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5.4 Conservation of Linear Momentum

8 min readmay 18, 2023

Kanya Shah

Kanya Shah

Daniella Garcia-Loos

Daniella Garcia-Loos

Kanya Shah

Kanya Shah

Daniella Garcia-Loos

Daniella Garcia-Loos

Types of Collisions

If the net force acting on an object is zero, then its is conserved. Some forces in one direction may be zero while in another direction will have several forces. For example, a ball in free fall would have a net force of zero in the x direction while the y direction’s net force is not equal to zero in that specific case. 

Let’s talk about types of collisions and whether is conserved in each collision respectively. A collision refers to a situation in which two objects strike each other and whether they stick together or rebound. The net external force is either zero or negligibly small. Even if is conserved during a collision, the may not be conserved because during the collision, KE can be converted into other forms of energy.

Inelastic Collisions

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2Fin.PNG?alt=media&token=1920edc7-8f63-4778-8e0c-08582584ff9a

Image Credit: sciencenotes.org

If the initial equals the final , then the objects went through an (rebound) since KE was conserved and so was . In situations where KE isn’t conserved during a collision due to losses associated with heat, sound, etc, that type is known as an (stick together). In either type of collision, will be conserved but the KE may not be so it’s important to be able to identify which collision is occurring in the problem. In completely inelastic collisions, the objects stick together after the collision after colliding. 

Example: Two carts of the same mass lie on a table. The first cart is moving and the second cart is at rest. At the end of the collision, the two carts were stuck together. What is the final speed of the system?  To find the initial of two carts, do = m1*v1 + m2*v2. To find the final of two carts that stick together after the collision, do pfinal = (m1 + m2) *vfinal. You can set the initial equation equal to the pfinal equation since is conserved. Finally, solve for the final velocity.

Elastic Collisions

For elastic collisions, you can also use the formula to solve the problems since KE and will be conserved. K initial = Kfinal so use a similar process to the example and solve for what the question is asking for. The final velocity of the system can be found using KE as well but remember to use the total mass in KE final. 

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2Fec2.PNG?alt=media&token=30b88f51-4cf9-469f-a383-e8cbb821ccab

Image Credit: sciencenotes.com

If the problem is a 2D question, consider both directions separately and then solve for the final KE, , or velocity. If you aren’t given the net force or the vector sum of the velocity, you may have to use the angle given to solve for those values individually. 

Example Problem 1:

Two carts, each with a mass of 2 kg, collide on a frictionless track. Cart A is moving to the right with a velocity of 3 m/s, and cart B is moving to the left with a velocity of 4 m/s. After the collision, both carts move to the right with a velocity of 1 m/s. What is the final velocity of each cart?

Solution:

In an , the total of the system is conserved, but the total is not. This means that the total of the two carts before and after the collision must be the same, but the of the two carts after the collision will be less than before the collision.

Before the collision, the of cart A is given by the formula: = mass * velocity

The mass of cart A is 2 kg, and its velocity is 3 m/s.

Therefore, the of cart A is: = 2 kg * 3 m/s = 6 kg*m/s

Before the collision, the of cart B is given by the formula: = mass * velocity

The mass of cart B is 2 kg, and its velocity is -4 m/s (since it is moving to the left).

Therefore, the of cart B is: = 2 kg * (-4 m/s) = -8 kg*m/s

Before the collision, the total of the two carts is: 6 kgm/s + (-8 kgm/s) = -2 kg*m/s

After the collision, the of cart A is given by the formula: = mass * velocity

The mass of cart A is 2 kg, and its velocity is 1 m/s.

Therefore, the of cart A is: = 2 kg * 1 m/s = 2 kg*m/s

After the collision, the of cart B is given by the formula: = mass * velocity

The mass of cart B is 2 kg, and its velocity is 1 m/s.

Therefore, the of cart B is: = 2 kg * 1 m/s = 2 kg*m/s

After the collision, the total of the two carts is: 2 kgm/s + 2 kgm/s = 4 kg*m/s

Since the total of the two carts must be conserved, the total before and after the collision must be equal: -2 kgm/s = 4 kgm/s

This equation is not true, so the solution is not possible. This means that the final velocities of the carts cannot be determined.

Center of Mass

Center of Mass of a system of masses is the point where the system can be balanced in a uniform gravitational field. For example, consider a hammer that has a heavy head but less mass in the handle. The center of the mass of the object would be closest to the heaviest side or the head of the hammer. To calculate the center of mass, you can use the formula xcm=m1x1+m2x2+ ...(m1+m2+...) to find the x coordinate for the center of mass. In the equation, x1 is the distance from one end of the object. The same formula can be used for the Y direction. You can manipulate the formula slightly if you have more than two objects by adding m3x3 and more depending on how many objects there are in the system. 

