A real spring is a physical object made of metal, whereas an ideal spring is a concept used in physics calculations. The main difference between the two is that an ideal spring is an idealized model that assumes certain ideal characteristics, while a real spring is subject to the limitations of material properties and physical constraints.
1. Material properties: A real spring is made of a physical material such as steel or plastic, which has its own unique characteristics such as modulus of elasticity and maximum elastic limit. An ideal spring, on the other hand, is assumed to be made of a perfectly elastic material that can deform without any loss of energy.
2. Physical dimensions: A real spring has a specific length, diameter, and shape, which determine its physical properties such as stiffness or spring constant. An ideal spring, however, is assumed to have zero dimensions and zero mass.
3. Hooke's law: Both real and ideal springs follow Hooke's law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. However, in real springs, this proportionality is only valid within the elastic limit, beyond which the material may deform plastically or even fracture. Ideal springs, on the other hand, assume a perfect linear relationship between force and displacement regardless of the magnitude.
4. Damping and energy loss: Real springs often exhibit damping and energy losses due to factors such as internal friction and air resistance. This means that when a real spring is compressed or stretched, some of the energy is lost as heat. Ideal springs, on the other hand, are assumed to have zero damping and preserve all the energy stored in them.
5. Non-linear behavior: Real springs may exhibit non-linear behavior under certain conditions, especially when compressed or stretched beyond their elastic limits. On the other hand, ideal springs are assumed to have a linear force-displacement relationship, making them simpler to analyze in physics calculations.
In summary, a real spring is a physical object with specific material properties and physical characteristics, subject to limitations and real-world constraints. An ideal spring, on the other hand, is an idealized model that assumes perfect physical properties, such as zero dimensions, zero mass, perfect elasticity, and no energy losses.
Compare and contrast a real spring and an ideal spring
23 answers
State Hooke’s Law both in words and equation form.
Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
In equation form, Hooke's Law can be stated as:
F = -kx
Where:
F is the force exerted by the spring (in Newtons, N)
k is the spring constant (also referred to as the stiffness or force constant, measured in Newtons per meter, N/m)
x is the displacement from the equilibrium position (in meters, m)
The negative sign indicates that the force exerted by the spring is opposite in direction to the displacement.
In equation form, Hooke's Law can be stated as:
F = -kx
Where:
F is the force exerted by the spring (in Newtons, N)
k is the spring constant (also referred to as the stiffness or force constant, measured in Newtons per meter, N/m)
x is the displacement from the equilibrium position (in meters, m)
The negative sign indicates that the force exerted by the spring is opposite in direction to the displacement.
Compare and contrast restoring force and applied force.
Restoring Force:
1. Definition: Restoring force refers to the force that acts on an object or system to bring it back to its equilibrium position when it is displaced from that position.
2. Origin: Restoring force arises due to the elastic properties of the object or system, which try to restore it to its original state.
3. Direction: The restoring force always acts in the opposite direction to the displacement from the equilibrium position.
4. Magnitude: The magnitude of the restoring force depends on the displacement and obeys Hooke's law for linear springs.
5. Example: When a spring is compressed or stretched, the restoring force is exerted to bring the spring back to its equilibrium length.
Applied Force:
1. Definition: Applied force refers to an external force that is applied to an object or system from an outside source.
2. Origin: Applied force can arise from physical interactions, such as pushing or pulling an object, or from the effect of other forces acting on the object or system.
3. Direction: The direction of the applied force depends on the external influence or interaction causing it.
4. Magnitude: The magnitude of the applied force can vary and is not necessarily related to the displacement or the elastic properties of the object or system.
5. Example: A person pushing a box or a gravitational force acting on an object are examples of applied forces.
In summary, the main difference between restoring force and applied force is that restoring force acts due to the elastic properties and displacement of an object or system, always opposing the displacement to restore equilibrium. In contrast, applied force refers to external forces acting on an object or system, which may or may not have a direct relationship with the displacement or elastic properties.
1. Definition: Restoring force refers to the force that acts on an object or system to bring it back to its equilibrium position when it is displaced from that position.
2. Origin: Restoring force arises due to the elastic properties of the object or system, which try to restore it to its original state.
3. Direction: The restoring force always acts in the opposite direction to the displacement from the equilibrium position.
4. Magnitude: The magnitude of the restoring force depends on the displacement and obeys Hooke's law for linear springs.
5. Example: When a spring is compressed or stretched, the restoring force is exerted to bring the spring back to its equilibrium length.
