A 170g cue ball is moving at 3.0 m/s when it strikes a 160g stationary numbered ball (object ball). After the collision the balls move off as shown. Determine the speed of the cue ball and the numbered (object) ball after the collision.

Question

The diagram illustrates the cue ball moving 60’ N of E. The object ball after the collision is 30' s of E.

Answer 1

you have several equations you can use.

Momenum in the original direction
Momentum in the 90 degrees direction
Conservation of energy.

Those three equations will do it.

Answer 2

hI bobpursley,
tHANK YOu for your prompt response. Is it possible for you to show the calculations as I've tried to answer it but to no prevail.

X 0.5

Answer 3

x: 0.51 kg m/s + 0= 0.17kg Vsin60' +0.16kg Vsin30'

y: 0+0= 0.17kgVcos60' +0.16kg cos30'
-0.17kgVcos60'=0.16kgcos30'

Answer 4

*but to no avail.

Answer 5

Ok, you have the momentum equations but you have failed to give each of the balls different speeds.

x:0.51 kg m/s + 0= 0.17kg Vc*sin60' +0.16kg Vo*sin30'

y: -0.17kgVc*cos60'=0.16kg*Vo*cos30'

Well, it is too easy, two equations, two unknowns. You need a teacher who made you calculate one of the angles.

Put the equations in this form...

a*Vo + b*Vc= c
d*Vo + e*Vc= d

then solve it directly ( your calculator can do it for you).

Answer 6

Hi,
Apologies again. But what would be the values of a, b, d and e, as well as the Vo and Vc. Our teacher had provided us with an example of one unknown value- so, we haven't been exposed to that formulae and so, am unsure as to how to use it.