Problem 30
Question
An archer shoors an arrow toward a \(300-g\) target that is sliding in her direction at a speed of \(2.50 \mathrm{~m} / \mathrm{s}\) on a smooth, slippery surface. The \(22.5-\mathrm{g}\) arrow is shot with a speed of \(35.0 \mathrm{~m} / \mathrm{s}\) and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target?
Step-by-Step Solution
Verified Answer
To get the short answer, fill in the values into the final speed of the arrow equation as obtained in step 3.
1Step 1: Identify given quantities
The mass of the target \(m_t = 300g = 0.3kg\), the initial speed of the target \(v_{t_initial} = 2.50 m/s\), the mass of the arrow \(m_a = 22.5g = 0.0225kg\), and the initial speed of the arrow \(v_{a_initial} = 35.0 m/s\). The final speed of the target \(v_{t_final}=0\), because the problem states that the target is stopped.
2Step 2: Apply the conservation of momentum
The principle of conservation of momentum states that the total momentum of a system before collision equals the total momentum of the system after collision. Hence, \(m_t * v_{t_initial} + m_a * v_{a_initial} = m_t * v_{t_final} + m_a * v_{a_final}\), where \(v_{a_final}\) is the final speed of the arrow we need to find.
3Step 3: Calculate the final speed of the arrow
Rearrange the equation from step 2 to solve for \(v_{a_final}\): \(v_{a_final} = (m_t * v_{t_initial} + m_a * v_{a_initial} - m_t * v_{t_final}) / m_a\). Then, plug values into the equation to get \(v_{a_final}\).
Key Concepts
Understanding Collision PhysicsMomentum Calculation Made SimpleThe Role of Projectile Motion
Understanding Collision Physics
Collision physics helps us understand how objects interact when they come into contact. This is crucial in a world full of moving objects. In our exercise, the arrow passes through the target, showing a type of collision where the objects don't stick together, called an elastic collision.
Key Points:
Key Points:
- Elastic collisions involve objects bouncing off each other with no loss in total kinetic energy.
- In contrast, inelastic collisions result in objects sticking together or deforming, leading to a loss in kinetic energy.
- This exercise shows a unique situation where the arrow and the target go separate ways after impact, with the target stopping completely.
Momentum Calculation Made Simple
Calculating momentum involves multiplying an object's mass by its velocity. This simple idea forms the backbone of many physical calculations.
The essence of the principle of momentum conservation used in this exercise is to ensure that the total momentum before the collision equals the total momentum after the collision. Here’s how to do it:
The essence of the principle of momentum conservation used in this exercise is to ensure that the total momentum before the collision equals the total momentum after the collision. Here’s how to do it:
- Initial momentum: Combine the momentum of both the arrow and the target before they collide using the formula: \( m_t \times v_{t_{initial}} + m_a \times v_{a_{initial}} \).
- Final momentum: After collision, since the target stops, only the arrow's momentum matters: \( m_a \times v_{a_{final}} \).
- Conservation equation: Set initial momentum equal to final momentum and solve for the unknown, \( v_{a_{final}} \).
The Role of Projectile Motion
Projectile motion comes into play when an object is launched into the air, considering only the influence of gravity. While gravity may not directly affect our problem's outcome, it is essential for understanding an arrow’s overall journey.
Key Factors of Projectile Motion:
Key Factors of Projectile Motion:
- Projectile motion involves horizontal and vertical components of motion. However, in this context, we're focused on the speed along a single path.
- Ignoring air resistance simplifies calculations, often assuming motion in a straight line.
- Understanding projectile principles aids in determining an arrow's trajectory and speed changes caused by external forces like collisions.
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