To calculate the kinetic energy (KE) of each object just before colliding with the ground, we use the formula:
\[ KE = \frac{1}{2} mv^2 \]
where:
- \( m \) is the mass in kilograms (kg),
- \( v \) is the velocity in meters per second (m/s).
Let's calculate the kinetic energy for each object:
-
10 grams of dirt falling at 2 m/s:
- Mass \( m = 10 \text{ g} = 0.01 \text{ kg} \)
- Velocity \( v = 2 \text{ m/s} \)
- \( KE = \frac{1}{2} \times 0.01 \times (2)^2 = \frac{1}{2} \times 0.01 \times 4 = 0.02 \text{ J} \)
-
100 grams of water falling at 10 m/s:
- Mass \( m = 100 \text{ g} = 0.1 \text{ kg} \)
- Velocity \( v = 10 \text{ m/s} \)
- \( KE = \frac{1}{2} \times 0.1 \times (10)^2 = \frac{1}{2} \times 0.1 \times 100 = 5 \text{ J} \)
-
10 grams of water falling at 1 m/s:
- Mass \( m = 10 \text{ g} = 0.01 \text{ kg} \)
- Velocity \( v = 1 \text{ m/s} \)
- \( KE = \frac{1}{2} \times 0.01 \times (1)^2 = \frac{1}{2} \times 0.01 \times 1 = 0.005 \text{ J} \)
-
100 grams of dirt falling at 15 m/s:
- Mass \( m = 100 \text{ g} = 0.1 \text{ kg} \)
- Velocity \( v = 15 \text{ m/s} \)
- \( KE = \frac{1}{2} \times 0.1 \times (15)^2 = \frac{1}{2} \times 0.1 \times 225 = 11.25 \text{ J} \)
Now, let's summarize the computed kinetic energies:
- 10 grams of dirt at 2 m/s: 0.02 J
- 100 grams of water at 10 m/s: 5 J
- 10 grams of water at 1 m/s: 0.005 J
- 100 grams of dirt at 15 m/s: 11.25 J
Conclusion: The object that results in the least kinetic energy when it collides with the ground is 10 grams of water falling at 1 m/s (0.005 J).