Applied Force:
1. Definition: Applied force refers to an external force that is applied to an object or system from an outside source.
2. Origin: Applied force can arise from physical interactions, such as pushing or pulling an object, or from the effect of other forces acting on the object or system.
3. Direction: The direction of the applied force depends on the external influence or interaction causing it.
4. Magnitude: The magnitude of the applied force can vary and is not necessarily related to the displacement or the elastic properties of the object or system.
5. Example: A person pushing a box or a gravitational force acting on an object are examples of applied forces.
In summary, the main difference between restoring force and applied force is that restoring force acts due to the elastic properties and displacement of an object or system, always opposing the displacement to restore equilibrium. In contrast, applied force refers to external forces acting on an object or system, which may or may not have a direct relationship with the displacement or elastic properties.
Spring A is harder to stretch than Spring B. Explain how the spring force constants of each spring would compare in this situation.
If Spring A is harder to stretch than Spring B, it means that Spring A has a greater stiffness or spring constant than Spring B. The spring constant (k) represents the measure of the amount of force required to stretch or compress a spring by a certain amount.
In this situation, the spring force constant of Spring A would be larger than the spring force constant of Spring B. This means that for the same displacement or stretch, Spring A would exert a greater force compared to Spring B. The force exerted by the spring can be calculated using Hooke's Law:
F = -kx
If Spring A requires a greater force to stretch it by a certain displacement (x) compared to Spring B, it implies that the spring constant (k) of Spring A is higher than that of Spring B. Mathematically, this can be represented as:
k_A > k_B
Where k_A represents the spring constant of Spring A and k_B represents the spring constant of Spring B.
In summary, if Spring A is harder to stretch than Spring B, it indicates that Spring A has a higher spring constant, meaning it requires a greater force to achieve the same displacement compared to Spring B.
In this situation, the spring force constant of Spring A would be larger than the spring force constant of Spring B. This means that for the same displacement or stretch, Spring A would exert a greater force compared to Spring B. The force exerted by the spring can be calculated using Hooke's Law:
F = -kx
If Spring A requires a greater force to stretch it by a certain displacement (x) compared to Spring B, it implies that the spring constant (k) of Spring A is higher than that of Spring B. Mathematically, this can be represented as:
k_A > k_B
Where k_A represents the spring constant of Spring A and k_B represents the spring constant of Spring B.
In summary, if Spring A is harder to stretch than Spring B, it indicates that Spring A has a higher spring constant, meaning it requires a greater force to achieve the same displacement compared to Spring B.
Using the image below as an aid, describe the energy conversions a spring undergoes during simple harmonic motion as it moves from the point of maximum compression to maximum stretch in a frictionless environment. Be sure to indicate the points at which there will be
maximum speed.
minimum speed
minimum acceleration
maximum speed.
minimum speed
minimum acceleration
In a frictionless environment, during simple harmonic motion, a spring undergoes energy conversions as it moves from the point of maximum compression to maximum stretch. To describe these energy conversions, I will refer to the image below.
During simple harmonic motion, the spring has the most potential energy when it is at the points of maximum compression and maximum stretch. As the spring moves from the point of maximum compression to the equilibrium position (middle point), the potential energy of the spring gets converted into kinetic energy. At the equilibrium position, the spring has the minimal potential energy and the maximum kinetic energy. This is the point of maximum speed, as all the potential energy has been converted into kinetic energy.
As the spring moves from the equilibrium position to the point of maximum stretch, the kinetic energy gets converted back into potential energy. At the point of maximum stretch, the spring has the most potential energy again. The kinetic energy is minimal at this point and the speed is also minimal, as the energy has been almost completely converted back into potential energy.
At the points of maximum compression and maximum stretch, when the spring briefly stops before changing direction, the velocity is momentarily zero. Therefore, at these points, the speed is zero and the acceleration is at a maximum, changing the direction of motion. The acceleration is maximum when the displacement is zero, as it is the point where the direction of motion changes.
In summary, as the spring moves from maximum compression to maximum stretch during simple harmonic motion in a frictionless environment:
- Maximum speed occurs at the equilibrium position, where all potential energy has been converted into kinetic energy.
- Minimum speed occurs at the points of maximum compression and maximum stretch, where the energy is primarily in the form of potential energy.
- Minimum acceleration occurs at the points of maximum compression and maximum stretch, where the velocity momentarily becomes zero and the direction of motion changes.
