Collisions

Collisions In any collision between two bodies in an “isolated system” ( Fnet system = 0) the total __________________ of the system remains constant or “conserved”.

Collisions In any collision between two bodies in an “isolated system” ( Fnet system = 0) the total momentum of the system remains constant or “conserved”.

Collisions In any collision between two bodies in an “isolated system” ( Fnet system = 0) the total momentum of the system remains constant or “conserved”. Therefore, in any collision in an isolated system , psys (before collision) = ____________

Collisions In any collision between two bodies in an “isolated system” ( Fnet system = 0) the total momentum of the system remains constant or “conserved”. Therefore, in any collision in an “isolated system” , psys (before collision) = psys (after collision)

Collisions where total kinetic energy is conserved

Collisions where total kinetic energy is conserved A collision in an isolated system in which the total kinetic energy of the system before the collision equals the total kinetic energy of the system after the collision is called an ______________ collision.

Collisions where total kinetic energy is conserved A collision in an isolated system in which the total kinetic energy of the system before the collision equals the total kinetic energy of the system after the collision is called an elastic collision.

Collisions where total kinetic energy is conserved A collision in an isolated system in which the total kinetic energy of the system before the collision equals the total kinetic energy of the system after the collision is called an elastic collision. We can write two equations for “elastic collisions”:

Collisions where total kinetic energy is conserved A collision in an isolated system in which the total kinetic energy of the system before the collision equals the total kinetic energy of the system after the collision is called an elastic collision. We can write two equations for “elastic collisions”: 1) psys (before collision) = psys (after collision)

Collisions where total kinetic energy is conserved A collision in an isolated system in which the total kinetic energy of the system before the collision equals the total kinetic energy of the system after the collision is called an elastic collision. We can write two equations for “elastic collisions”: 1) psys (before collision) = psys (after collision) 2)

Collisions where total kinetic energy is conserved A collision in an isolated system in which the total kinetic energy of the system before the collision equals the total kinetic energy of the system after the collision is called an elastic collision. We can write two equations for “elastic collisions”: 1) psys (before collision) = psys (after collision) 2) Ek sys (before collision) = Ek sys (after collision)

Collisions where total kinetic energy is conserved A collision in an isolated system in which the total kinetic energy of the system before the collision equals the total kinetic energy of the system after the collision is called an elastic collision. We can write two equations for “elastic collisions”: 1) psys (before collision) = psys (after collision) 2) Ek sys (before collision) = Ek sys (after collision) During an elastic collision, no energy is ______ to the system in the form of heat, light, sound etc.

Collisions where total kinetic energy is conserved A collision in an isolated system in which the total kinetic energy of the system before the collision equals the total kinetic energy of the system after the collision is called an elastic collision. We can write two equations for “elastic collisions”: 1) psys (before collision) = psys (after collision) 2) Ek sys (before collision) = Ek sys (after collision) During an elastic collision, no energy is lost to the system in the form of heat, light, sound etc.

Collisions where total kinetic energy is conserved A collision in an isolated system in which the total kinetic energy of the system before the collision equals the total kinetic energy of the system after the collision is called an elastic collision. We can write two equations for “elastic collisions”: 1) psys (before collision) = psys (after collision) 2) Ek sys (before collision) = Ek sys (after collision) During an elastic collision, no energy is lost to the system in the form of heat, light, sound etc. Examples: Most collisions are not elastic, but intermolecular collisions are often elastic.

Collisions where total kinetic energy is not conserved During some collisions, energy is lost to the surroundings in the form of heat, light and sound. For these types of collisions, how should the kinetic energy of the system before the collision compare with the kinetic energy of the system after the collision?

Collisions where total kinetic energy is not conserved During some collisions, energy is lost to the surroundings in the form of heat, light and sound. For these types of collisions, the kinetic energy of the system before the collision is larger than the kinetic energy of the system after the collision.

Collisions where total kinetic energy is not conserved During some collisions, energy is lost to the surroundings in the form of heat, light and sound. For these types of collisions, the kinetic energy of the system before the collision is larger than the kinetic energy of the system after the collision. This type of collision is called an ________ collision.

