Question

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

Answer 1

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

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

where \( m \) is the mass of the object and \( v \) is its velocity.

Let's calculate the kinetic energy for each of the given options:

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

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

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

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

Now, let's summarize the kinetic energies:

- 100 g of water at 10 m/s: **5 Joules**
- 100 g of dirt at 15 m/s: **11.25 Joules**
- 10 g of water at 1 m/s: **0.005 Joules**
- 10 g of dirt at 2 m/s: **0.02 Joules**

The falling object with the least kinetic energy when it collides with the ground is:

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