Question

Which falling object will result in the least kinetic energy collides with the ground A 100 g of dirt falling at 15 m/s B 10 g of dirt falling at 2 m/s C 100 g of water falling at 10 m/s D 10 g of water falling at 1 m/s

Answer 1

To find which falling object results in the least kinetic energy upon colliding with the ground, we can use the kinetic energy formula:

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

where \( KE \) is the kinetic energy, \( m \) is the mass in kilograms, and \( v \) is the velocity in meters per second.

Let's calculate the kinetic energy for each option:

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

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

**C. 100 g of water falling at 10 m/s**
- Mass \( m = 100 \) g = 0.1 kg
- Velocity \( v = 10 \) m/s
\[
KE_C = \frac{1}{2} \times 0.1 \times (10)^2 = \frac{1}{2} \times 0.1 \times 100 = 5 \text{ J}
\]

**D. 10 g of water falling at 1 m/s**
- Mass \( m = 10 \) g = 0.01 kg
- Velocity \( v = 1 \) m/s
\[
KE_D = \frac{1}{2} \times 0.01 \times (1)^2 = \frac{1}{2} \times 0.01 \times 1 = 0.005 \text{ J}
\]

Now, let's compare the kinetic energies:
- \( KE_A = 11.25 \) J
- \( KE_B = 0.02 \) J
- \( KE_C = 5 \) J
- \( KE_D = 0.005 \) J

The object that results in the least kinetic energy upon colliding with the ground is **D. 10 g of water falling at 1 m/s**, with a kinetic energy of **0.005 J**.