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:
-
100 grams of dirt falling at 15 m/s:
- Mass \( m = 0.1 , \text{kg} \)
- Velocity \( v = 15 , \text{m/s} \)
- \( KE = \frac{1}{2} \times 0.1 \times (15^2) = \frac{1}{2} \times 0.1 \times 225 = 11.25 , \text{J} \)
-
10 grams of dirt falling at 2 m/s:
- Mass \( m = 0.01 , \text{kg} \)
- Velocity \( v = 2 , \text{m/s} \)
- \( KE = \frac{1}{2} \times 0.01 \times (2^2) = \frac{1}{2} \times 0.01 \times 4 = 0.02 , \text{J} \)
-
100 grams of water falling at 10 m/s:
- Mass \( m = 0.1 , \text{kg} \)
- Velocity \( v = 10 , \text{m/s} \)
- \( KE = \frac{1}{2} \times 0.1 \times (10^2) = \frac{1}{2} \times 0.1 \times 100 = 5 , \text{J} \)
-
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 \( 0.005 , \text{J} \).
1 answer