CHAPTER 6. MOTIONIN TWO DIMENSIONS 6.4
pbefore = pafter
m 1 vi 1 + m 2 vi 2 = m 1 vf 1 + m 2 vf 2
�
150
1000
�
(2) +
�
150
1000
�
(− 1 ,5) =
�
150
1000
�
(− 1 ,5) +
�
150
1000
�
(vf 2 )
0 , 3 − 0 ,225 =− 0 ,225 + 0, 15 vf 2
vf 2 = 3 m· s−^1
So after the collision, ball 2 moves with a velocity of 3 m· s−^1.
Step 4 : Elastic collisions
The fact that characterises an elastic collisionis that the total
kinetic energy of the particles before the collision is the same as
the total kinetic energyof the particles after thecollision. This
means that if we can show that the initial kinetic energy is equal
to the final kinetic energy, we have shown that the collision is
elastic.
Step 5 : Calculating the initial total kinetic energy
EKbefore =
1
2
m 1 v^2 i 1 +
1
2
m 2 v^2 i 2
=
�
1
2
�
(0,15)(2)^2 +
�
1
2
�
(0,15)(− 1 ,5)^2
= 0. 469 ....J
Step 6 : Calculating the final total kinetic energy
EKafter =
1
2
m 1 v^2 f 1 +
1
2
m 2 v^2 f 2
=
�
1
2
�
(0,15)(− 1 ,5)^2 +
�
1
2
�
(0,15)(2)^2
= 0. 469 ....J
So EKbefore= EKafterand hence the collisionis elastic.
