A 100-kg machine gun fires a 0.5-kg bullet with a velocity of 400 m/s relative to the ground. What is the recoil speed of the machine gun? 2.0 m/s 20 m/s 1.0 m/s 4.0 m/s 6.0 m/s
A 100-kg machine gun fires a 0.5-kg bullet with a velocity of 400 m/s relative to the ground. What is the recoil speed of the machine gun? 2.0 m/s 20 m/s 1.0 m/s 4.0 m/s 6.0 m/s
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Chapter1: Units, Trigonometry. And Vectors
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![**Physics Problem: Recoil Speed Calculation**
*A 100-kg machine gun fires a 0.5-kg bullet with a velocity of 400 m/s relative to the ground. What is the recoil speed of the machine gun?*
**Answer Options:**
- 2.0 m/s
- 20 m/s
- 1.0 m/s
- 4.0 m/s
- 6.0 m/s
This question requires understanding of the law of conservation of momentum, which states that the total momentum before an event must equal the total momentum after the event in an isolated system.
**To solve:**
1. **Calculate the momentum of the bullet:**
\[
p_{\text{bullet}} = m_{\text{bullet}} \times v_{\text{bullet}} = 0.5 \, \text{kg} \times 400 \, \text{m/s} = 200 \, \text{kg}\cdot\text{m/s}
\]
2. **Equate the momentum of the system before and after the gun fires (initially the system is at rest, so total initial momentum is 0):**
\[
0 = m_{\text{bullet}} \times v_{\text{bullet}} + m_{\text{gun}} \times v_{\text{gun}}
\]
3. **Rearrange to solve for the velocity of the gun \( v_{\text{gun}} \):**
\[
v_{\text{gun}} = -\frac{m_{\text{bullet}} \times v_{\text{bullet}}}{m_{\text{gun}}}
\]
\[
v_{\text{gun}} = -\frac{200 \, \text{kg}\cdot\text{m/s}}{100 \, \text{kg}} = -2.0 \, \text{m/s}
\]
The negative sign indicates the direction of the recoil is opposite to the bullet's velocity. Hence, the recoil speed of the machine gun is **2.0 m/s**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdcc69741-2aee-44ad-99d7-267602eb5865%2Ff9c4c2fb-226a-4a9f-95a8-21719d2a1dfd%2Fcbtva3k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Physics Problem: Recoil Speed Calculation**
*A 100-kg machine gun fires a 0.5-kg bullet with a velocity of 400 m/s relative to the ground. What is the recoil speed of the machine gun?*
**Answer Options:**
- 2.0 m/s
- 20 m/s
- 1.0 m/s
- 4.0 m/s
- 6.0 m/s
This question requires understanding of the law of conservation of momentum, which states that the total momentum before an event must equal the total momentum after the event in an isolated system.
**To solve:**
1. **Calculate the momentum of the bullet:**
\[
p_{\text{bullet}} = m_{\text{bullet}} \times v_{\text{bullet}} = 0.5 \, \text{kg} \times 400 \, \text{m/s} = 200 \, \text{kg}\cdot\text{m/s}
\]
2. **Equate the momentum of the system before and after the gun fires (initially the system is at rest, so total initial momentum is 0):**
\[
0 = m_{\text{bullet}} \times v_{\text{bullet}} + m_{\text{gun}} \times v_{\text{gun}}
\]
3. **Rearrange to solve for the velocity of the gun \( v_{\text{gun}} \):**
\[
v_{\text{gun}} = -\frac{m_{\text{bullet}} \times v_{\text{bullet}}}{m_{\text{gun}}}
\]
\[
v_{\text{gun}} = -\frac{200 \, \text{kg}\cdot\text{m/s}}{100 \, \text{kg}} = -2.0 \, \text{m/s}
\]
The negative sign indicates the direction of the recoil is opposite to the bullet's velocity. Hence, the recoil speed of the machine gun is **2.0 m/s**.
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