If a system has a net force of zero, then it follows that the acceleration of the center of the mass will be zero. Likewise, if a mass is moving initially, the center of mass will move with the same velocity. When is non zero, the center of mass of a system accelerates precisely as if it were a point particle of mass M acted upon by the force

Example Problem 1:

A stick of length 1 meter is balanced on a fulcrum at its midpoint. A mass of 2 kg is attached to one end of the stick, and a mass of 3 kg is attached to the other end. What is the distance from the fulcrum to the center of mass of the system?

Solution:

The center of mass of a system is the point at which the system behaves as if all of its mass were concentrated. For a system of two masses, the center of mass is given by the formula: center of mass = (mass1 * distance1 + mass2 * distance2) / (mass1 + mass2)

In this problem, the distance from the fulcrum to the 2 kg mass is 0.5 meters, and the distance from the fulcrum to the 3 kg mass is 0.5 meters.

Therefore, the center of mass of the system is: center of mass = (2 kg * 0.5 m + 3 kg * 0.5 m) / (2 kg + 3 kg) = 1.25 m

This means that the center of mass of the system is 1.25 meters from the fulcrum.

Example Problem 2:

A ladder of length 5 meters is leaning against a wall at an angle of 60 degrees. The ladder has a mass of 10 kg, and its center of mass is located at a distance of 3 meters from the bottom of the ladder. What is the horizontal force exerted on the ladder by the wall?

Solution:

The horizontal force exerted on the ladder by the wall is equal to the weight of the ladder times the cosine of the angle between the ladder and the wall.

The weight of the ladder is given by the formula: weight = mass * acceleration due to gravity

The mass of the ladder is 10 kg, and the acceleration due to gravity is 9.8 m/s^2.

Therefore, the weight of the ladder is: weight = 10 kg * 9.8 m/s^2 = 98 N

The horizontal force exerted on the ladder by the wall is therefore: horizontal force = weight * cos(angle) = 98 N * cos(60 degrees) = 49 N

This means that the horizontal force exerted on the ladder by the wall is 49 N.

Example Problem 3:

A bar of length 4 meters is balanced on a fulcrum at its midpoint. A mass of 2 kg is attached to one end of the bar, and a mass of 3 kg is attached to the other end. The mass of the bar itself is 5 kg. What is the distance from the fulcrum to the center of mass of the system?

Solution:

The center of mass of a system is the point at which the system behaves as if all of its mass were concentrated. For a system of three masses, the center of mass is given by the formula: center of mass = (mass1 * distance1 + mass2 * distance2 + mass3 * distance3) / (mass1 + mass2 + mass3)

In this problem, the distance from the fulcrum to the 2 kg mass is 0.5 meters, the distance from the fulcrum to the 3 kg mass is 0.5 meters, and the distance from the fulcrum to the bar itself is 2 meters.

Therefore, the center of mass of the system is: center of mass = (2 kg * 0.5 m + 3 kg * 0.5 m + 5 kg * 2 m) / (2 kg + 3 kg + 5 kg) = 1.44 m

This means that the center of mass of the system is 1.44 meters from the fulcrum.

Note: When solving for the center of mass of a system, it is important to take into account not only the masses and distances of the objects in the system, but also the mass and distance of the system itself (in this case, the bar). This is because the center of mass of a system is the point at which the system behaves as if all of its mass were concentrated, including the mass of the system itself.

🎥Watch: AP Physics 1 - Conservation of Momentum

Key Terms to Review (8)

Conserved Momentum

: Conserved momentum means that the total momentum of a system remains constant if no external forces act on it. In other words, when objects interact within a closed system, their combined momenta before and after remain unchanged.

Elastic Collision

: An elastic collision is a type of collision where kinetic energy is conserved. In this type of collision, the objects bounce off each other without any deformation.

Fnet, ext

: Fnet, ext refers to the net external force acting on an object. It is the vector sum of all the external forces acting on an object.

Inelastic Collision

: An inelastic collision is a type of collision where kinetic energy is not conserved. The objects involved stick together after the collision and move as one.

Key Term: K initial = K final

: Definition: The principle of conservation of kinetic energy, which states that the total amount of kinetic energy in a system remains constant before and after an event or interaction.

Kinetic Energy

: Kinetic energy is the energy an object possesses due to its motion. It depends on both the mass and velocity of the object.

Momentum

: Momentum is a property of moving objects that depends on both their mass and velocity. It describes how difficult it is to stop an object in motion.

Pinitial

: Pinitial refers to the initial total momentum before a collision occurs. It takes into account both the mass and velocity of all objects involved in the system.

