Question

Which falling object will result in the least kinetic energy when it collides with the ground?(1 point) Responses 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 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

Answer 1

To determine which falling object will have the least kinetic energy upon collision with the ground, we can use the kinetic energy formula:

\[
KE = \frac{1}{2}mv^2
\]

where \( m \) is the mass of the object in kilograms and \( v \) is the velocity in meters per second.

Let's calculate the kinetic energy for each option:

1. **10 grams of water falling at 1 m/s**:
\[
m = 10 \text{ grams} = 0.01 \text{ kg}, \quad v = 1 \text{ m/s}
\]
\[
KE = \frac{1}{2} \times 0.01 \times (1)^2 = 0.005 \text{ J}
\]

2. **100 grams of dirt falling at 15 m/s**:
\[
m = 100 \text{ grams} = 0.1 \text{ kg}, \quad v = 15 \text{ m/s}
\]
\[
KE = \frac{1}{2} \times 0.1 \times (15)^2 = 11.25 \text{ J}
\]

3. **10 grams of dirt falling at 2 m/s**:
\[
m = 10 \text{ grams} = 0.01 \text{ kg}, \quad v = 2 \text{ m/s}
\]
\[
KE = \frac{1}{2} \times 0.01 \times (2)^2 = 0.02 \text{ J}
\]

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

Now, let's summarize the kinetic energies calculated:
- **10 grams of water falling at 1 m/s**: 0.005 J
- **100 grams of dirt falling at 15 m/s**: 11.25 J
- **10 grams of dirt falling at 2 m/s**: 0.02 J
- **100 grams of water falling at 10 m/s**: 5 J

Comparing these values, the falling 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)**.