Question
A ball with a mass of 5.0 g is moving at a speed of 2.0 m/s. Would doubling the mass or doubling the speed have a greater effect on the kinetic energy of the ball? Explain.
Answers
Answered by
GPT-4o mini
To determine whether doubling the mass or doubling the speed has a greater effect on the kinetic energy of the ball, we need to use the formula for kinetic energy (KE), which is given by:
\[
KE = \frac{1}{2} m v^2
\]
where:
- \( m \) is the mass of the object,
- \( v \) is the speed of the object.
Given:
- The mass \( m = 5.0 \, \text{g} = 0.005 \, \text{kg} \) (converted to kilograms),
- The speed \( v = 2.0 \, \text{m/s} \).
First, let's calculate the initial kinetic energy:
\[
KE_{\text{initial}} = \frac{1}{2} (0.005 \, \text{kg}) (2.0 \, \text{m/s})^2
\]
\[
= \frac{1}{2} (0.005) (4)
\]
\[
= 0.01 \, \text{J}
\]
**Case 1: Doubling the Mass**
If we double the mass:
\[
m_{\text{new}} = 2 \times 0.005 \, \text{kg} = 0.01 \, \text{kg}
\]
The speed remains the same at \( 2.0 \, \text{m/s} \):
\[
KE_{\text{new}} = \frac{1}{2} (0.01 \, \text{kg}) (2.0 \, \text{m/s})^2
\]
\[
= \frac{1}{2} (0.01) (4)
\]
\[
= 0.02 \, \text{J}
\]
So, doubling the mass increases the kinetic energy from \( 0.01 \, \text{J} \) to \( 0.02 \, \text{J} \) (an increase of \( 0.01 \, \text{J} \)).
**Case 2: Doubling the Speed**
If we double the speed:
\[
v_{\text{new}} = 2 \times 2.0 \, \text{m/s} = 4.0 \, \text{m/s}
\]
The mass remains the same at \( 0.005 \, \text{kg} \):
\[
KE_{\text{new}} = \frac{1}{2} (0.005 \, \text{kg}) (4.0 \, \text{m/s})^2
\]
\[
= \frac{1}{2} (0.005) (16)
\]
\[
= 0.04 \, \text{J}
\]
So, doubling the speed increases the kinetic energy from \( 0.01 \, \text{J} \) to \( 0.04 \, \text{J} \) (an increase of \( 0.03 \, \text{J} \)).
**Conclusion**
Comparing the results:
- Doubling the mass results in an increase of \( 0.01 \, \text{J} \).
- Doubling the speed results in an increase of \( 0.03 \, \text{J} \).
Thus, doubling the speed has a greater effect on the kinetic energy of the ball than doubling the mass. This is because kinetic energy is proportional to the square of the speed, making changes in speed much more impactful compared to changes in mass.
\[
KE = \frac{1}{2} m v^2
\]
where:
- \( m \) is the mass of the object,
- \( v \) is the speed of the object.
Given:
- The mass \( m = 5.0 \, \text{g} = 0.005 \, \text{kg} \) (converted to kilograms),
- The speed \( v = 2.0 \, \text{m/s} \).
First, let's calculate the initial kinetic energy:
\[
KE_{\text{initial}} = \frac{1}{2} (0.005 \, \text{kg}) (2.0 \, \text{m/s})^2
\]
\[
= \frac{1}{2} (0.005) (4)
\]
\[
= 0.01 \, \text{J}
\]
**Case 1: Doubling the Mass**
If we double the mass:
\[
m_{\text{new}} = 2 \times 0.005 \, \text{kg} = 0.01 \, \text{kg}
\]
The speed remains the same at \( 2.0 \, \text{m/s} \):
\[
KE_{\text{new}} = \frac{1}{2} (0.01 \, \text{kg}) (2.0 \, \text{m/s})^2
\]
\[
= \frac{1}{2} (0.01) (4)
\]
\[
= 0.02 \, \text{J}
\]
So, doubling the mass increases the kinetic energy from \( 0.01 \, \text{J} \) to \( 0.02 \, \text{J} \) (an increase of \( 0.01 \, \text{J} \)).
**Case 2: Doubling the Speed**
If we double the speed:
\[
v_{\text{new}} = 2 \times 2.0 \, \text{m/s} = 4.0 \, \text{m/s}
\]
The mass remains the same at \( 0.005 \, \text{kg} \):
\[
KE_{\text{new}} = \frac{1}{2} (0.005 \, \text{kg}) (4.0 \, \text{m/s})^2
\]
\[
= \frac{1}{2} (0.005) (16)
\]
\[
= 0.04 \, \text{J}
\]
So, doubling the speed increases the kinetic energy from \( 0.01 \, \text{J} \) to \( 0.04 \, \text{J} \) (an increase of \( 0.03 \, \text{J} \)).
**Conclusion**
Comparing the results:
- Doubling the mass results in an increase of \( 0.01 \, \text{J} \).
- Doubling the speed results in an increase of \( 0.03 \, \text{J} \).
Thus, doubling the speed has a greater effect on the kinetic energy of the ball than doubling the mass. This is because kinetic energy is proportional to the square of the speed, making changes in speed much more impactful compared to changes in mass.