Collisions where total kinetic energy is not conserved During some collisions, energy is lost to the surroundings in the form of heat, light and sound. For these types of collisions, the kinetic energy of the system before the collision is larger than the kinetic energy of the system after the collision. This type of collision is called an inelastic collision.

Collisions where total kinetic energy is not conserved During some collisions, energy is lost to the surroundings in the form of heat, light and sound. For these types of collisions, the kinetic energy of the system before the collision is larger than the kinetic energy of the system after the collision. This type of collision is called an inelastic collision. Inelastic collisions can be described using one equation and one inequality:

Collisions where total kinetic energy is not conserved During some collisions, energy is lost to the surroundings in the form of heat, light and sound. For these types of collisions, the kinetic energy of the system before the collision is larger than the kinetic energy of the system after the collision. This type of collision is called an inelastic collision. Inelastic collisions can be described using one equation and one inequality: 1) ?

Collisions where total kinetic energy is not conserved During some collisions, energy is lost to the surroundings in the form of heat, light and sound. For these types of collisions, the kinetic energy of the system before the collision is larger than the kinetic energy of the system after the collision. This type of collision is called an inelastic collision. Inelastic collisions can be described using one equation and one inequality: 1) Ek sys (before collision) > Ek sys (after collision)

Collisions where total kinetic energy is not conserved During some collisions, energy is lost to the surroundings in the form of heat, light and sound. For these types of collisions, the kinetic energy of the system before the collision is larger than the kinetic energy of the system after the collision. This type of collision is called an inelastic collision. Inelastic collisions can be described using one equation and one inequality: 1) Ek sys (before collision) > Ek sys (after collision) 2) psys (before collision) = psys (after collision)

A special inelastic collision

A special inelastic collision A collision in which the colliding bodies lock or couple together as one body after the collision is called a _______________ ___________ collision.

A special inelastic collision A collision in which the colliding bodies lock or couple together as one body after the collision is called a perfectly inelastic collision.

Head-on Elastic Collisions

Head-on Elastic Collisions m1 m2 the centers of mass of the colliding bodies are lined up in a head-on collision. V1 = + V1 stationary - +

Head-on Elastic Collisions the centers of mass of the colliding bodies are lined up in a head-on collision. (not glancing) m1 m2 V1 = + V1 stationary The masses rebound along the same line after the collision, so head-on collisions are 1-d collisions - +

Head-on Elastic Collisions the centers of mass of the colliding bodies are lined up in a head-on collision. (not glancing) m1 m2 V1 = + V1 stationary The masses rebound along the same line after the collision, so head-on collisions are 1-d collisions - + We will assume that Fnet system = 0, so what is conserved in this and any such collision?

Head-on Elastic Collisions the centers of mass of the colliding bodies are lined up in a head-on collision. (not glancing) m1 m2 V1 = + V1 stationary The masses rebound along the same line after the collision, so head-on collisions are 1-d collisions - + We will assume that Fnet system = 0, so what is conserved in this and any such collision? psys (before collision) = psys (after collision)

Head-on Elastic Collisions the centers of mass of the colliding bodies are lined up in a head-on collision. (not glancing) m1 m2 V1 = + V1 stationary The masses rebound along the same line after the collision, so head-on collisions are 1-d collisions - + We will assume that Fnet system = 0, so what is conserved in this and any such collision? psys (before collision) = psys (after collision) ? = ?

Head-on Elastic Collisions the centers of mass of the colliding bodies are lined up in a head-on collision. (not glancing) m1 m2 V1 = + V1 stationary The masses rebound along the same line after the collision, so head-on collisions are 1-d collisions - + We will assume that Fnet system = 0, so what is conserved in this and any such collision? psys (before collision) = psys (after collision) m1 V1 + 0 = ?

