Question

Which falling object will result in the least kinetic energy when it collides with the ground?(1 point) Responses 10 grams of dirt falling at 2 m/s 10 grams of dirt falling at 2 m/s 100 grams of water falling at 10 m/s 100 grams of water falling at 10 m/s 10 grams of water falling at 1 m/s 10 grams of water falling at 1 m/s 100 grams of dirt falling at 15 m/s

Answer 1

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:

1. **10 grams of dirt falling at 2 m/s**:
- Mass \( m = 10 \text{ g} = 0.01 \text{ kg} \)
- Velocity \( v = 2 \text{ m/s} \)
- \( KE = \frac{1}{2} \times 0.01 \times (2)^2 = \frac{1}{2} \times 0.01 \times 4 = 0.02 \text{ J} \)

2. **100 grams of water falling at 10 m/s**:
- Mass \( m = 100 \text{ g} = 0.1 \text{ kg} \)
- Velocity \( v = 10 \text{ m/s} \)
- \( KE = \frac{1}{2} \times 0.1 \times (10)^2 = \frac{1}{2} \times 0.1 \times 100 = 5 \text{ J} \)

3. **10 grams of water falling at 1 m/s**:
- Mass \( m = 10 \text{ g} = 0.01 \text{ kg} \)
- Velocity \( v = 1 \text{ m/s} \)
- \( KE = \frac{1}{2} \times 0.01 \times (1)^2 = \frac{1}{2} \times 0.01 \times 1 = 0.005 \text{ J} \)

4. **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)**.