5.4 Conservation of Linear Momentum

If the net force acting on an object is zero, then its momentum 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 momentum 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 momentum is conserved during a collision, the kinetic energy may not be conserved because during the collision, KE can be converted into other forms of energy.

If the initial kinetic energy equals the final kinetic energy, then the objects went through an elastic collision (rebound) since KE was conserved and so was momentum. In situations where KE isn’t conserved during a collision due to losses associated with heat, sound, etc, that type is known as an inelastic collision (stick together). In either type of collision, momentum 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 momentum of two carts, do pinitial = m1*v1 + m2*v2. To find the final momentum 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 momentum is conserved. Finally, solve for the final velocity.

For elastic collisions, you can also use the Kinetic Energy formula to solve the problems since KE and momentum 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. 

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?

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

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

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

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

Before the collision, the momentum of cart B is given by the formula: momentum = 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 momentum of cart B is: momentum = 2 kg * (-4 m/s) = -8 kg*m/s

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

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

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

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

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

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

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

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

Since the total momentum of the two carts must be conserved, the total momentum 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 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. 

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?

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.

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?

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.

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?

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.

🎥Watch: AP Physics 1 - Conservation of Momentum