Problem 1
Question
A ball of mass \(0.4 \mathrm{~kg}\) hits a wall at right angles with a speed of \(12 \mathrm{~ms}^{-1}\) and bounces off, again at right angles to the wall, with a speed of \(8 \mathrm{~ms}^{-1}\). The impulse exerted by the wall on the ball is: (a) \(1.6 \mathrm{Ns}\) (b) \(20 \mathrm{Ns}\) (c) \(4 \mathrm{Ns}\) (d) \(8 \mathrm{Ns}\).
Step-by-Step Solution
Verified Answer
8 Ns (Option d)
1Step 1: Understand Impulse
Impulse is the change in momentum of an object when a force is applied over time. It is given by the formula: \[ \text{Impulse} = \text{Change in momentum} \]
2Step 2: Calculate Initial Momentum
Momentum is defined as the product of mass and velocity. The initial momentum of the ball is given by: \[ p_{\text{initial}} = m \times v_{\text{initial}} = 0.4 \text{ kg} \times 12 \text{ m/s} = 4.8 \text{ kg} \text{ m/s} \]
3Step 3: Calculate Final Momentum
The final momentum of the ball after bouncing off the wall is: \[ p_{\text{final}} = m \times v_{\text{final}} = 0.4 \text{ kg} \times (-8 \text{ m/s}) = -3.2 \text{ kg} \text{ m/s} \] Note: The velocity is negative because the direction is reversed after the bounce.
4Step 4: Determine Change in Momentum
The change in momentum (impulse) is the final momentum minus the initial momentum: \[ \text{Change in momentum} = p_{\text{final}} - p_{\text{initial}} = -3.2 \text{ kg} \text{ m/s} - 4.8 \text{ kg} \text{ m/s} = -8 \text{ kg} \text{ m/s} \]
5Step 5: Calculate Impulse Magnitude
Since impulse is the magnitude of change in momentum, we take the absolute value: \[ \text{Impulse} = |-\text{Change in momentum}| = |-8 \text{ kg} \text{ m/s}| = 8 \text{ Ns} \]
6Step 6: Select the Correct Answer
The correct answer to the problem based on the calculated impulse is: (d) 8 Ns.
Key Concepts
momentumimpulsemass and velocity relationshipNewton's laws of motion
momentum
Momentum refers to the quantity of motion that an object has. It is a crucial concept in physics, particularly when analyzing motion and collisions. The formula to calculate momentum is given by \( p = mv \), where \( p \) represents momentum, \( m \) denotes mass, and \( v \) is the velocity of the object.
In our exercise, the initial and final momentum of the ball are calculated to understand how the ball’s motion changes as it bounces off the wall. It is essential to note that momentum is a vector quantity, meaning it has both magnitude and direction.
For example, the initial momentum when the ball is moving towards the wall is calculated as:
\[ p_{\text{initial}} = 0.4 \text{ kg} \times 12 \text{ m/s} = 4.8 \text{ kg} \text{ m/s} \]
After the ball bounces back, its velocity is in the opposite direction, so the final momentum is:
\[ p_{\text{final}} = 0.4 \text{ kg} \times (-8 \text{ m/s}) = -3.2 \text{ kg} \text{ m/s} \]
This change in momentum is key to calculating impulse.
In our exercise, the initial and final momentum of the ball are calculated to understand how the ball’s motion changes as it bounces off the wall. It is essential to note that momentum is a vector quantity, meaning it has both magnitude and direction.
For example, the initial momentum when the ball is moving towards the wall is calculated as:
\[ p_{\text{initial}} = 0.4 \text{ kg} \times 12 \text{ m/s} = 4.8 \text{ kg} \text{ m/s} \]
After the ball bounces back, its velocity is in the opposite direction, so the final momentum is:
\[ p_{\text{final}} = 0.4 \text{ kg} \times (-8 \text{ m/s}) = -3.2 \text{ kg} \text{ m/s} \]
This change in momentum is key to calculating impulse.
impulse
Impulse describes the change in momentum resulting from a force applied over a period of time. It is given by the formula:
\[ \text{Impulse} = \text{Change in momentum} \]
In the exercise, to find the impulse exerted by the wall, we need to calculate how the ball’s momentum changes from the moment it hits the wall to the moment it bounces back.
