A 80-kg archer stands at rest on frictionless ice and shoots a 36-g arrow. The arrow's speed is 75 m/s. What is the recoil speed of the archer? 1.6 m/s 75 m/s 3.4 cm/s 26 cm/s

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**Physics Problem: Conservation of Momentum**

An 80-kg archer stands at rest on frictionless ice and shoots a 36-g arrow. The arrow's speed is 75 m/s. What is the recoil speed of the archer?

**Answer Choices:**

- 1.6 m/s
- 0
- 75 m/s
- 3.4 cm/s
- 26 cm/s

**Explanation:**

This problem involves the conservation of momentum principle. Since the archer and the arrow system is initially at rest, the total momentum is zero. After the arrow is shot, the momentum of the arrow and the archer must still sum to zero. 

Calculate the recoil speed of the archer using the formula:

\[ m_{\text{archer}} \times v_{\text{archer}} + m_{\text{arrow}} \times v_{\text{arrow}} = 0 \]

\[ 80 \, \text{kg} \times v_{\text{archer}} + 0.036 \, \text{kg} \times 75 \, \text{m/s} = 0 \]

Solve for \( v_{\text{archer}} \):

\[ v_{\text{archer}} = - \frac{0.036 \, \text{kg} \times 75 \, \text{m/s}}{80 \, \text{kg}} \] 

\[ v_{\text{archer}} = -0.03375 \, \text{m/s} \]

Converting to cm/s:

\[ v_{\text{archer}} = -3.375 \, \text{cm/s} \]

The negative sign indicates the direction opposite to the arrow's motion. 
The closest answer is 3.4 cm/s (since directions are not specified in the answer options, take magnitude).
Transcribed Image Text:**Physics Problem: Conservation of Momentum** An 80-kg archer stands at rest on frictionless ice and shoots a 36-g arrow. The arrow's speed is 75 m/s. What is the recoil speed of the archer? **Answer Choices:** - 1.6 m/s - 0 - 75 m/s - 3.4 cm/s - 26 cm/s **Explanation:** This problem involves the conservation of momentum principle. Since the archer and the arrow system is initially at rest, the total momentum is zero. After the arrow is shot, the momentum of the arrow and the archer must still sum to zero. Calculate the recoil speed of the archer using the formula: \[ m_{\text{archer}} \times v_{\text{archer}} + m_{\text{arrow}} \times v_{\text{arrow}} = 0 \] \[ 80 \, \text{kg} \times v_{\text{archer}} + 0.036 \, \text{kg} \times 75 \, \text{m/s} = 0 \] Solve for \( v_{\text{archer}} \): \[ v_{\text{archer}} = - \frac{0.036 \, \text{kg} \times 75 \, \text{m/s}}{80 \, \text{kg}} \] \[ v_{\text{archer}} = -0.03375 \, \text{m/s} \] Converting to cm/s: \[ v_{\text{archer}} = -3.375 \, \text{cm/s} \] The negative sign indicates the direction opposite to the arrow's motion. The closest answer is 3.4 cm/s (since directions are not specified in the answer options, take magnitude).
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