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

Paul Robinson

Initial Conditions

Block 1, of mass m1, moves across a frictionless surface with speed v1i. It collides elastically with block 2, of mass m2, which is at rest. After the collision, block 1 moves with speed v1f, while block 2 moves with speed v2f. What are v1f and v2f?

Part 1: Isolate v2f

Using the conservation of momentum, we isolate v2f in terms of the other variables.

( )

1 1 1 1 2 2

2 2 1 1 1 1

12 1 1

2

i f f

f f i

f f i

m v m v m v

m v m v m v

mv v v

m

= +

= −

= −

Part 2: Solve for v1f

Take our v2f expression, and plug it into the conservation of kinetic energy.

( )

2 2 21 1 1 1 2 2

222 2 1

1 1 1 1 2 1 122

i f f

i f f i

m v m v m v

mm v m v m v v

m

= +

⎡ ⎤= + −⎢ ⎥

⎣ ⎦

Part 2 cont…

Now, take the updated KE equation and solve for v1f in terms of the given constants. First we cancel out, then group the v1

2 on one side, and then factor and cancel out again.

( )2

22 2 11 1 1 1 1 1 22

2

22 2 1

1 1 1 1 22

i f f i

i f

mm v m v v v m

m

mm v m v

m

= + −

= + ( )21 1 2f iv v m−

1m ( )2

2 2 11 1i f

mv v− = ( )

( )

2

1 12

1 1

f i

i f

v vm

v v

−

− ( ) ( ) 211 1 1 1

2i f f i

mv v v v

m+ = −

( )

1 2

11

1 11 2

1 1 12

i f f i

f i

m

m

mv

v

vm

v v vm

m

−

+ = −

=+

M

Solve for v2f

Take the v1f

expression and plug into the conservation of momentum equation. Then, simply solve for v2f

1 1 1 1 2 2

1 21 1 1 1 2 2

1 2

1 22 2 1 1 1 1

1 2

1 2

12 1

1

2

2

2 1 1 11 2

2

i f f

f

i i f

f i

i

i

f i

m v m v m v

m mm v m v m v

m m

m mm v m

mv v

v m vm m

m mm v v

m

m

m

m

m m

= +

⎛ ⎞−= +⎜ ⎟+⎝ ⎠

⎛ ⎞−= − ⎜ ⎟+⎝ ⎠

⎛ ⎞⎛ ⎞−= −⎜ ⎟⎜ ⎟⎜ ⎟+⎝

⎛ ⎞=⎜ ⎟+⎝ ⎠

⎠⎝ ⎠M