Peoples Physics Book Version-3

http://www.ck12.org Chapter 7. Momentum Conservation

Answer: We know the equation for conservation of momentum, along with the masses of the objects in question as
well two of the three velocities. Therefore all we need to do is manipulate the conservation of momentum equation
so that it is solved for the velocity of the cue ball after the collision and then plug in the known values to get the
velocity of the cue ball.

mcvic+mevie=mcvf c+mevf e

vf c=

mcvic+mevie−mevf e

mc

=

.17kg× 2 .0m/s+.17kg×0m/s−.17kg× 1 .2m/s

.17kg

=.80m/s

Now we want to find the direction of the cue ball. To do this we will use the diagram below.

We know that the momentum in theydirection of the two balls is equal. Therefore we can say that the velocity in
theydirection is also equal because the masses of the two balls are equal.

mcvcy=mevey→vcy=vey

Given this and the diagram, we can find the direction of the cue ball. After 1 second, the 8 ball will have traveled
1 .2m. Therefore we can find the distance it has traveled in theydirection.

sin 25o=

opposite

hypotenuse

=

x

1 .2m

→x=sin 25× 1 .2m=.51m

Therefore, in one second the cue ball will have traveled.51m in theydirection as well. We also know how far in
total the cue ball travels in one second(.80m). Thus we can find the direction of the cue ball.

sin−^1

opposite

hypotenuse

=sin−^1

.51m

.80m

= 40 o