Head-on Elastic Collisions the centers of mass of the colliding bodies are lined up in a head-on collision. (not glancing) m1 m2 V1 = + V1 stationary The masses rebound along the same line after the collision, so head-on collisions are 1-d collisions - + We will assume that Fnet system = 0, so what is conserved in this and any such collision? psys (before collision) = psys (after collision) m1 V1 + 0 = m1 V1’ + m2 V2’

Head-on Elastic Collisions the centers of mass of the colliding bodies are lined up in a head-on collision. (not glancing) m1 m2 V1 = + V1 stationary The masses rebound along the same line after the collision, so head-on collisions are 1-d collisions - + We will assume that Fnet system = 0, so what is conserved in this and any such collision? psys (before collision) = psys (after collisiion) m1 V1 + 0 = m1 V1’ + m2 V2’ V1’ = velocity of m1 after V2’ = velocity of m2 after

Head-on Elastic Collisions the centers of mass of the colliding bodies are lined up in a head-on collision. (not glancing) m1 m2 V1 = + V1 stationary The masses rebound along the same line after the collision, so head-on collisions are 1-d collisions - + We will assume that Fnet system = 0, so what is conserved in this and any such collision? psys (before collision) = psys (after collisiion) m1 V1 + 0 = m1 V1’ + m2 V2’ V1’ = velocity of m1 after eq #1 m1 V1 = m1 V1’ + m2 V2’ V2’ = velocity of m2 after

Head-on Elastic Collisions the centers of mass of the colliding bodies are lined up in a head-on collision. (not glancing) m1 m2 V1 = + V1 stationary The masses rebound along the same line after the collision, so head-on collisions are 1-d collisions - + We will assume that Fnet system = 0, so what is conserved in this and any such collision? psys (before collision) = psys (after collisiion) m1 V1 + 0 = m1 V1’ + m2 V2’ V1’ = velocity of m1 after eq #1 m1 V1 = m1 V1’ + m2 V2’ V2’ = velocity of m2 after This collision is also elastic. What is conserved before and after any elastic collisions?

Head-on Elastic Collisions the centers of mass of the colliding bodies are lined up in a head-on collision. (not glancing) m1 m2 V1 = + V1 stationary The masses rebound along the same line after the collision, so head-on collisions are 1-d collisions - + We will assume that Fnet system = 0, so what is conserved in this and any such collision? psys (before collision) = psys (after collisiion) m1 V1 + 0 = m1 V1’ + m2 V2’ V1’ = velocity of m1 after eq #1 m1 V1 = m1 V1’ + m2 V2’ V2’ = velocity of m2 after This collision is also elastic. What is conserved before and after any elastic collisions? Ek sys (before collision) = Ek sys (after collision)

Head-on Elastic Collisions the centers of mass of the colliding bodies are lined up in a head-on collision. (not glancing) m1 m2 V1 = + V1 stationary The masses rebound along the same line after the collision, so head-on collisions are 1-d collisions - + We will assume that Fnet system = 0, so what is conserved in this and any such collision? psys (before collision) = psys (after collisiion) m1 V1 + 0 = m1 V1’ + m2 V2’ V1’ = velocity of m1 after eq #1 m1 V1 = m1 V1’ + m2 V2’ V2’ = velocity of m2 after This collision is also elastic. What is conserved before and after any elastic collisions? Ek sys (before collision) = Ek sys (after collision) ? = ?

Head-on Elastic Collisions the centers of mass of the colliding bodies are lined up in a head-on collision. (not glancing) m1 m2 V1 = + V1 stationary The masses rebound along the same line after the collision, so head-on collisions are 1-d collisions - + We will assume that Fnet system = 0, so what is conserved in this and any such collision? psys (before collision) = psys (after collisiion) m1 V1 + 0 = m1 V1’ + m2 V2’ V1’ = velocity of m1 after eq #1 m1 V1 = m1 V1’ + m2 V2’ V2’ = velocity of m2 after This collision is also elastic. What is conserved before and after any elastic collisions? Ek sys (before collision) = Ek sys (after collision) m1 V12/2 + 0 = ?

Head-on Elastic Collisions the centers of mass of the colliding bodies are lined up in a head-on collision. (not glancing) m1 m2 V1 = + V1 stationary The masses rebound along the same line after the collision, so head-on collisions are 1-d collisions - + We will assume that Fnet system = 0, so what is conserved in this and any such collision? psys (before collision) = psys (after collisiion) m1 V1 + 0 = m1 V1’ + m2 V2’ V1’ = velocity of m1 after eq #1 m1 V1 = m1 V1’ + m2 V2’ V2’ = velocity of m2 after This collision is also elastic. What is conserved before and after any elastic collisions? Ek sys (before collision) = Ek sys (after collision) m1 V12/2 + 0 = m1 V1’2/2 + m2 V2’2 /2

