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An expert archer demonstrates his ability by hitting tennis balls thrown by an assistant. A 2 -oz tennis ball has a velocity of \((32 \mathrm{ft} / \mathrm{s}) \mathrm{i}-(7 \mathrm{ft} / \mathrm{s}) \mathrm{j}\) and is \(33 \mathrm{ft}\) above the ground when it is hit by \(\mathrm{a} 1.2-\mathrm{oz}\) arrow traveling with a velocity of \((165 \mathrm{ft} / \mathrm{s}) \mathrm{j}+(230 \mathrm{ft} / \mathrm{s}) \mathrm{k}\) where \(\mathrm{j}\) is directed upwards. Determine the position \(P\) where the ball and arrow will hit the ground, relative to point \(O\) located directly under the point of impact.

Short Answer

Expert verified
Position \( P \) is calculated using conservation of momentum and kinematic equations. The results give coordinates \( (x, 0, z) \) relative to point \( O \).

Step by step solution

01

Analyze Given Information

Understand the given values: The tennis ball has a velocity \( (32 \mathrm{ft} / \mathrm{s}) \mathrm{i} - (7 \mathrm{ft} / \mathrm{s}) \mathrm{j} \) and mass of 2 oz (convert to 0.125 lb). It is \(33 \mathrm{ft}\) above ground. The arrow has a velocity \( (165 \mathrm{ft} / \mathrm{s}) \mathrm{j} + (230 \mathrm{ft} / \mathrm{s}) \mathrm{k} \) and mass of 1.2 oz (convert to 0.075 lb). The impact height is at \(33 \mathrm{ft}\) above ground.
02

Use Conservation of Momentum

Apply the conservation of momentum because the total momentum before impact equals the total momentum after impact. Calculate total momentum:Before impact:\[\text{Ball momentum} = m_b \cdot \vec{v}_b = 0.125 \cdot ((32)\hat{i} - (7)\hat{j}) \\text{Arrow momentum} = m_a \cdot \vec{v}_a = 0.075 \cdot ((165)\hat{j} + (230)\hat{k}) \\text{Total momentum before} = 0.125 \cdot ((32)\hat{i} - (7)\hat{j}) + 0.075 \cdot ((165)\hat{j} + (230)\hat{k})\]
03

Calculate Velocity After Impact

To find the velocity after the impact, we assume the final velocity \( \vec{v}_f \) has components \( v_{xf}\hat{i} + v_{yf}\hat{j} + v_{zf}\hat{k} \). Use the formula \( \text{Total momentum before} = \text{Total mass} \times \vec{v}_f \) where \( \text{Total mass} = 0.125 + 0.075 = 0.2 \, \text{lb}\).Solve to find: \[ v_{xf} = 32 \, \text{ft/s}, \, v_{yf} = \frac{-7 \times 0.125 + 165 \times 0.075}{0.2}, \, v_{zf} = \frac{230 \times 0.075}{0.2}\]
04

Determine Time to Hit Ground

Use the vertical velocity to find the time \( t \) it takes for both to hit the ground. Starting at 33 ft high:\[- (33) + v_{yf} \cdot t - \frac{1}{2}(32.2) t^2 = 0\]Solve this quadratic equation for \( t \).
05

Calculate Horizontal and Perpendicular Distance

Use the horizontal and perpendicular velocity components \( v_{xf} \) and \( v_{zf} \) to find the displacement along \( i \) and \( k \):\[x = v_{xf} \cdot t, \, z = v_{zf} \cdot t\]
06

Determine Position P

Combine the results to find position \( P \) with respect to point \( O \):\[\text{Position } P = (x \hat{i} + 0 \hat{j} + z \hat{k}) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Projectile Motion
Projectile motion is the motion of an object thrown or projected into the air, subject only to acceleration as a result of gravity. This concept helps us understand the movement of two objects like our tennis ball and arrow when they interact mid-air. The flight path of these objects can be predicted using the laws of physics, which are determined by their initial velocities and positions.

It's important to understand these components:
  • Horizontal motion occurs at a constant velocity because no acceleration (other than minor air resistance, which we're considering negligible here) acts in the horizontal direction.
  • Vertical motion is affected by gravity, slowing upward motion and accelerating the object downwards.
For both our tennis ball and arrow,
  • Their initial positions and velocities determine their motion paths immediately after the impact.
  • The simplest way to calculate their paths is by using conservation of momentum, which simplifies the problem to one of projectile motion post-impact.
By breaking down the motion into horizontal and vertical components, we can calculate the final destination of both the ball and arrow after their interaction.
Momentum Calculation
Momentum is a fundamental concept in physics, represented as the product of an object's mass and velocity. It helps us understand how different objects, like our tennis ball and arrow, affect each other on impact. In this exercise, we make use of the principle of conservation of momentum. This principle states that the total momentum of a system remains constant if the system is not influenced by external forces.

To perform momentum calculations:
  • First, identify the masses and velocities of both the ball and arrow before impact, expressed as vectors to account for directions.
  • Next, calculate their individual momenta (mass multiplied by velocity) and then sum these to get the total momentum before impact.
  • This total pre-impact momentum must equal the total post-impact momentum, allowing us to solve for the final velocities of the combined system.
During the collision, the initial momentum of each object contributes to the final motion post-impact. By rearranging the components of momenta, the velocities after an impact are found. This is a crucial step for understanding how the combined object will move forward post-impact.
Impact Physics
Impact physics examines what happens when two objects collide, such as our tennis ball and arrow. The dynamics of this collision are influenced by several factors, including their masses, velocities, and the angle at which they collide.

In our problem:
  • The arrow strikes the tennis ball while both are in motion, creating a complex interaction of forces at a height above the ground.
  • The conservation of momentum is crucial here as it predicts the motion of both objects post-collision.
The impact forces the two objects to combine their momenta and results in a new pathway for their movement.
  • Using the velocities derived from momentum calculations, one can determine how the objects will travel and where they will land on the ground by calculating the time taken to hit the ground.
  • The sphere of impact will direct them simultaneously downward and in their horizontal direction until gravity brings them to rest.
This comprehensive understanding of impact physics allows us to predict the behavior of objects after collisions accurately, which can be applied to a range of real-world activities, from sports science to engineering applications.

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Most popular questions from this chapter

A 300 -kg space vehicle traveling with a velocity \(\mathbf{v}_{0}=(360 \mathrm{m} / \mathrm{s})\) i passes through the origin \(O\) at \(t=0 .\) Explosive charges then separate the vehicle into three parts \(A, B,\) and \(C,\) with mass, respectively, \(150 \mathrm{kg}, 100 \mathrm{kg},\) and \(50 \mathrm{kg}\). Knowing that at \(t=4 \mathrm{s}\), the positions of parts \(A\) and \(B\) are observed to be \(A(1170 \mathrm{m},-290 \mathrm{m},-585 \mathrm{m})\) and \(B(1975 \mathrm{m}, 365 \mathrm{m}, 800 \mathrm{m}),\) determine the corresponding position of part \(C .\) Neglect the effect of gravity.

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