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

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
100 grams of dirt falling at 15 m/s

100 grams of water falling at 10 m/s
100 grams of water falling at 10 m/s

Answer 1

To determine which falling object will result in the least kinetic energy when it collides with the ground, we can use the kinetic energy formula:

\[
KE = \frac{1}{2} m v^2
\]

where \( KE \) is kinetic energy, \( m \) is mass in kilograms, and \( v \) is velocity in meters per second.

Now, let's calculate the kinetic energy for each option:

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

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

3. **100 grams of dirt falling at 15 m/s**:
- Mass \( m = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Velocity \( v = 15 \, \text{m/s} \)
- \( KE = \frac{1}{2} (0.1) (15^2) = \frac{1}{2} (0.1) (225) = 11.25 \, \text{J} \)

4. **100 grams of water falling at 10 m/s**:
- Mass \( m = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Velocity \( v = 10 \, \text{m/s} \)
- \( KE = \frac{1}{2} (0.1) (10^2) = \frac{1}{2} (0.1) (100) = 5 \, \text{J} \)

Now, compare the kinetic energies:

- 10 grams of dirt at 2 m/s: \( 0.02 \, \text{J} \)
- 10 grams of water at 1 m/s: \( 0.005 \, \text{J} \)
- 100 grams of dirt at 15 m/s: \( 11.25 \, \text{J} \)
- 100 grams of water at 10 m/s: \( 5 \, \text{J} \)

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** with \( 0.005 \, \text{J} \).