One Dimensional Elastic Collision.

W
2 m/s 1 m/s
4 kg 1 kg
A
spring
B
X
The two boxes are initially moving in the positive
X direction. The ideal spring is connected to A.
The collision between the boxes is assumed to be
elastic since the spring is ideal (no kinetic
converted to internal energy in spring).
(a) Find the speed of each box after the collision.
Since the collision is elastic the total momentum before and after the collision
are the same and the total kinetic energy before and after are the same. Let v A
and v B be the x components of the boxes after collision.
Momentum
Kinetic
:
Energy
4
( 2) + 1( 1)
:
1
2
= 4v
A
1
2
+ 1v
B
⇒
1
2
= 9 − 4v
1
2
( 4)( 2
2 ) + ()( 1 1
2 ) = ( 4) v
2 + () 1 v
2
When v B is substituted into the second equation we obtain:
5v 2 −18v
+ 16=
0 ⇒ v = 2 and 1. 6
A
A
A
v
B
A
m
( )
s
A
B
The first equation gives the corresponding values: v B = 1 and 2.6 (m/s). The
solution pair (2.0 and 1.0) is impossible since it would mean that "A" had passed
through "B". The correct solution is v A = 1.60 m/s and v B = 2.60 (m/s).
(b) Find the speed of the boxes when the spring has maximum compression.
When the spring has its maximum compression box "A" must be at rest relative to
the spring and therefore to "B". Let v be the common speed of box boxes at this
instant. Momentum conservation gives:
( 2) + 11 ( ) = 4v + 1v
⇒ v 1. 80( m
)
4 =
(c) Find the maximum compression of the spring if k = 100 (N/m).
We apply the Energy Principle with "i" being just before the collision and "f"
at maximum spring compression:
1
2 1 2 1 2 1 2
( K − K ) + ( U −U
) = 0 ⇒ ( 4+
11.8 ) − ( 4) 2 − () 11 + ( 100) x −0
f
i
S f
Si
2
Solving we obtain: x f = .0894 (m).
2
s
2
2
f
1

2 m/s 1 m/s
4 1
initial
x
final
(d) Find the speed of both boxes when the spring
compression is “x” (0 < x < .0894 m).
v A
and v B
are the X components of the velocities when spring is compressed by “x”.
4 2 11 = 4v + 1v
⇒ v = 9 − 4v
+ 1
Momentum conservation: ( ) ( )
A B B
A
Energy Conservation:
ΔK
1
2
A
+ ΔK
1
2
+ ΔU
= 0
1
2
2
2
2
2
2
( 4) v − ( 4) 2 + () 1 v − () 11 + ( 100) x − 0 = 0
A
B
s
B
1
2
1
2
2
Substitution of v B
from (1) into (2) gives:
5v
v
A
2
A
−18v
18 ±
=
A
+
2
( 16 + 25x
)
4 − 500x
10
2
= 0
(
m
s )
While the spring is being compressed the
speed of “A” is:
v
A
18 +
=
4 − 500x
10
2
(
m
s )
1
While the spring is returning to its
uncompressed length the speed of “A” is:
v
A
18 −
=
4 − 500x
10
2
(
m
s )
2
Check of our solutions:
As spring starts to compress we use (1) and x = 0: v A
= 2.00 (m/s) ;v B
= 9-4v A
=1.00 (m/s).
Just after spring returns to x = 0 we use (2): v A
= 1.6 (m/s) ; v B
= 9-4v A
= 2.60 (m/s).
When x = .0894 (m) we can use either (1) or (2): v A
= 1.80 (m/s) and v B
= (9-4v A
)= 1.80 (m/s)
2

Speeds calculated during the collision using:
v
A
18 ±
=
4 − 500x
10
2
(
m
s )
2.8
2.6
after
speed
2.4
2.2
2
1.8
before
A
1.6
1.4
1.2
1
after
B
before
0 0.02 0.04 0.06 0.08 0.1
spring compression
9
8
7
total energy
kinetic energy
6
energy
5
4
3
2
1
spring potential energy
0
0 0.02 0.04 0.06 0.08 0.1
spring compression
3

5 m/s
2kg
5 m/s
2 kg
A
spring
B
We let “x” be the amount spring is compressed. For an elastic collision the total
momentum and total energy ( kinetic + spring potential energy) are conserved
momentum (kg.m/s)
10
8
p a
6
4
2
0
0
-2
0.02 0.04 0.06 0.08 0.1
-4
-6
-8
-10
Energies (J)
50
45
40
35
30
p total 25
20
15
10
U s
5
p b
0
0 0.02 0.04 0.06 0.08 0.1
X (m)
X (m)
K
E total
4