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):

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

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

Let's calculate the kinetic energy for each object:

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

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

3. **100 grams of water falling at 10 m/s:**
- Mass \( m = 0.1 \, \text{kg} \)
- Velocity \( v = 10 \, \text{m/s} \)
- \( KE = \frac{1}{2} \times 0.1 \times (10^2) = \frac{1}{2} \times 0.1 \times 100 = 5 \, \text{J} \)

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

Now, we can compare the kinetic energies:

- 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 J
- 10 grams of water at 1 m/s: 0.005 J

The object that will result in the least kinetic energy when it collides with the ground is **10 grams of water falling at 1 m/s**, with \( 0.005 \, \text{J} \).