1
Spring 2014
Instructor: Steven Detweiler
Assume that we have two masses M
1 and M
2 moving with initial speeds v
1 i and v
2 i in the same direction—if the objects are actually moving towards each other then just assume that v
2
, for example, is negative.
The conservation of momentum implies that
M
1 v
1 i
+ M
2 v
2 i
= M
1 v
1 f
+ M
2 v
2 f where v
1 f and v
2 f are the speeds of the masses after the collision.
The conservation of kinetic energy implies that
1
2
M
1 v
2
1 i
+
1
2
M
2 v
2
2 i
=
1
2
M
1 v
2
1 f
+
1
2
M
2 v
2
2 f
.
First we change these equations by putting the all of the M
1 terms on the left-hand side and the M
2 terms on the right-hand side to obtain
M
1
( v
1 i
M
1
( v
2
1 i
− v
1 f
) = M
2
( v
2 f
− v
2
1 i
) = M
2
( v
2
2 f
− v
1 f
)
− v
2
2 i
) .
(1)
(2)
Now, we divide the first of these equations by the second so that the masses cancel each other out, and we have v 2
1 i
− v 2
1 f v
1 i
− v
1 f
= v 2
2 i v
2 i
− v 2
2 f
− v
2 f
.
But, we know that a 2 − b 2 = ( a − b )( a + b ), so our equation above is equivalent to
(3) v
1 i
+ v
1 f
= v
2 f
+ v
2 i
.
(4)
At this point the algebra gets a bit challenging. But, we choose any two of the three equations (1), (2) or (4) to solve for the two unknowns,
For example first use v
1 f and v
2 f
.
v
2 f
= v
1 i
+ v
1 f
− v
2 i
, (5) from equation (4), and use this equation to substitute for v
2 f solve for v
1 f into equation (1), and then v
1 f
=
M
1
M
1
− M
2
+ M
2 v
1 i
+
M
1
2 M
2
+ M
2 v
2 i
.
By switching the subscripts 1 and 2, this becomes
(6) v
2 f
=
M
2
M
1
− M
1
+ M
2 v
2 i
+
M
1
2 M
1
+ M
2 v
1 i
.
(7)
All of the above equations, become much easier to interpret, if the situation that you have has one of the masses initially at rest.