Question

analyzing the relationship between mass and ke read through the scenarios below and calculate the predicted change in kinetic energy of the object compared to a 100 kg ball traveling at 5 m/s. a 1,000 kg ball traveling at 5 m/s would have kinetic energy. a 10 kg ball traveling at 5 m/s would have kinetic energy. a 100 kg person falling at 5 m/s would have kinetic energy.

Explanation:

To solve this, we use the kinetic energy formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity. First, calculate the KE of the 100 kg ball at 5 m/s. Then compare with other scenarios.

Step 1: Calculate KE of the reference object (100 kg, 5 m/s)

The formula for kinetic energy is \( KE = \frac{1}{2}mv^2 \). For the 100 kg ball moving at 5 m/s:
\[
KE_{\text{reference}} = \frac{1}{2} \times 100 \, \text{kg} \times (5 \, \text{m/s})^2
\]
\[
KE_{\text{reference}} = \frac{1}{2} \times 100 \times 25
\]
\[
KE_{\text{reference}} = 50 \times 25 = 1250 \, \text{J}
\]

Step 2: Analyze the 1,000 kg ball at 5 m/s

Using the same velocity (\( v = 5 \, \text{m/s} \)) but mass \( m = 1000 \, \text{kg} \):
\[
KE_1 = \frac{1}{2} \times 1000 \, \text{kg} \times (5 \, \text{m/s})^2
\]
\[
KE_1 = \frac{1}{2} \times 1000 \times 25
\]
\[
KE_1 = 500 \times 25 = 12500 \, \text{J}
\]
Compare to \( KE_{\text{reference}} = 1250 \, \text{J} \). Since \( 12500 = 10 \times 1250 \), the 1,000 kg ball has 10 times more kinetic energy.

Step 3: Analyze the 10 kg ball at 5 m/s

Using \( m = 10 \, \text{kg} \) and \( v = 5 \, \text{m/s} \):
\[
KE_2 = \frac{1}{2} \times 10 \, \text{kg} \times (5 \, \text{m/s})^2
\]
\[
KE_2 = \frac{1}{2} \times 10 \times 25
\]
\[
KE_2 = 5 \times 25 = 125 \, \text{J}
\]
Compare to \( KE_{\text{reference}} = 1250 \, \text{J} \). Since \( 125 = \frac{1}{10} \times 1250 \), the 10 kg ball has 1/10th (or 10 times less) kinetic energy.

Step 4: Analyze the 100 kg person falling at 5 m/s

Here, \( m = 100 \, \text{kg} \) and \( v = 5 \, \text{m/s} \), same as the reference object. Using the KE formula:
\[
KE_3 = \frac{1}{2} \times 100 \, \text{kg} \times (5 \, \text{m/s})^2
\]
\[
KE_3 = \frac{1}{2} \times 100 \times 25 = 1250 \, \text{J}
\]
This is equal to \( KE_{\text{reference}} \), so the 100 kg person has the same kinetic energy.

Final Answers:

• A 1,000 kg ball traveling at 5 m/s would have \(\boldsymbol{10}\) times more kinetic energy.
• A 10 kg ball traveling at 5 m/s would have \(\boldsymbol{\frac{1}{10}}\) (or 10 times less) kinetic energy.
• A 100 kg person falling at 5 m/s would have \(\boldsymbol{\text{the same}}\) kinetic energy.

Answer:

To solve this, we use the kinetic energy formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity. First, calculate the KE of the 100 kg ball at 5 m/s. Then compare with other scenarios.

Step 1: Calculate KE of the reference object (100 kg, 5 m/s)

The formula for kinetic energy is \( KE = \frac{1}{2}mv^2 \). For the 100 kg ball moving at 5 m/s:
\[
KE_{\text{reference}} = \frac{1}{2} \times 100 \, \text{kg} \times (5 \, \text{m/s})^2
\]
\[
KE_{\text{reference}} = \frac{1}{2} \times 100 \times 25
\]
\[
KE_{\text{reference}} = 50 \times 25 = 1250 \, \text{J}
\]

Step 2: Analyze the 1,000 kg ball at 5 m/s

Using the same velocity (\( v = 5 \, \text{m/s} \)) but mass \( m = 1000 \, \text{kg} \):
\[
KE_1 = \frac{1}{2} \times 1000 \, \text{kg} \times (5 \, \text{m/s})^2
\]
\[
KE_1 = \frac{1}{2} \times 1000 \times 25
\]
\[
KE_1 = 500 \times 25 = 12500 \, \text{J}
\]
Compare to \( KE_{\text{reference}} = 1250 \, \text{J} \). Since \( 12500 = 10 \times 1250 \), the 1,000 kg ball has 10 times more kinetic energy.

Step 3: Analyze the 10 kg ball at 5 m/s

Using \( m = 10 \, \text{kg} \) and \( v = 5 \, \text{m/s} \):
\[
KE_2 = \frac{1}{2} \times 10 \, \text{kg} \times (5 \, \text{m/s})^2
\]
\[
KE_2 = \frac{1}{2} \times 10 \times 25
\]
\[
KE_2 = 5 \times 25 = 125 \, \text{J}
\]
Compare to \( KE_{\text{reference}} = 1250 \, \text{J} \). Since \( 125 = \frac{1}{10} \times 1250 \), the 10 kg ball has 1/10th (or 10 times less) kinetic energy.

Step 4: Analyze the 100 kg person falling at 5 m/s

Here, \( m = 100 \, \text{kg} \) and \( v = 5 \, \text{m/s} \), same as the reference object. Using the KE formula:
\[
KE_3 = \frac{1}{2} \times 100 \, \text{kg} \times (5 \, \text{m/s})^2
\]
\[
KE_3 = \frac{1}{2} \times 100 \times 25 = 1250 \, \text{J}
\]
This is equal to \( KE_{\text{reference}} \), so the 100 kg person has the same kinetic energy.

Final Answers:

• A 1,000 kg ball traveling at 5 m/s would have \(\boldsymbol{10}\) times more kinetic energy.
• A 10 kg ball traveling at 5 m/s would have \(\boldsymbol{\frac{1}{10}}\) (or 10 times less) kinetic energy.
• A 100 kg person falling at 5 m/s would have \(\boldsymbol{\text{the same}}\) kinetic energy.