Head-on Elastic Collisions the centers of mass of the colliding bodies are lined up in a head-on collision. (not glancing) m1 m2 V1 = + V1 stationary The masses rebound along the same line after the collision, so head-on collisions are 1-d collisions - + We will assume that Fnet system = 0, so what is conserved in this and any such collision? psys (before collision) = psys (after collisiion) m1 V1 + 0 = m1 V1’ + m2 V2’ V1’ = velocity of m1 after eq #1 m1 V1 = m1 V1’ + m2 V2’ V2’ = velocity of m2 after This collision is also elastic. What is conserved before and after any elastic collisions? Ek sys (before collision) = Ek sys (after collision) m1 V12/2 + 0 = m1 V1’2/2 + m2 V2’2 /2 eq #2 m1 V12/2 = m1 V1’2/2 + m2 V2’2 /2

Head-on Elastic Collisions m1 m2 V1 = + V1 stationary eq #1 m1 V1 = m1 V1’ + m2 V2’ eq #2 m1 V12/2 = m1 V1’2/2 + m2 V2’2 /2 - +

Head-on Elastic Collisions Using simple algebra, we can solve these two equations for V1’ and V2’ in terms of m1 , m2 and V1 m1 m2 V1 = + V1 stationary eq #1 m1 V1 = m1 V1’ + m2 V2’ eq #2 m1 V12/2 = m1 V1’2/2 + m2 V2’2 /2 - +

Head-on Elastic Collisions Using simple algebra, we can solve these two equations for V1’ and V2’ in terms of m1 , m2 and V1 m1 m2 V1 = + V1 stationary eq #1 m1 V1 = m1 V1’ + m2 V2’ eq #2 m1 V12/2 = m1 V1’2/2 + m2 V2’2 /2 - + V1’ = (m1 - m2 ) V1 / (m1 + m2 )

Head-on Elastic Collisions Using simple algebra, we can solve these two equations for V1’ and V2’ in terms of m1 , m2 and V1 m1 m2 V1 = + V1 stationary eq #1 m1 V1 = m1 V1’ + m2 V2’ eq #2 m1 V12/2 = m1 V1’2/2 + m2 V2’2 /2 - + V1’ = (m1 - m2 ) V1 / (m1 + m2 ) V2’ = (2 m1 ) V1 / (m1 + m2 )

Head-on Elastic Collisions Using simple algebra, we can solve these two equations for V1’ and V2’ in terms of m1 , m2 and V1 m1 m2 V1 = + V1 stationary eq #1 m1 V1 = m1 V1’ + m2 V2’ eq #2 m1 V12/2 = m1 V1’2/2 + m2 V2’2 /2 - + V1’ = (m1 - m2 ) V1 / (m1 + m2 ) V2’ = (2 m1 ) V1 / (m1 + m2 ) These equations are know as the head-on elastic collision equations. Memorize and know where they come from.

Example: A 2. 0 kg ball moving at 3 Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball. What are the velocities of each ball after the collision? m1 = 2.0 kg m2 = 4.0 kg V1 = + 3.0 m/s stationary - +

Example: A 2. 0 kg ball moving at 3 Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball. What are the velocities of each ball after the collision? m1 = 2.0 kg m2 = 4.0 kg V1’ = (m1 - m2 ) V1 / (m1 + m2 ) = (2 – 4) (3) / ( 2 + 4) = -1.0 = 1.0 m/s [left] V1 = + 3.0 m/s stationary V2’ = (2 m1 ) V1 / (m1 + m2 ) = (2 X 2 ) (3) / (2 + 4 ) = +2.0 = 2.0 m/s [right] - +

Example: A 2. 0 kg ball moving at 3 Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball. What are the velocities of each ball after the collision? m1 = 2.0 kg m2 = 4.0 kg V1’ = (m1 - m2 ) V1 / (m1 + m2 ) = (2 – 4) (3) / ( 2 + 4) = -1.0 = 1.0 m/s [left] V1 = + 3.0 m/s stationary V2’ = (2 m1 ) V1 / (m1 + m2 ) = (2 X 2 ) (3) / (2 + 4 ) = +2.0 = 2.0 m/s [right] - + Now check that total momentum and kinetic energy of the system are conserved before and after the collision.

