Elastic Collision of Two Bodies in One Dimension: The Generalized Case Paul Robinson

Initial Conditions Block 1, of mass m 1 , moves across a frictionless surface with speed v 1i . It collides elastically with block 2, of mass m 2 , which is at rest. After the collision, block 1 moves with speed v 1f , while block 2 moves with speed v 2f . What are v 1f and v 2f ?

Part 1: Isolate v 2f Using the conservation of momentum, we isolate v 2f variables.

in terms of the other

m v

1 1

i m v

2 2

f



m v

1 1

f



m v

1 1

f



m v

2 2

f



m v

1 1

i v

2

f



m

1

m

2 

v

1

f



v

1

i



Part 2: Solve for v 1f Take our v into the conservation of kinetic energy.

2f expression, and plug it

m v

1 1

i

2 

m v

1 1

f

2 

m v

2 2

f

2

m v

1 1

i

2 

m v

1 1

f

2 

m

2  

m

1

m

2 2 2 

v

1

f



v

1

i

 2  

Part 2 cont… Now, take the updated KE equation and solve for v group the v 1 2 1f in terms of the given constants. First we cancel out, then on one side, and then factor and cancel out again.

m v

1 1

i

2 

m v

1 1

f

2 

m

1 2

m

2 2 

v

1

f



v

1

i

 2

m

2

m v

1 1

i

2 

m v

1 1

f

2 

m

1 2

m

2 2

v

1

f



v

1

i m

1

m

1  

m

2

m

2 

v

1

f



v

1

i

 2

m

2

m

1 

v

1

i

2 

v

1

f

2  

m

1 2

m

2 

v

1

i



v

1

f

 

v

1

i



v

1

f

  

v

1

f m

1

m

2 

v

1

i



v

1

f

 2 

v

1

i



v

1

i



v

1

f



m

1

m

2 

v

1

f



v

1

i

 2

Solve for v 2f Take the v 1f expression and plug into the conservation of momentum equation. Then, simply solve for v 2f

m v

1 1

i



m v

1 1

f



m v

2 2

f m v

1 1

i m v

2 2

f f



m

1  

m m

1 1  

m

2

m

2  

v

1

i



m v

2 2

f



m v i



m

1  

m m

1 1 

m

2 

m

2 

v

1

i

 

m

1 

m

1  

m

1

m

1 

m

2 

m

2  

v

1

i

   

v

2

f

   

m

1 2

m

1 

m

2  

v

1

i