Question

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

Answer 1

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

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

where \( KE \) is the kinetic energy, \( m \) is the mass of the object, and \( v \) is its velocity just before impact.

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

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

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

3. **100 grams of water falling at 10 m/s:**
- Mass, \( m = 0.1 \, kg \)
- Velocity, \( v = 10 \, m/s \)
\[
KE = \frac{1}{2} \times 0.1 \times 10^2 = \frac{1}{2} \times 0.1 \times 100 = 5.0 \, J
\]

4. **10 grams of water falling at 1 m/s:**
- Mass, \( m = 0.01 \, kg \)
- Velocity, \( v = 1 \, m/s \)
\[
KE = \frac{1}{2} \times 0.01 \times 1^2 = \frac{1}{2} \times 0.01 \times 1 = 0.005 \, J
\]

Now we can compare the kinetic energies calculated:

- 100 grams of dirt at 15 m/s: **11.25 J**
- 10 grams of dirt at 2 m/s: **0.02 J**
- 100 grams of water at 10 m/s: **5.0 J**
- 10 grams of water at 1 m/s: **0.005 J**

Among these, the object with the least kinetic energy when it collides with the ground is:

**10 grams of water falling at 1 m/s** (0.005 J).