Example: A 2. 0 kg ball moving at 3 Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball. What are the velocities of each ball after the collision? m1 = 2.0 kg m2 = 4.0 kg V1’ = (m1 - m2 ) V1 / (m1 + m2 ) = (2 – 4) (3) / ( 2 + 4) = -1.0 = 1.0 m/s [left] V1 = + 3.0 m/s stationary V2’ = (2 m1 ) V1 / (m1 + m2 ) = (2 X 2 ) (3) / (2 + 4 ) = +2.0 = 2.0 m/s [right] - + Now check that total momentum and kinetic energy of the system are conserved before and after the collision. eq #1 m1 V1 = m1 V1’ + m2 V2’ (2)(3) = (2)(-1) + (4) (+2) 6 = -2 + 8 6 = 6 yes, p conserved eq #2 m1 V12/2 = m1 V1’2/2 + m2 V2’2 /2 (2)(3)2/2 = (2)(1)2/2 + (4)(2)2 /2 9 = 1 + 8 9 = 9 yes Ek conserved

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? m1 = 2.0 kg 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg - +

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? m1 = 2.0 kg Why can’t we use the head-on elastic collision equations to solve this problem? 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg - +

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? m1 = 2.0 kg We can’t use the head-on elastic collision equations to solve this problem because m2 should be at rest for the equations to be valid. 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg - +

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? m1 = 2.0 kg We can’t use the head-on elastic collision equations to solve this problem because m2 should be at rest for the equations to be valid. How can modify the system so that we can use the equations? 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg - +

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? m1 = 2.0 kg We can’t use the head-on elastic collision equations to solve this problem because m2 should be at rest for the equations to be valid. How can modify the system so that we can use the equations? 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg - + We can switch our “frame of reference”. Let’s imagine you are riding along with mass m2 .

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? m1 = 2.0 kg We can’t use the head-on elastic collision equations to solve this problem because m2 should be at rest for the equations to be valid. How can modify the system so that we can use the equations? 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg - + How fast are you moving with respect to m2 ‘s frame of reference? We can switch our “frame of reference”. Let’s imagine you are riding along with mass m2 .

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? m1 = 2.0 kg We can’t use the head-on elastic collision equations to solve this problem because m2 should be at rest for the equations to be valid. How can modify the system so that we can use the equations? 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg - + In m2 ‘s frame of reference, you and m2 are not moving. We can switch our “frame of reference”. Let’s imagine you are riding along with mass m2 .

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? m1 = 2.0 kg We can’t use the head-on elastic collision equations to solve this problem because m2 should be at rest for the equations to be valid. How can modify the system so that we can use the equations? 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg - + In m2 ‘s frame of reference, you and m2 are not moving. We can switch our “frame of reference”. Let’s imagine you are riding along with mass m2 . So in m2 ‘s frame of reference: V2 = ?

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? m1 = 2.0 kg We can’t use the head-on elastic collision equations to solve this problem because m2 should be at rest for the equations to be valid. How can modify the system so that we can use the equations? 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg - + In m2 ‘s frame of reference, you and m2 are not moving. We can switch our “frame of reference”. Let’s imagine you are riding along with mass m2 . So in m2 ‘s frame of reference: V2 = 0 m/s

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? m1 = 2.0 kg We can’t use the head-on elastic collision equations to solve this problem because m2 should be at rest for the equations to be valid. How can modify the system so that we can use the equations? 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg - + In m2 ‘s frame of reference, you and m2 are not moving. We can switch our “frame of reference”. Let’s imagine you are riding along with mass m2 . So in m2 ‘s frame of reference: V2 = 0 m/s V1 = ?

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? m1 = 2.0 kg We can’t use the head-on elastic collision equations to solve this problem because m2 should be at rest for the equations to be valid. How can modify the system so that we can use the equations? 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg - + In m2 ‘s frame of reference, you and m2 are not moving. We can switch our “frame of reference”. Let’s imagine you are riding along with mass m2 . So in m2 ‘s frame of reference: V2 = 0 m/s V1 = 3 + 1 = 4.0 m/s [right] (head-on collision)

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? So in m2 ‘s frame of reference: V2 = 0 m/s V1 = 3 + 1 = 4.0 m/s [right] (head-on collision) m1 = 2.0 kg 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg Now we can use the head-on elastic collision equations because m2 is at rest. - +

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? So in m2 ‘s frame of reference: V2 = 0 m/s V1 = 3 + 1 = 4.0 m/s [right] (head-on collision) m1 = 2.0 kg 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg Now we can use the head-on elastic collision equations because m2 is at rest. - + V1’ = (m1 - m2 ) V1 / (m1 + m2 ) = ?