During simple harmonic motion, the spring has the most potential energy when it is at the points of maximum compression and maximum stretch. As the spring moves from the point of maximum compression to the equilibrium position (middle point), the potential energy of the spring gets converted into kinetic energy. At the equilibrium position, the spring has the minimal potential energy and the maximum kinetic energy. This is the point of maximum speed, as all the potential energy has been converted into kinetic energy.
As the spring moves from the equilibrium position to the point of maximum stretch, the kinetic energy gets converted back into potential energy. At the point of maximum stretch, the spring has the most potential energy again. The kinetic energy is minimal at this point and the speed is also minimal, as the energy has been almost completely converted back into potential energy.
At the points of maximum compression and maximum stretch, when the spring briefly stops before changing direction, the velocity is momentarily zero. Therefore, at these points, the speed is zero and the acceleration is at a maximum, changing the direction of motion. The acceleration is maximum when the displacement is zero, as it is the point where the direction of motion changes.
In summary, as the spring moves from maximum compression to maximum stretch during simple harmonic motion in a frictionless environment:
- Maximum speed occurs at the equilibrium position, where all potential energy has been converted into kinetic energy.
- Minimum speed occurs at the points of maximum compression and maximum stretch, where the energy is primarily in the form of potential energy.
- Minimum acceleration occurs at the points of maximum compression and maximum stretch, where the velocity momentarily becomes zero and the direction of motion changes.
A mass sits on a level surface, connected to a spring. The spring is anchored to a wall. Locations along the surface are labelled as maximum compression, maximum stretch and equilibrium
A mass of 4.0 kg is in motion along a horizontal, frictionless surface with a speed of 2.5 m/s when it makes contact with a horizontal spring anchored to a wall. The spring compresses and brings the mass to a momentary rest. The spring constant is 1250 N/m.
Determine the maximum distance the spring is compressed.
Determine the speed of the block when the spring is compressed a distance of 9.5 cm
A mass of 4.0 kg is in motion along a horizontal, frictionless surface with a speed of 2.5 m/s when it makes contact with a horizontal spring anchored to a wall. The spring compresses and brings the mass to a momentary rest. The spring constant is 1250 N/m.
Determine the maximum distance the spring is compressed.
Determine the speed of the block when the spring is compressed a distance of 9.5 cm
To determine the maximum distance the spring is compressed, we can use the principle of conservation of mechanical energy. At the moment the mass comes to a momentary rest, all its kinetic energy is converted into potential energy stored in the compressed spring.
1. Find the potential energy stored in the compressed spring:
Potential energy (PE) = 1/2 k x^2
Where k is the spring constant and x is the compression distance.
Given:
Spring constant (k) = 1250 N/m
Since the spring brings the mass to rest, the kinetic energy of the mass is completely converted into potential energy.
Potential energy (PE) = Kinetic energy (KE)
1/2 m v^2 = 1/2 k x^2
Rearranging the equation to solve for x:
x = sqrt((m v^2) / k)
Given:
Mass (m) = 4.0 kg
Speed (v) = 2.5 m/s
Spring constant (k) = 1250 N/m
x = sqrt((4.0 kg * (2.5 m/s)^2) / 1250 N/m)
x = sqrt(10 J / 1250 N/m)
x = sqrt(0.008 m^2)
x ≈ 0.089 m
Therefore, the maximum distance the spring is compressed is approximately 0.089 m or 8.9 cm.
To determine the speed of the block when the spring is compressed a distance of 9.5 cm, we can use the principle of conservation of mechanical energy again. At this point, the potential energy stored in the spring is equal to the initial kinetic energy of the mass.
2. Find the speed of the block:
Potential energy (PE) = Kinetic energy (KE)
1/2 k x^2 = 1/2 m v^2
Rearranging the equation to solve for v:
v = sqrt((k x^2) / m)
Given:
Spring constant (k) = 1250 N/m
Compression distance (x) = 9.5 cm = 0.095 m
Mass (m) = 4.0 kg
v = sqrt((1250 N/m * (0.095 m)^2) / 4.0 kg)
v = sqrt(14.21875 J / 4.0 kg)
v = sqrt(3.55468 m^2/s^2)
v ≈ 1.884 m/s
Therefore, the speed of the block when the spring is compressed a distance of 9.5 cm is approximately 1.884 m/s.
1. Find the potential energy stored in the compressed spring:
Potential energy (PE) = 1/2 k x^2
Where k is the spring constant and x is the compression distance.
Given:
Spring constant (k) = 1250 N/m
Since the spring brings the mass to rest, the kinetic energy of the mass is completely converted into potential energy.