5.4 Conservation of Linear Momentum

8 min readmay 18, 2023

Kanya Shah

Kanya Shah

Daniella Garcia-Loos

Daniella Garcia-Loos

Kanya Shah

Kanya Shah

Daniella Garcia-Loos

Daniella Garcia-Loos

Types of Collisions

If the net force acting on an object is zero, then its is conserved. Some forces in one direction may be zero while in another direction will have several forces. For example, a ball in free fall would have a net force of zero in the x direction while the y direction’s net force is not equal to zero in that specific case. 

Let’s talk about types of collisions and whether is conserved in each collision respectively. A collision refers to a situation in which two objects strike each other and whether they stick together or rebound. The net external force is either zero or negligibly small. Even if is conserved during a collision, the may not be conserved because during the collision, KE can be converted into other forms of energy.

Inelastic Collisions

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2Fin.PNG?alt=media&token=1920edc7-8f63-4778-8e0c-08582584ff9a

Image Credit: sciencenotes.org

If the initial equals the final , then the objects went through an (rebound) since KE was conserved and so was . In situations where KE isn’t conserved during a collision due to losses associated with heat, sound, etc, that type is known as an (stick together). In either type of collision, will be conserved but the KE may not be so it’s important to be able to identify which collision is occurring in the problem. In completely inelastic collisions, the objects stick together after the collision after colliding. 

Example: Two carts of the same mass lie on a table. The first cart is moving and the second cart is at rest. At the end of the collision, the two carts were stuck together. What is the final speed of the system?  To find the initial of two carts, do = m1*v1 + m2*v2. To find the final of two carts that stick together after the collision, do pfinal = (m1 + m2) *vfinal. You can set the initial equation equal to the pfinal equation since is conserved. Finally, solve for the final velocity.

Elastic Collisions

For elastic collisions, you can also use the formula to solve the problems since KE and will be conserved. K initial = Kfinal so use a similar process to the example and solve for what the question is asking for. The final velocity of the system can be found using KE as well but remember to use the total mass in KE final. 

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2Fec2.PNG?alt=media&token=30b88f51-4cf9-469f-a383-e8cbb821ccab

Image Credit: sciencenotes.com

If the problem is a 2D question, consider both directions separately and then solve for the final KE, , or velocity. If you aren’t given the net force or the vector sum of the velocity, you may have to use the angle given to solve for those values individually. 

Example Problem 1:

Two carts, each with a mass of 2 kg, collide on a frictionless track. Cart A is moving to the right with a velocity of 3 m/s, and cart B is moving to the left with a velocity of 4 m/s. After the collision, both carts move to the right with a velocity of 1 m/s. What is the final velocity of each cart?

Solution:

In an , the total of the system is conserved, but the total is not. This means that the total of the two carts before and after the collision must be the same, but the of the two carts after the collision will be less than before the collision.

Before the collision, the of cart A is given by the formula: = mass * velocity

The mass of cart A is 2 kg, and its velocity is 3 m/s.

Therefore, the of cart A is: = 2 kg * 3 m/s = 6 kg*m/s

Before the collision, the of cart B is given by the formula: = mass * velocity

The mass of cart B is 2 kg, and its velocity is -4 m/s (since it is moving to the left).

Therefore, the of cart B is: = 2 kg * (-4 m/s) = -8 kg*m/s

Before the collision, the total of the two carts is: 6 kgm/s + (-8 kgm/s) = -2 kg*m/s

After the collision, the of cart A is given by the formula: = mass * velocity

The mass of cart A is 2 kg, and its velocity is 1 m/s.

Therefore, the of cart A is: = 2 kg * 1 m/s = 2 kg*m/s

After the collision, the of cart B is given by the formula: = mass * velocity

The mass of cart B is 2 kg, and its velocity is 1 m/s.

Therefore, the of cart B is: = 2 kg * 1 m/s = 2 kg*m/s

After the collision, the total of the two carts is: 2 kgm/s + 2 kgm/s = 4 kg*m/s

Since the total of the two carts must be conserved, the total before and after the collision must be equal: -2 kgm/s = 4 kgm/s

This equation is not true, so the solution is not possible. This means that the final velocities of the carts cannot be determined.

Center of Mass

Center of Mass of a system of masses is the point where the system can be balanced in a uniform gravitational field. For example, consider a hammer that has a heavy head but less mass in the handle. The center of the mass of the object would be closest to the heaviest side or the head of the hammer. To calculate the center of mass, you can use the formula xcm=m1x1+m2x2+ ...(m1+m2+...) to find the x coordinate for the center of mass. In the equation, x1 is the distance from one end of the object. The same formula can be used for the Y direction. You can manipulate the formula slightly if you have more than two objects by adding m3x3 and more depending on how many objects there are in the system. 

