To determine which object will have 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 of the object in kilograms, and \( v \) is the velocity in meters per second.
Let's calculate the kinetic energy for each option:
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10 grams of water falling at 1 m/s:
- Mass = 10 grams = 0.01 kg
- Velocity = 1 m/s
- Kinetic Energy = \( \frac{1}{2} (0.01) (1^2) = 0.005 \) J
-
100 grams of dirt falling at 15 m/s:
- Mass = 100 grams = 0.1 kg
- Velocity = 15 m/s
- Kinetic Energy = \( \frac{1}{2} (0.1) (15^2) = \frac{1}{2} (0.1) (225) = 11.25 \) J
-
10 grams of dirt falling at 2 m/s:
- Mass = 10 grams = 0.01 kg
- Velocity = 2 m/s
- Kinetic Energy = \( \frac{1}{2} (0.01) (2^2) = \frac{1}{2} (0.01) (4) = 0.02 \) J
-
100 grams of water falling at 10 m/s:
- Mass = 100 grams = 0.1 kg
- Velocity = 10 m/s
- Kinetic Energy = \( \frac{1}{2} (0.1) (10^2) = \frac{1}{2} (0.1) (100) = 5 \) J
Now, summarizing the kinetic energy calculations:
- 10 grams of water at 1 m/s: \( 0.005 \) J
- 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
The object with the least kinetic energy when colliding with the ground is:
10 grams of water falling at 1 m/s (0.005 J).
1 answer