Potential energy (PE) = Kinetic energy (KE)
1/2 m v^2 = 1/2 k x^2
Rearranging the equation to solve for x:
x = sqrt((m v^2) / k)
Given:
Mass (m) = 4.0 kg
Speed (v) = 2.5 m/s
Spring constant (k) = 1250 N/m
x = sqrt((4.0 kg * (2.5 m/s)^2) / 1250 N/m)
x = sqrt(10 J / 1250 N/m)
x = sqrt(0.008 m^2)
x ≈ 0.089 m
Therefore, the maximum distance the spring is compressed is approximately 0.089 m or 8.9 cm.
To determine the speed of the block when the spring is compressed a distance of 9.5 cm, we can use the principle of conservation of mechanical energy again. At this point, the potential energy stored in the spring is equal to the initial kinetic energy of the mass.
2. Find the speed of the block:
Potential energy (PE) = Kinetic energy (KE)
1/2 k x^2 = 1/2 m v^2
Rearranging the equation to solve for v:
v = sqrt((k x^2) / m)
Given:
Spring constant (k) = 1250 N/m
Compression distance (x) = 9.5 cm = 0.095 m
Mass (m) = 4.0 kg
v = sqrt((1250 N/m * (0.095 m)^2) / 4.0 kg)
v = sqrt(14.21875 J / 4.0 kg)
v = sqrt(3.55468 m^2/s^2)
v ≈ 1.884 m/s
Therefore, the speed of the block when the spring is compressed a distance of 9.5 cm is approximately 1.884 m/s.
A toy spring launcher is used to launch a 10.5 g object into the air. The spring of the launcher is initially compressed 6.0 cm. The spring has a spring constant of 5.2 x 10² N/m.
Determine the elastic potential energy of the compressed spring.
Determine the speed of the object as it is ejected upward from the launcher.
Determine the elastic potential energy of the compressed spring.
Determine the speed of the object as it is ejected upward from the launcher.
To determine the elastic potential energy of the compressed spring, we can use the formula for potential energy stored in a spring:
Potential energy (PE) = 1/2 k x^2
Where k is the spring constant and x is the compression distance.
Given:
Spring constant (k) = 5.2 x 10² N/m
Compression distance (x) = 6.0 cm = 0.06 m
PE = 1/2 * (5.2 x 10² N/m) * (0.06 m)^2
PE = 0.936 J
Therefore, the elastic potential energy of the compressed spring is 0.936 Joules.
To determine the speed of the object as it is ejected upward from the launcher, we can use the principle of conservation of mechanical energy. The elastic potential energy stored in the compressed spring is converted into kinetic energy of the object.
Potential energy (PE) = Kinetic energy (KE)
1/2 k x^2 = 1/2 m v^2
Since the mass is given as 10.5 g, we first need to convert it to kilograms:
Mass (m) = 10.5 g = 0.0105 kg
Now we can solve for the speed (v):
1/2 * (5.2 x 10² N/m) * (0.06 m)^2 = 1/2 * (0.0105 kg) * v^2
Rearranging the equation to solve for v:
v^2 = [(5.2 x 10² N/m) * (0.06 m)^2] / (0.0105 kg)
v^2 = (18.72 N/m) / (0.0105 kg)
v^2 ≈ 1782.86 m^2/s^2
v ≈ sqrt(1782.86 m^2/s^2)
v ≈ 42.25 m/s
Therefore, the speed of the object as it is ejected upward from the launcher is approximately 42.25 m/s.
Potential energy (PE) = 1/2 k x^2
Where k is the spring constant and x is the compression distance.
Given:
Spring constant (k) = 5.2 x 10² N/m
Compression distance (x) = 6.0 cm = 0.06 m
PE = 1/2 * (5.2 x 10² N/m) * (0.06 m)^2
PE = 0.936 J
Therefore, the elastic potential energy of the compressed spring is 0.936 Joules.
To determine the speed of the object as it is ejected upward from the launcher, we can use the principle of conservation of mechanical energy. The elastic potential energy stored in the compressed spring is converted into kinetic energy of the object.
