Question

A .06 kg tennis ball, moving at 2.50 m/s collides with a .09 kg ball moving away from it at 1.15 m/s. Assuming a perfectly elastic collision, what are the speeds and directions of the balls after the collision?

Please explain each step or why you do what you do!

Answer 1

m1=.06kg
m2=.09kg
u1=2.5m/s
u2=1.15m/s
v=velocity after collision

m1u1+m2u2=m1v1+m2v2
(.06*2.5)+(.09*1.15)=(.06)v1+(.09)v2
.2535=(.06)v1+(.09)v2

In a perfectly elastic, one-dimensional collision, u1-u2=v2-v1
(2.5-1.15)=v2-v1
1.35=v2-v1

~~~ v2=v1+1.35 ~~~

Plug this into .2535=(.06)v1+(.09)v2

.2535=(.06)v1+(.09)(v1+1.35)
.2535=(.06)v1+(.09)v1+.1215
.132=.15v1
v1=.88m/s in the original direction

Plug that into v2=v1+1.35

v2=.88+1.35
v2= 2.23m/s in the original direction