As calculated previously:
Initial momentum: \[ 4.8 \text{ kg} \text{ m/s} \]
Final momentum: \[ -3.2 \text{ kg} \text{ m/s} \]
The change in momentum is calculated as:
\[ \text{Change in momentum} = p_{\text{final}} - p_{\text{initial}} = -3.2 \text{ kg} \text{ m/s} - 4.8 \text{ kg} \text{ m/s} = -8 \text{ kg} \text{ m/s} \]
Since impulse is the magnitude of this change, we take the absolute value:
\[ \text{Impulse} = |-8 \text{ kg} \text{ m/s}| = 8 \text{ Ns} \]
This value shows the impact of the wall force on the ball.
\[ \text{Impulse} = \text{Change in momentum} \]
In the exercise, to find the impulse exerted by the wall, we need to calculate how the ball’s momentum changes from the moment it hits the wall to the moment it bounces back.
As calculated previously:
Initial momentum: \[ 4.8 \text{ kg} \text{ m/s} \]
Final momentum: \[ -3.2 \text{ kg} \text{ m/s} \]
The change in momentum is calculated as:
\[ \text{Change in momentum} = p_{\text{final}} - p_{\text{initial}} = -3.2 \text{ kg} \text{ m/s} - 4.8 \text{ kg} \text{ m/s} = -8 \text{ kg} \text{ m/s} \]
Since impulse is the magnitude of this change, we take the absolute value:
\[ \text{Impulse} = |-8 \text{ kg} \text{ m/s}| = 8 \text{ Ns} \]
This value shows the impact of the wall force on the ball.
mass and velocity relationship
The relationship between mass and velocity plays a pivotal role in understanding momentum and impulse. According to the formula \( p = mv \), both mass and velocity are directly proportional to momentum.
For example, if the mass of an object increases while its velocity remains constant, the momentum also increases. Conversely, with a decrease in mass, momentum decreases, assuming a constant velocity.
In our given problem, the ball’s mass stays constant at 0.4 kg. However, its velocity changes from 12 m/s to -8 m/s as it bounces off the wall:
Initial momentum: \[ p_{\text{initial}} = 0.4 \text{ kg} \times 12 \text{ m/s} = 4.8 \text{ kg} \text{ m/s} \]
Final momentum: \[ p_{\text{final}} = 0.4 \text{ kg} \times (-8 \text{ m/s}) = -3.2 \text{ kg} \text{ m/s} \]
The change in momentum results in the impulse exerted by the wall. Understanding how mass and velocity contribute to these changes provides a comprehensive picture of the physical forces involved.
For example, if the mass of an object increases while its velocity remains constant, the momentum also increases. Conversely, with a decrease in mass, momentum decreases, assuming a constant velocity.
In our given problem, the ball’s mass stays constant at 0.4 kg. However, its velocity changes from 12 m/s to -8 m/s as it bounces off the wall:
Initial momentum: \[ p_{\text{initial}} = 0.4 \text{ kg} \times 12 \text{ m/s} = 4.8 \text{ kg} \text{ m/s} \]
Final momentum: \[ p_{\text{final}} = 0.4 \text{ kg} \times (-8 \text{ m/s}) = -3.2 \text{ kg} \text{ m/s} \]
The change in momentum results in the impulse exerted by the wall. Understanding how mass and velocity contribute to these changes provides a comprehensive picture of the physical forces involved.
Newton's laws of motion
Newton's laws of motion form the foundational principles governing the behavior of objects in motion. They are crucial in understanding momentum and impulse.
**First Law:** An object remains at rest or in uniform motion unless acted upon by an external force. This is significant in explaining why a moving ball changes its state when a force (impulse) is applied by the wall.
**Second Law:** The acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass (\( F = ma \)). This helps understand how the ball’s velocity changes upon hitting the wall.
**Third Law:** For every action, there is an equal and opposite reaction. This principle applies directly to the impulse experienced by the ball. When the ball exerts force on the wall, the wall exerts an equal and opposite force on the ball, leading to the calculated impulse.
The combined understanding of these laws enables us to analyze the ball’s motion accurately and understand the fundamental physics involved.
**First Law:** An object remains at rest or in uniform motion unless acted upon by an external force. This is significant in explaining why a moving ball changes its state when a force (impulse) is applied by the wall.
**Second Law:** The acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass (\( F = ma \)). This helps understand how the ball’s velocity changes upon hitting the wall.
**Third Law:** For every action, there is an equal and opposite reaction. This principle applies directly to the impulse experienced by the ball. When the ball exerts force on the wall, the wall exerts an equal and opposite force on the ball, leading to the calculated impulse.
The combined understanding of these laws enables us to analyze the ball’s motion accurately and understand the fundamental physics involved.
Other exercises in this chapter
Problem 1
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