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? So in m2 ‘s frame of reference: V2 = 0 m/s V1 = 3 + 1 = 4.0 m/s [right] (head-on collision) m1 = 2.0 kg 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg Now we can use the head-on elastic collision equations because m2 is at rest. - + V1’ = (m1 - m2 ) V1 / (m1 + m2 ) = (2 – 4) (4) / (2 + 4) = ?

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? So in m2 ‘s frame of reference: V2 = 0 m/s V1 = 3 + 1 = 4.0 m/s [right] (head-on collision) m1 = 2.0 kg 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg Now we can use the head-on elastic collision equations because m2 is at rest. - + V1’ = (m1 - m2 ) V1 / (m1 + m2 ) = (2 – 4) (4) / (2 + 4) = - 1.3 m/s or 1.3 m/s [left]

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? So in m2 ‘s frame of reference: V2 = 0 m/s V1 = 3 + 1 = 4.0 m/s [right] (head-on collision) m1 = 2.0 kg 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg Now we can use the head-on elastic collision equations because m2 is at rest. - + V1’ = (m1 - m2 ) V1 / (m1 + m2 ) = (2 – 4) (4) / (2 + 4) = - 1.3 m/s or 1.3 m/s [left] But this velocity is with respect to m2. How can we find V1’ relative to our original ground frame of reference? V1’ (w.r.t. ground) = ?

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? So in m2 ‘s frame of reference: V2 = 0 m/s V1 = 3 + 1 = 4.0 m/s [right] (head-on collision) m1 = 2.0 kg 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg Now we can use the head-on elastic collision equations because m2 is at rest. - + V1’ = (m1 - m2 ) V1 / (m1 + m2 ) = (2 – 4) (4) / (2 + 4) = - 1.3 m/s or 1.3 m/s [left] But this velocity is with respect to m2. How can we find V1’ relative to our original ground frame of reference? V1’ (w.r.t. ground) = -1.3 -1.0 = ?

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? So in m2 ‘s frame of reference: V2 = 0 m/s V1 = 3 + 1 = 4.0 m/s [right] (head-on collision) m1 = 2.0 kg 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg Now we can use the head-on elastic collision equations because m2 is at rest. - + V1’ = (m1 - m2 ) V1 / (m1 + m2 ) = (2 – 4) (4) / (2 + 4) = - 1.3 m/s or 1.3 m/s [left] But this velocity is with respect to m2. How can we find V1’ relative to our original ground frame of reference? V1’ (w.r.t. ground) = -1.3 -1.0 = -2.3 or ?

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? So in m2 ‘s frame of reference: V2 = 0 m/s V1 = 3 + 1 = 4.0 m/s [right] (head-on collision) m1 = 2.0 kg 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg Now we can use the head-on elastic collision equations because m2 is at rest. - + V1’ = (m1 - m2 ) V1 / (m1 + m2 ) = (2 – 4) (4) / (2 + 4) = - 1.3 m/s or 1.3 m/s [left] But this velocity is with respect to m2. How can we find V1’ relative to our original ground frame of reference? V1’ (w.r.t. ground) = -1.3 -1.0 = -2.3 or 2.3 m/s [left]

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? So in m2 ‘s frame of reference: V2 = 0 m/s V1 = 3 + 1 = 4.0 m/s [right] (head-on collision) m1 = 2.0 kg 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg Now we can use the head-on elastic collision equations because m2 is at rest. - + V1’ = (m1 - m2 ) V1 / (m1 + m2 ) = (2 – 4) (4) / (2 + 4) = - 1.3 m/s or 1.3 m/s [left] But this velocity is with respect to m2. How can we find V1’ relative to our original ground frame of reference? V1’ (w.r.t. ground) = -1.3 -1.0 = -2.3 or 2.3 m/s [left] V2’ = (2 m1 ) V1 / (m1 + m2 ) = ?