Potential energy (PE) = Kinetic energy (KE)
1/2 k x^2 = 1/2 m v^2
Since the mass is given as 10.5 g, we first need to convert it to kilograms:
Mass (m) = 10.5 g = 0.0105 kg
Now we can solve for the speed (v):
1/2 * (5.2 x 10² N/m) * (0.06 m)^2 = 1/2 * (0.0105 kg) * v^2
Rearranging the equation to solve for v:
v^2 = [(5.2 x 10² N/m) * (0.06 m)^2] / (0.0105 kg)
v^2 = (18.72 N/m) / (0.0105 kg)
v^2 ≈ 1782.86 m^2/s^2
v ≈ sqrt(1782.86 m^2/s^2)
v ≈ 42.25 m/s
Therefore, the speed of the object as it is ejected upward from the launcher is approximately 42.25 m/s.
State the names of the two classifications of collisions covered in Unit 2.
The two classifications of collisions covered in Unit 2 are:
1. Elastic collisions: In an elastic collision, both kinetic energy and momentum are conserved. The objects involved bounce off each other without any loss of energy. This type of collision is characterized by a preservation of kinetic energy and a change in direction of motion.
2. Inelastic collisions: In an inelastic collision, only momentum is conserved while kinetic energy is not conserved. Some kinetic energy is lost during the collision, often converted into other forms such as heat, sound, or deformation. This type of collision is characterized by a deformation or sticking together of the objects involved.
1. Elastic collisions: In an elastic collision, both kinetic energy and momentum are conserved. The objects involved bounce off each other without any loss of energy. This type of collision is characterized by a preservation of kinetic energy and a change in direction of motion.
2. Inelastic collisions: In an inelastic collision, only momentum is conserved while kinetic energy is not conserved. Some kinetic energy is lost during the collision, often converted into other forms such as heat, sound, or deformation. This type of collision is characterized by a deformation or sticking together of the objects involved.
Define each type of collision from Part 1
Sure!
1. Elastic collisions: An elastic collision is a type of collision between two objects where both momentum and kinetic energy are conserved. In an elastic collision, the total kinetic energy of the system remains the same before and after the collision. The objects involved in an elastic collision rebound off each other without any loss of kinetic energy. This type of collision is often seen when objects collide without deforming or sticking together, such as billiard balls or bouncing balls.
2. Inelastic collisions: An inelastic collision is a type of collision between two objects where momentum is conserved, but kinetic energy is not. In an inelastic collision, some or all of the kinetic energy of the system is lost during the collision and transformed into other forms of energy, such as heat, sound, or deformation. The objects involved in an inelastic collision may stick together or deform upon impact. Examples of inelastic collisions include a car hitting a wall, where the kinetic energy is dissipated as deformation and sound, or a ball of clay hitting another ball and sticking to it.
1. Elastic collisions: An elastic collision is a type of collision between two objects where both momentum and kinetic energy are conserved. In an elastic collision, the total kinetic energy of the system remains the same before and after the collision. The objects involved in an elastic collision rebound off each other without any loss of kinetic energy. This type of collision is often seen when objects collide without deforming or sticking together, such as billiard balls or bouncing balls.
2. Inelastic collisions: An inelastic collision is a type of collision between two objects where momentum is conserved, but kinetic energy is not. In an inelastic collision, some or all of the kinetic energy of the system is lost during the collision and transformed into other forms of energy, such as heat, sound, or deformation. The objects involved in an inelastic collision may stick together or deform upon impact. Examples of inelastic collisions include a car hitting a wall, where the kinetic energy is dissipated as deformation and sound, or a ball of clay hitting another ball and sticking to it.
From the examples below choose only two and indicate the type of collision each represents. Justify each choice using concepts from this unit.
Two billiard balls are moving toward each other along an essentially frictionless surface and collide head-on.
While distracted by the sun, a driver heading eastbound doesn’t see a stop sign and proceeds to drive through an intersection. Meanwhile, another car travelling southbound enters the intersection and collides with the eastbound vehicle. Both cars remain together after the initial contact occurs.
Standing at the baseline of a tennis court, a tennis player tosses a tennis ball over her head swings at the ball. The racquet hits the tossed ball and travels to the other side of the tennis court.
A rubber ball is thrown at an aluminum can. Hitting the can, the ball causes a dent in the can.
Two billiard balls are moving toward each other along an essentially frictionless surface and collide head-on.
While distracted by the sun, a driver heading eastbound doesn’t see a stop sign and proceeds to drive through an intersection. Meanwhile, another car travelling southbound enters the intersection and collides with the eastbound vehicle. Both cars remain together after the initial contact occurs.
Standing at the baseline of a tennis court, a tennis player tosses a tennis ball over her head swings at the ball. The racquet hits the tossed ball and travels to the other side of the tennis court.
