Question

---Spring Question---

Answer 1

Answer:

The distance from the edge of the table at which the ball will hit the ground is approximately 0.49 meters

Explanation:

The mass of the ball = 0.84 kg

The spring constant k = 142 N/m

The compression of the spring = 0.06

Therefore, by conservation of energy, we have;

The potential energy of the spring = The kinetic energy given to the ball

The potential energy of the spring = 1/2 × k × x² = 1/2 × 142 × 0.06² = 0.2556

The potential energy of the spring = 0.2556 J

The kinetic energy given to the ball = 1/2 × m × v²

Where;

v = The final velocity of the ball

Substituting gives;

1/2 × 0.84 × v² = 0.2556

∴ v = √(0.2556/(1/2 × 0.84)) ≈ 0.78

The final velocity of the ball = v ≈ 0.78 m/s

The height of the ball above the ground, h = 2.0 m

Therefore, we have;

The time, t, it takes the ball to touch the ground from the 2.0 m height, h, is given as follows;

h = 1/2·g·t²

t = √(2·h/g)

Where;

g = The acceleration due to gravity = 9.81 m/s²

∴ By substitution, we have;

t = √(2 × 2/9.81) ≈ 0.63855

t ≈ 0.63855 s

The distance from the edge of the table at which the ball will hit the ground = The horizontal distance covered during before the ball hits the ground

The horizontal distance covered during before the ball hits the ground = The (horizontal) velocity of the ball × The time it takes the ball to touch the ground

∴ The horizontal distance covered during before the ball hits the ground = 0.78 × 0.62855 ≈ 0.490269 ≈ 0.49

The distance from the edge of the table at which the ball will hit the ground = The horizontal distance covered during before the ball hits the ground  ≈ 0.49 m