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? So in m2 ‘s frame of reference: V2 = 0 m/s V1 = 3 + 1 = 4.0 m/s [right] (head-on collision) m1 = 2.0 kg 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg Now we can use the head-on elastic collision equations because m2 is at rest. - + V1’ = (m1 - m2 ) V1 / (m1 + m2 ) = (2 – 4) (4) / (2 + 4) = - 1.3 m/s or 1.3 m/s [left] But this velocity is with respect to m2. How can we find V1’ relative to our original ground frame of reference? V1’ (w.r.t. ground) = -1.3 -1.0 = -2.3 or 2.3 m/s [left] V2’ = (2 m1 ) V1 / (m1 + m2 ) = ( 2 X 2 ) ( 4 ) / ( 2 + 4) = ?

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? So in m2 ‘s frame of reference: V2 = 0 m/s V1 = 3 + 1 = 4.0 m/s [right] (head-on collision) m1 = 2.0 kg 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg Now we can use the head-on elastic collision equations because m2 is at rest. - + V1’ = (m1 - m2 ) V1 / (m1 + m2 ) = (2 – 4) (4) / (2 + 4) = - 1.3 m/s or 1.3 m/s [left] But this velocity is with respect to m2. How can we find V1’ relative to our original ground frame of reference? V1’ (w.r.t. ground) = -1.3 -1.0 = -2.3 or 2.3 m/s [left] V2’ = (2 m1 ) V1 / (m1 + m2 ) = ( 2 X 2 ) ( 4 ) / ( 2 + 4) = 2.7 m/s or 2.7 m/s [right]

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? So in m2 ‘s frame of reference: V2 = 0 m/s V1 = 3 + 1 = 4.0 m/s [right] (head-on collision) m1 = 2.0 kg 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg Now we can use the head-on elastic collision equations because m2 is at rest. - + V1’ = (m1 - m2 ) V1 / (m1 + m2 ) = (2 – 4) (4) / (2 + 4) = - 1.3 m/s or 1.3 m/s [left] But this velocity is with respect to m2. How can we find V1’ relative to our original ground frame of reference? V1’ (w.r.t. ground) = -1.3 -1.0 = -2.3 or 2.3 m/s [left] V2’ = (2 m1 ) V1 / (m1 + m2 ) = ( 2 X 2 ) ( 4 ) / ( 2 + 4) = 2.7 m/s or 2.7 m/s [right] But this velocity is with respect to m2. How can we find V2’ relative to our original ground frame of reference? V2’ (w.r.t. ground) = ?

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? So in m2 ‘s frame of reference: V2 = 0 m/s V1 = 3 + 1 = 4.0 m/s [right] (head-on collision) m1 = 2.0 kg 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg Now we can use the head-on elastic collision equations because m2 is at rest. - + V1’ = (m1 - m2 ) V1 / (m1 + m2 ) = (2 – 4) (4) / (2 + 4) = - 1.3 m/s or 1.3 m/s [left] But this velocity is with respect to m2. How can we find V1’ relative to our original ground frame of reference? V1’ (w.r.t. ground) = -1.3 -1.0 = -2.3 or 2.3 m/s [left] V2’ = (2 m1 ) V1 / (m1 + m2 ) = ( 2 X 2 ) ( 4 ) / ( 2 + 4) = 2.7 m/s or 2.7 m/s [right] But this velocity is with respect to m2. How can we find V2’ relative to our original ground frame of reference? V2’ (w.r.t. ground) = 2.7 -1.0 = ?

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? So in m2 ‘s frame of reference: V2 = 0 m/s V1 = 3 + 1 = 4.0 m/s [right] (head-on collision) m1 = 2.0 kg 1.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg Now we can use the head-on elastic collision equations because m2 is at rest. - + V1’ = (m1 - m2 ) V1 / (m1 + m2 ) = (2 – 4) (4) / (2 + 4) = - 1.3 m/s or 1.3 m/s [left] But this velocity is with respect to m2. How can we find V1’ relative to our original ground frame of reference? V1’ (w.r.t. ground) = -1.3 -1.0 = -2.3 or 2.3 m/s [left] V2’ = (2 m1 ) V1 / (m1 + m2 ) = ( 2 X 2 ) ( 4 ) / ( 2 + 4) = 2.7 m/s or 2.7 m/s [right] But this velocity is with respect to m2. How can we find V2’ relative to our original ground frame of reference? V2’ (w.r.t. ground) = 2.7 -1.0 = 1.7 or 1.7 m/s [right]