A rubber ball is thrown at an aluminum can. Hitting the can, the ball causes a dent in the can.
The two examples and their corresponding types of collisions are:
1. Two billiard balls colliding head-on: This represents an elastic collision. In an elastic collision, kinetic energy and momentum are conserved. When the billiard balls collide, they rebound off each other without any loss of kinetic energy. The conservation of momentum dictates that the total momentum before the collision is equal to the total momentum after the collision. Therefore, this collision is elastic.
2. Two cars colliding and remaining together: This represents an inelastic collision. In an inelastic collision, momentum is conserved, but kinetic energy is not. When the two cars collide and remain together, their kinetic energy is transformed into other forms of energy, such as deformation, sound, or heat. The cars stick together after the collision, indicating that kinetic energy has been converted into deformation energy and possibly sound energy. Thus, this collision is inelastic.
1. Two billiard balls colliding head-on: This represents an elastic collision. In an elastic collision, kinetic energy and momentum are conserved. When the billiard balls collide, they rebound off each other without any loss of kinetic energy. The conservation of momentum dictates that the total momentum before the collision is equal to the total momentum after the collision. Therefore, this collision is elastic.
2. Two cars colliding and remaining together: This represents an inelastic collision. In an inelastic collision, momentum is conserved, but kinetic energy is not. When the two cars collide and remain together, their kinetic energy is transformed into other forms of energy, such as deformation, sound, or heat. The cars stick together after the collision, indicating that kinetic energy has been converted into deformation energy and possibly sound energy. Thus, this collision is inelastic.
From the examples below choose only two and indicate the type of collision each represents. Justify each choice using concepts from this unit.
Two billiard balls are moving toward each other along an essentially frictionless surface and collide head-on.
While distracted by the sun, a driver heading eastbound doesn’t see a stop sign and proceeds to drive through an intersection. Meanwhile, another car travelling southbound enters the intersection and collides with the eastbound vehicle. Both cars remain together after the initial contact occurs.
Standing at the baseline of a tennis court, a tennis player tosses a tennis ball over her head swings at the ball. The racquet hits the tossed ball and travels to the other side of the tennis court.
A rubber ball is thrown at an aluminum can. Hitting the can, the ball causes a dent in the can.
Two billiard balls are moving toward each other along an essentially frictionless surface and collide head-on.
While distracted by the sun, a driver heading eastbound doesn’t see a stop sign and proceeds to drive through an intersection. Meanwhile, another car travelling southbound enters the intersection and collides with the eastbound vehicle. Both cars remain together after the initial contact occurs.
Standing at the baseline of a tennis court, a tennis player tosses a tennis ball over her head swings at the ball. The racquet hits the tossed ball and travels to the other side of the tennis court.
A rubber ball is thrown at an aluminum can. Hitting the can, the ball causes a dent in the can.
The two examples and their corresponding types of collisions are:
1. Two billiard balls colliding head-on: This represents an elastic collision. In an elastic collision, both momentum and kinetic energy are conserved. When the billiard balls collide, they rebound off each other without any loss of kinetic energy. This type of collision is characterized by a preservation of kinetic energy and a change in direction of motion.
2. A rubber ball hitting an aluminum can and causing a dent: This represents an inelastic collision. In an inelastic collision, momentum is conserved, but kinetic energy is not. When the rubber ball hits the aluminum can and causes a dent, some of the kinetic energy is transferred to deformation energy. The objects stick together momentarily as they undergo deformation. This type of collision is characterized by a loss of kinetic energy and a change in shape or deformation of the objects involved.
1. Two billiard balls colliding head-on: This represents an elastic collision. In an elastic collision, both momentum and kinetic energy are conserved. When the billiard balls collide, they rebound off each other without any loss of kinetic energy. This type of collision is characterized by a preservation of kinetic energy and a change in direction of motion.
2. A rubber ball hitting an aluminum can and causing a dent: This represents an inelastic collision. In an inelastic collision, momentum is conserved, but kinetic energy is not. When the rubber ball hits the aluminum can and causes a dent, some of the kinetic energy is transferred to deformation energy. The objects stick together momentarily as they undergo deformation. This type of collision is characterized by a loss of kinetic energy and a change in shape or deformation of the objects involved.
In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The collision is cushioned by a spring (k = 1200 N/m).
Determine the velocity of each cart after the collision.
Determine the maximum compression of the spring.
Determine the velocity of each cart after the collision.
Determine the maximum compression of the spring.
To determine the velocity of each cart after the collision, we can use the principle of conservation