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A ball with an initial velocity of \(10 \mathrm{m} / \mathrm{s}\) moves at an angle \(60^{\circ}\) above the \(+x\) -direction. The ball hits a vertical wall and bounces off so that it is moving \(60^{\circ}\) above the \(-x\) -direction with the same speed. What is the impulse delivered by the wall?

Short Answer

Expert verified
The magnitude of the impulse is \( |\vec{J}| = 2mv\cos(60^\circ) \), where \( m \) is the mass of the ball and \( v \) is the initial speed, \( 10 \mathrm{m/s} \). Since the mass is not given, the answer is in terms of \( m \).

Step by step solution

01

Determine the initial momentum

The initial momentum of the ball can be found by using the formula \( \vec{p}_{\text{initial}} = m \vec{v}_{\text{initial}} \), where \( m \) is the mass of the ball and \( \vec{v}_{\text{initial}} \) is the initial velocity vector. Given the speed is \( 10 \mathrm{m/s} \) and the angle is \( 60^\circ \), decompose the velocity into horizontal \( (v_x) \) and vertical components \( (v_y) \) using trigonometric functions: \( v_x = v \cos(60^\circ) \), \( v_y = v \sin(60^\circ) \).
02

Determine the final momentum

After the collision, the ball moves at the same speed but in the \(-x\)-direction at an angle of \(60^\circ\) above it. As in Step 1, decompose the final velocity into horizontal \( (v'_x) \) and vertical components \( (v'_y) \) again using trigonometry. Now \( v'_x = -v \cos(60^\circ) \) because it's in the opposite direction, but \( v'_y = v \sin(60^\circ) \) remains the same because the vertical movement is unaffected by the wall.
03

Calculate the change in momentum (impulse)

The impulse delivered by the wall \( \vec{J} \), is defined as the change in momentum \( \Delta \vec{p} \). Since momentum is a vector quantity and the mass of the ball doesn't change, it can be found by subtracting the initial momentum vector from the final momentum vector: \( \vec{J} = \vec{p}_{\text{final}} - \vec{p}_{\text{initial}} = m(\vec{v}_{\text{final}} - \vec{v}_{\text{initial}}) \). Compute the differences of the horizontal and vertical components separately to find \( \vec{J} \).
04

Express the impulse in vector form

Using the values found for the horizontal and vertical components of the final and initial velocities, express the impulse in vector form. \( J_x = m(v'_x - v_x) \) and \( J_y = m(v'_y - v_y) = m(0) \) since the vertical component hasn't changed.
05

Calculate the magnitude of the impulse

Since there's no change in the vertical component of momentum, the impulse only has a horizontal component. Find the magnitude of the horizontal impulse using the formula \( |\vec{J}| = |J_x| = m|v'_x - v_x| \), where \( m \) is the mass, which can be simplified as \( |\vec{J}| = 2mv\cos(60^\circ) \) due to the symmetry of the problem. You can now calculate the impulse given the mass of the ball.

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Key Concepts

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

Momentum Conservation
In a closed system, the total momentum before an event, such as a collision, is the same as the total momentum after the event; this is known as momentum conservation. When discussing a situation like a ball hitting and bouncing off a wall, it's important to consider that the wall exerts a force on the ball, thus changing its velocity and therefore its momentum.

However, because the system (ball plus wall) is assumed to be closed, and the wall is much more massive than the ball, the wall essentially 'absorbs' the reaction and does not have noticeable movement itself. Ultimately, the law of conservation of momentum ensures that the initial momentum (in the absence of external forces) will be the same as the final momentum.

In the exercise analyzed, momentum conservation plays a key role in understanding that despite the direction of the ball changing, the magnitude of the velocity (speed) stays the same. Therefore, the changes in momentum are entirely due to changes in direction, not speed, since the collision is perfectly elastic and no kinetic energy is lost. We calculate the impulse to quantify the change in the ball's momentum.
Collision in Physics
A collision in physics refers to any event where two or more objects come into contact and exert forces on one another in a relatively short time span. Collisions can be classified into two categories: elastic and inelastic.

In an elastic collision, both momentum and kinetic energy are conserved. The ball-wall collision depicted in our exercise is a good example of an elastic collision because the ball's speed remains the same before and after the impact, indicating no kinetic energy has been lost. However, its direction changes, implying a change in momentum.

The wall applies an impulse to the ball, which rapidly changes the ball's momentum, demonstrating Newton's third law, which states that for every action, there is an equal and opposite reaction. The ball impacts the wall with a certain force and momentum and is then pushed off with the same amount of force and momentum in the opposite direction. Understanding the mechanics of the collision helps us to calculate the impulse provided by the wall, revealing the wall's influence on the ball's motion post-collision.
Vector Decomposition
The concept of vector decomposition is crucial in physics because it allows us to break down a vector into its components along specified axes, most commonly into horizontal and vertical components. This is particularly useful when dealing with motion at an angle.

In our scenario, vector decomposition is used to find the components of the ball’s initial and final velocities along the x (horizontal) and y (vertical) axes. By applying trigonometric functions such as cosine and sine to the given angle, we can determine these vector components.

Initial Velocity Decomposition

For the initial velocity, the horizontal component is found using the cosine function, while the vertical component is found using the sine function. The equations used are:
\( v_x = v \cos(\theta) \) and \( v_y = v \sin(\theta) \).

Final Velocity Decomposition

After the collision, the horizontal component of the final velocity is negative, as the ball has reversed direction, but the vertical component remains unaffected. The equations become:
\( v'_x = -v \cos(\theta) \) and \( v'_y = v \sin(\theta) \).

By decomposing the initial and final velocity vectors, we gain the necessary information to calculate the impulse by analyzing the components separately. Decomposing vectors in such a manner allows for precise and simplified computations for solving physics problems involving angles and multi-directional forces.

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