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? V1’ (w.r.t. ground) = 2.3 m/s [left] or 7/3 m/s [left] m1 = 2.0 kg 1.0 m/s V2’ (w.r.t. ground) = 1.7 m/s [right] Or 5/3 m/s [right] V1 = + 3.0 m/s m2 = 4.0 kg Now check that total momentum and kinetic energy of the system are conserved before and after the collision. - +

Harder Example: A 2. 0 kg ball moving at 3 Harder Example: A 2.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 1.0 m/s. What are the velocities of each ball after the collision? V1’ (w.r.t. ground) = 2.3 m/s [left] or 7/3 m/s [left] m1 = 2.0 kg 1.0 m/s V2’ (w.r.t. ground) = 1.7 m/s [right] Or 5/3 m/s [right] V1 = + 3.0 m/s m2 = 4.0 kg Now check that total momentum and kinetic energy of the system are conserved before and after the collision. - + eq #2 m1 V12/2 + m2 V22/2 = m1 V1’2/2 + m2 V2’2 /2 (2) (3)2/2 + (4)(1)2/2 = (2) (7/3)2/2 + (4) (5/3)2 /2 9 + 2 = 49/9 + 50/9 11 = 99/9 11 = 11 Yes kinetic energy is conserved eq #1 m1 V1 + m2 V2 = m1 V1’ + m2 V2’ (2) (3) + (4) (-1) = (2) (-7/3) + 4 ( 5/3) 6 - 4 = - 14 /3 + 20 / 3 2 = 6/3 2 = 2 Yes momentum is conserved

Try this: A 5. 0 kg ball moving at 3 Try this: A 5.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 8.0 m/s. What are the velocities of each ball after the collision? Leave answers as fractions. Check that total momentum and kinetic energy of the system are conserved before and after the collision. m1 = 5.0 kg 8.0 m/s V1 = + 3.0 m/s m2 = 4.0 kg - +

+ So in m2 ‘s frame of reference: V2 = 0 m/s Try this: A 5.0 kg ball moving at 3.0 m/s [right] makes a head-on elastic collision with a stationary 4.0 kg ball moving to the left at 8.0 m/s. What are the velocities of each ball after the collision? Leave answers as fractions. Check that total momentum and kinetic energy of the system are conserved before and after the collision. + So in m2 ‘s frame of reference: V2 = 0 m/s V1 = 3 + 8 = 11 m/s [right] (head-on collision) m1 = 5.0 kg 8.0 m/s V1 = + 3.0 m/s V1’ = (m1 - m2 ) V1 / (m1 + m2 ) = (5 – 4) (11) / (5 + 4) = 11/9 m/s [right] m2 = 4.0 kg eq #1 m1 V1 + m2 V2 = m1 V1’ + m2 V2’ (5) (3) + (4) (-8) = (5) (-61/9) + 4 (38/9) 15 -32 = -305/9 +152/9 -17 = -17 Momentum is conserved V1’ (w.r.t. ground) = 11/9 -8= -61/9 m/s = 61/9 m/s [left] V2’ = (2 m1 ) V1 / (m1 + m2 ) = ( 2 X 5 ) ( 11 ) / ( 5 + 4) = 110/9 m/s [right] V2’ (w.r.t. ground) = 110/9 -8 = 38/9 m/s [right] eq #2 m1 V12/2 + m2 V22/2 = m1 V1’2/2 + m2 V2’2 /2 (5) (3)2/2 + (4)(8)2/2 = (5) (61/9)2/2 + (4) (38/9)2 /2 22.5 + 128 = 114.845679 + 35.65432 150.5 = 150.5 Yes kinetic energy is conserved

Homework Do head-on elastic collisions handout. New textbook: Read through p240 –p244 Be sure to explain the special cases p243 The book has a more complex formula. Why do both formulas have validity for head-on elastic collision formulas? Try Q1,2 p243 q3 p248