CHAPTER 6. MOTIONIN TWO DIMENSIONS 6.4
Momentum is conserved. Therefore:
pi = pf
pi 1 + pi 2 = pf 1 + pf 2
m 1 vi 1 + m 2 vi 2 = m 1 vf 1 + m 2 vf 2
(0,05)(0) + (0,1)(3) = (0,05)vf 1 + (0,1)vf 2
0 ,3 = 0, 05 vf 1 + 0, 1 vf 2 (6.5)
Energy is also conserved. Therefore:
KEi = KEf
KEi 1 + KEi 2 = KEf 1 + KEf 2
1
2
m 1 vi^21 +
1
2
m 2 v^2 i 2 =
1
2
m 1 vf^21 +
1
2
m 2 v^2 f 2
(
1
2
)(0,05)(0)^2 + (
1
2
)(0,1)(3)^2 =
1
2
(0,05)(vf 1 )^2 + (
1
2
)(0,1)(vf 2 )^2
0 ,45 = 0, 025 v^2 f 1 + 0, 05 v^2 f 2 (6.6)
Substitute Equation 6.5into Equation 6.6 and solve for vf 2.
m 2 v^2 i 2 = m 1 v^2 f 1 + m 2 vf^22
= m 1
�
m 2
m 1
(vi 2 − vf 2 )
� 2
+ m 2 v^2 f 2
= m 1
m^22
m^21
(vi 2 − vf 2 )^2 + m 2 v^2 f 2
=
m^22
m 1
(vi 2 − vf 2 )^2 + m 2 v^2 f 2
v^2 i 2 =
m 2
m 1
(vi 2 − vf 2 )^2 + v^2 f 2
=
m 2
m 1
�
vi^22 − 2 · vi 2 · vf 2 + v^2 f 2
�
+ v^2 f 2
0 =
�
m 2
m 1
− 1
�
v^2 i 2 − 2
m 2
m 1
vi 2 · vf 2 +
�
m 2
m 1
+ 1
�
v^2 f 2
=
�
0. 1
0. 05
− 1
�
(3)^2 − 2
0. 1
0. 05
(3)· vf 2 +
�
0. 1
0. 05
+ 1
�
v^2 f 2
= (2− 1)(3)^2 − 2 · 2(3)· vf 2 + (2 + 1)v^2 f 2
= 9− 12 vf 2 + 3v^2 f 2
= 3− 4 vf 2 + v^2 f 2
= (vf 2 − 3)(vf 2 − 1)
Substituting back into Equation 6.5, we get:
vf 1 =
m 2
m 1
(vi 2 − vf 2 )
=
0. 1
0. 05
(3− 3)
= 0 m· s−^1
or
vf 1 =
m 2
m 1
(vi 2 − vf 2 )
=
0. 1
0. 05
(3− 1)
= 4 m· s−^1
