A 4.8 kg rifle fires a 5.2 g bullet with muzzle velocity of 986 m/s. What is the recoil velocity, VR, of the rifle?
A 4.8 kg rifle fires a 5.2 g bullet with muzzle velocity of 986 m/s. What is the recoil velocity, VR, of the rifle?
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![A 4.8 kg rifle fires a 5.2 g bullet with a muzzle velocity of 986 m/s. What is the recoil velocity, \( V_R \), of the rifle?
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**Solution Explanation:**
To find the recoil velocity of the rifle, we can apply the principle of conservation of momentum. Before the rifle is fired, the total momentum is zero as neither the rifle nor the bullet is moving. After firing, the momentum of the bullet must be equal and opposite to the momentum of the rifle to conserve the total momentum of the system.
### Step-by-Step Solution:
1. **Convert Bullet Mass to Kilograms:**
\[
m_{\text{bullet}} = 5.2 \, \text{g} = 0.0052 \, \text{kg}
\]
2. **Calculate the Momentum of the Bullet:**
\[
p_{\text{bullet}} = m_{\text{bullet}} \times v_{\text{bullet}} = 0.0052 \, \text{kg} \times 986 \, \text{m/s} = 5.1272 \, \text{kg m/s}
\]
3. **Apply Conservation of Momentum:**
\[
0 = m_{\text{rifle}} \times (-V_R) + p_{\text{bullet}}
\]
\[
0 = 4.8 \, \text{kg} \times (-V_R) + 5.1272 \, \text{kg m/s}
\]
4. **Solve for Recoil Velocity (\( V_R \)):**
\[
4.8 \, \text{kg} \times V_R = 5.1272 \, \text{kg m/s}
\]
\[
V_R = \frac{5.1272 \, \text{kg m/s}}{4.8 \, \text{kg}} = 1.068 \, \text{m/s}
\]
**Conclusion:**
The recoil velocity \( V_R \) of the rifle is approximately 1.068 m/s in the opposite direction of the bullet's motion.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3738e16a-10fd-4378-b4a3-b8733d2d5bb5%2F17577c6a-1be3-4d8c-a449-05853c843c3b%2F6ekc27_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A 4.8 kg rifle fires a 5.2 g bullet with a muzzle velocity of 986 m/s. What is the recoil velocity, \( V_R \), of the rifle?
---
**Solution Explanation:**
To find the recoil velocity of the rifle, we can apply the principle of conservation of momentum. Before the rifle is fired, the total momentum is zero as neither the rifle nor the bullet is moving. After firing, the momentum of the bullet must be equal and opposite to the momentum of the rifle to conserve the total momentum of the system.
### Step-by-Step Solution:
1. **Convert Bullet Mass to Kilograms:**
\[
m_{\text{bullet}} = 5.2 \, \text{g} = 0.0052 \, \text{kg}
\]
2. **Calculate the Momentum of the Bullet:**
\[
p_{\text{bullet}} = m_{\text{bullet}} \times v_{\text{bullet}} = 0.0052 \, \text{kg} \times 986 \, \text{m/s} = 5.1272 \, \text{kg m/s}
\]
3. **Apply Conservation of Momentum:**
\[
0 = m_{\text{rifle}} \times (-V_R) + p_{\text{bullet}}
\]
\[
0 = 4.8 \, \text{kg} \times (-V_R) + 5.1272 \, \text{kg m/s}
\]
4. **Solve for Recoil Velocity (\( V_R \)):**
\[
4.8 \, \text{kg} \times V_R = 5.1272 \, \text{kg m/s}
\]
\[
V_R = \frac{5.1272 \, \text{kg m/s}}{4.8 \, \text{kg}} = 1.068 \, \text{m/s}
\]
**Conclusion:**
The recoil velocity \( V_R \) of the rifle is approximately 1.068 m/s in the opposite direction of the bullet's motion.
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