring is approximately 0.323 m.
Reasons:
The given parameters are;
Mass of the block attached to the spring, m₁ = 15.0 kg
Force constant of the spring, K = 575 N/m
Mass of the stone that strikes the block, m₂ = 3.00 kg
Speed of the stone, v₂ = 8.00 m/s
Speed with which the stone rebounds, v₃ = 2.00 m/s
a) The total initial momentum = 3 kg. × 8 m/s = 24 kg·m/s
The final momentum, just after collision = 3 × (-2) kg·m/s + 15 kg ×v₁
By conservation of momentum, we have;
24 kg·m/s = 3 × (-2) kg·m/s + 15 kg ×v₁
[tex]v_1 = \dfrac{24 \, kg \cdot m/s + 6 \, kg \cdot m/s}{15 \, kg} = 2 \, m/s[/tex]
The speed of the 15 kg. just after collision, v₁ = 2 m/s.
b) A collision is elastic when the kinetic energy of the collision is conserved
The initial kinetic energy, K.E.₁ = 0.5 × 3 kg. ×(8 m/s)² = 96 J
The sum of the final kinetic energy are;
0.5 × 3 kg. × (2 m/s)² + 0.5 × 15 kg × (2 m/s)² = 36 J
The initial kinetic energy ≠ The final kinetic energy
Therefore, the collision is not elastic
(c) The kinetic energy given by the block = The elastic potential energy gained by the spring
Kinetic energy of the block, K.E. = 0.5 × 15 kg × (2 m/s)² = 30 J
Elastic energy gained by the block = 0.5 × K × x² = 0.5 × 575 N/m × x²
Therefore;
0.5 × 575 N/m × x² = 30 J
[tex]x^2 = \dfrac{30 \, J}{0.5 \times 575 \, N/m} = \dfrac{12}{115} \, m^2[/tex]
[tex]x = 2 \cdot \sqrt{\dfrac{3}{115} } \approx 0.323[/tex]
The compression of the spring, x ≈ 0.323 m.
Learn more here:
https://brainly.com/question/7694106
Questions;
(a) The speed of the 15 kg mass immediately after the collision.
(b) Determine the type of collision; Elastic or inelastic collision.
(c) The distance to which the spring is compressed by the block.
