Question

Which falling object has the least kinetic energy when it collides on the ground?(1 point)
Responses

120 Kg of sand falling 12 m/s

12 Kg of sand falling at 8 m/s

12 Kg of water falling at 10 m/s

120 Kg of water falling at 12 m/s

Answer 1

Kinetic energy (KE) can be calculated using the formula:

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

Where:
- \( m \) is the mass of the object,
- \( v \) is the velocity of the object.

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

1. **120 kg of sand falling at 12 m/s**:
\[ KE = \frac{1}{2} \times 120 \, \text{kg} \times (12 \, \text{m/s})^2 \]
\[ KE = 60 \times 144 = 8640 \, \text{J} \]

2. **12 kg of sand falling at 8 m/s**:
\[ KE = \frac{1}{2} \times 12 \, \text{kg} \times (8 \, \text{m/s})^2 \]
\[ KE = 6 \times 64 = 384 \, \text{J} \]

3. **12 kg of water falling at 10 m/s**:
\[ KE = \frac{1}{2} \times 12 \, \text{kg} \times (10 \, \text{m/s})^2 \]
\[ KE = 6 \times 100 = 600 \, \text{J} \]

4. **120 kg of water falling at 12 m/s**:
\[ KE = \frac{1}{2} \times 120 \, \text{kg} \times (12 \, \text{m/s})^2 \]
\[ KE = 60 \times 144 = 8640 \, \text{J} \]

Now comparing the kinetic energies calculated:

- 120 kg of sand at 12 m/s: 8640 J
- 12 kg of sand at 8 m/s: 384 J
- 12 kg of water at 10 m/s: 600 J
- 120 kg of water at 12 m/s: 8640 J

The falling object with the least kinetic energy when it collides with the ground is **12 kg of sand falling at 8 m/s**, which has **384 J of kinetic energy**.