During an ice-skating extravaganza, Robin Hood on Ice, a 50.0 -kg archer is standing still on ice skates. Assume that the friction between the ice skates and the ice is negligible. The archer shoots a \(0.100-\mathrm{kg}\) arrow horizontally at a speed of \(95.0 \mathrm{~m} / \mathrm{s} .\) At what speed does the archer recoil?

Short Answer

Expert verified
Answer: The archer recoils at a speed of approximately 0.19 m/s in the opposite direction to the arrow's motion.

Step by step solution

01

Identify the knowns and the unknowns

In this exercise, the known quantities are: - The mass of the archer, \(m_1 = 50.0\,\text{kg}\) - The mass of the arrow, \(m_2 = 0.100\,\text{kg}\) - The speed of the arrow after being shot, \(v_2 = 95.0\,\text{m/s}\) We need to find the recoil speed of the archer, which we will denote as \(v_1\).
02

Apply the law of conservation of momentum

The total momentum of the system (archer + arrow) must remain constant before and after the arrow is shot. Initially, the momentum is zero, as both the archer and the arrow are stationary. After shooting, we can apply the equation: $$m_1 v_1 + m_2 v_2 = 0$$ We are trying to find the value of \(v_1\), the speed at which the archer recoils.
03

Solve the equation for the unknown

We can rearrange the equation from step 2 to isolate \(v_1\): $$v_1 = -\frac{m_2 v_2}{m_1}$$ Now, we can plug in the known values and calculate the recoil speed of the archer: $$v_1 = -\frac{(0.100\,\text{kg})(95.0\,\text{m/s})}{(50.0\,\text{kg})}$$
04

Calculate the recoil speed

After plugging in the known values, we can find the magnitude of the recoil speed: $$v_1 \approx -0.19\,\text{m/s}$$ We can observe that the value of \(v_1\) is negative which implies that the archer's recoil is in the opposite direction compared to the arrow's motion. Thus, the archer recoils at a speed of approximately \(0.19\,\text{m/s}\) in the opposite direction to the arrow's motion.

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