Question

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

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

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 \(v_1\).

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.

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: $$v_1 = -\frac{(0.100\,\text{kg})(95.0\,\text{m/s})}{(50.0\,\text{kg})}$$

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.

Key concepts

Recoil Speed

When thinking about the recoil speed, imagine a situation where forces are exchanged between two objects. In the context of our ice-skating scenario, the archer and the arrow experience an immediate but equal force in opposite directions.
As the arrow is pushed forward with force, the archer is pushed backward. This backward movement is what we call the recoil speed.

Recoil speed is a familiar concept in various everyday situations. Consider how a rubber band snaps your fingers back when released, or how pressing against a wall can push you backward. Here, we are looking at kinetic activity in the form of momentum, where one entity moves forward while the other simultaneously recoils.
For Robin Hood on Ice, the careful calculation of this recoil speed involves examining the relationship between the mass of the archer and the momentum transferred to the arrow.

• Observe how smaller mass objects move at higher speeds during recoil as compared to larger mass objects.
• Remember that the greater the speed and mass of the projectile, the more noticeable the recoil effect will be.

Momentum Equation

The momentum equation is all about measuring motion. In physics, momentum is the product of an object’s mass and its velocity.
It's what tells you how much force is needed to stop an object in motion. Our task is to calculate how Robin Hood reacts when letting fly an arrow.

Conservation of momentum is a fundamental principle. It tells us that in an isolated system, where no external forces are acting, total momentum remains constant before and after an event.
In our scenario, before Robin shoots the arrow, both he and the arrow are stationary. This means their combined momentum is zero.

• Total initial momentum = Total final momentum
• The equation: \[ m_1 v_1 + m_2 v_2 = 0 \] keeps things balanced

By following the steps, we can solve for the value of one unknown. Here, it's the archer's recoil speed, represented by \( v_1 \). The importance of this equation lies in its ability to showcase how kinetic dynamics operate within the system of archer and arrow.
Relating this back to the everyday experience, the achievement of force balance is how things resolve naturally in our surroundings.

Physics Problem Solving

Tackling physics problems effectively requires a structured approach. With our ice-skating archer, understanding and solving the problem comes down to clarity and following a few essential steps.

First, identifying what's known and what's needed. Here, we know the masses and velocities involved. The unknown is the archer's recoil speed.
This process of identifying knowns and unknowns is often the starting point for any physics problem.

• List out the given values. This helps in visualizing the problem better.
• Translate the problem into a mathematical form using relevant equations.

Another key point is recognizing the underlying physics principles at play. Conservation of momentum in this example is the guiding star.
Finally, use mathematical rigor to juggle the numbers correctly by solving the equations systematically.
Solving this physics problem successfully rests on a student’s ability to think logically and interpret these relationships.

In perspective, physics problem-solving is akin to piecing together a jigsaw puzzle, where each logical step reveals more of the complete picture.