If a system has a net force of zero, then it follows that the acceleration of the center of the mass will be zero. Likewise, if a mass is moving initially, the center of mass will move with the same velocity. When is non zero, the center of mass of a system accelerates precisely as if it were a point particle of mass M acted upon by the force

Example Problem 1:

A stick of length 1 meter is balanced on a fulcrum at its midpoint. A mass of 2 kg is attached to one end of the stick, and a mass of 3 kg is attached to the other end. What is the distance from the fulcrum to the center of mass of the system?

Solution:

The center of mass of a system is the point at which the system behaves as if all of its mass were concentrated. For a system of two masses, the center of mass is given by the formula: center of mass = (mass1 * distance1 + mass2 * distance2) / (mass1 + mass2)

In this problem, the distance from the fulcrum to the 2 kg mass is 0.5 meters, and the distance from the fulcrum to the 3 kg mass is 0.5 meters.

Therefore, the center of mass of the system is: center of mass = (2 kg * 0.5 m + 3 kg * 0.5 m) / (2 kg + 3 kg) = 1.25 m

This means that the center of mass of the system is 1.25 meters from the fulcrum.

Example Problem 2:

A ladder of length 5 meters is leaning against a wall at an angle of 60 degrees. The ladder has a mass of 10 kg, and its center of mass is located at a distance of 3 meters from the bottom of the ladder. What is the horizontal force exerted on the ladder by the wall?

Solution:

The horizontal force exerted on the ladder by the wall is equal to the weight of the ladder times the cosine of the angle between the ladder and the wall.

The weight of the ladder is given by the formula: weight = mass * acceleration due to gravity

The mass of the ladder is 10 kg, and the acceleration due to gravity is 9.8 m/s^2.

Therefore, the weight of the ladder is: weight = 10 kg * 9.8 m/s^2 = 98 N

The horizontal force exerted on the ladder by the wall is therefore: horizontal force = weight * cos(angle) = 98 N * cos(60 degrees) = 49 N

This means that the horizontal force exerted on the ladder by the wall is 49 N.

Example Problem 3:

A bar of length 4 meters is balanced on a fulcrum at its midpoint. A mass of 2 kg is attached to one end of the bar, and a mass of 3 kg is attached to the other end. The mass of the bar itself is 5 kg. What is the distance from the fulcrum to the center of mass of the system?

Solution:

The center of mass of a system is the point at which the system behaves as if all of its mass were concentrated. For a system of three masses, the center of mass is given by the formula: center of mass = (mass1 * distance1 + mass2 * distance2 + mass3 * distance3) / (mass1 + mass2 + mass3)

In this problem, the distance from the fulcrum to the 2 kg mass is 0.5 meters, the distance from the fulcrum to the 3 kg mass is 0.5 meters, and the distance from the fulcrum to the bar itself is 2 meters.

Therefore, the center of mass of the system is: center of mass = (2 kg * 0.5 m + 3 kg * 0.5 m + 5 kg * 2 m) / (2 kg + 3 kg + 5 kg) = 1.44 m

This means that the center of mass of the system is 1.44 meters from the fulcrum.

Note: When solving for the center of mass of a system, it is important to take into account not only the masses and distances of the objects in the system, but also the mass and distance of the system itself (in this case, the bar). This is because the center of mass of a system is the point at which the system behaves as if all of its mass were concentrated, including the mass of the system itself.

🎥Watch: AP Physics 1 - Conservation of Momentum

Key Terms to Review (8)

Conserved Momentum

: Conserved momentum means that the total momentum of a system remains constant if no external forces act on it. In other words, when objects interact within a closed system, their combined momenta before and after remain unchanged.

Elastic Collision

: An elastic collision is a type of collision where kinetic energy is conserved. In this type of collision, the objects bounce off each other without any deformation.

Fnet, ext

: Fnet, ext refers to the net external force acting on an object. It is the vector sum of all the external forces acting on an object.

Inelastic Collision

: An inelastic collision is a type of collision where kinetic energy is not conserved. The objects involved stick together after the collision and move as one.

Key Term: K initial = K final

: Definition: The principle of conservation of kinetic energy, which states that the total amount of kinetic energy in a system remains constant before and after an event or interaction.

Kinetic Energy

: Kinetic energy is the energy an object possesses due to its motion. It depends on both the mass and velocity of the object.

Momentum

: Momentum is a property of moving objects that depends on both their mass and velocity. It describes how difficult it is to stop an object in motion.

Pinitial

: Pinitial refers to the initial total momentum before a collision occurs. It takes into account both the mass and velocity of all objects involved in the system.


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AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.


© 2024 Fiveable Inc. All rights reserved.

AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.