Two Dimensional, Two Object Collision?

Question

Hey everyone, I am really stumped on this question for Grade 12 Physics. Just looking for a formula or a process which I can use to find the answer, thanks :)

A 0.4kg cue ball hits a stationary 0.30kg billiard ball so the cue ball deflects with a speed of 1.2 m/s at an angle of 30 degree from its original path. Calculate the original speed of the cue ball if the billiard ball ends up traveling at 1.5 m/s.

I have...
m1 = 0.4kg
m2 = 0.3kg

v'1 = 1.2 m/s
v'2 = 1.5 m/s

v1 = ?
v2 = 0 m/s

Thanks for help

Answer 1

Because the cue ball initially travels horizonally, the final horizonal momentum of the cue ball and the final horizonal momentum of the stationary ball will add up the the intial momentum of the cue ball.

find the final momentum of the cue ball
p = mv
p = (.4)(1.2)
p = .48 kgm/s

cos(30) .48 = .416 kgm/s
sin(30).48 = .24 kgm/s

find mometum of the stationary ball
p = mv
p= (.3)(1.5)
p=.45kgm/s

again, because the cue ball intially travels horizonaly, the final vertical momentum of the cue ball and the final vertical momentum of the stationary ball will be equal.

find horizonal momentum of the stationary ball.
a^2 + b^2 = c^2
.24^2 + b^2 = .45^2
b = .38 kgm/s

.38 + .416 = .796

p = mv
.796 = .4 v
v = 1.99 m/s or round it to 2m/s

more info: the stationary ball will travel at the angle of 32.23 degrees below the horizonal

Answer 2

Use conservation of momentum in two dimensions

in direction of motion:
(mass cue) (v cue initial) = (mass cue) x (vcue final) x cos (30) + (mass ball) x (vballfinal) x cos (? angl e)

in the transverse direction:

0 = (mass cue) x (vcue) x sin (30) + (mass ball) x (vballfinal) x sin (? angle)

You have 2 unknowns, the original speed and the angle at which the billiard ball rolled off. You have 2 equations. Do some algebra to solve for them.

Edit--the two answers above and the one two below are wrong. The problem is solvable as given. The one immediately below looks about right (tho I didn't check numbers).

Answer 3

I will call the cue c with mass mc
the billiard is b with mass mb

v1c=?
v2c=1.2 at 30

v1b=0
v2b=1.5 direction unknown


I will set the x axis as the original path of the cue ball.

Step one, determine the direction of the billiard ball

since the cue had no y momentum prior to collision and has sin(30)*1.2 after

then
0.4*sin(30)*1.2=
0.30*sin(th)*1.5

th=ASIN(sin(30)*.4*1.2/
(.3*1.5)

th=32.23 degrees below the x axis

Now we know the x value of the billiard ball after collision since

v1c*.4=.4*cos(30)*1.2+
0.30*cos(32.23)*1.5

v1c=2 m/s prior to collision

j

Answer 4

it's sort of an ill defined problem. i'll tell you below:
you can only solve it if the ball travels along the original direction of the cue's path. So along that direction, call it x-axis:
m1v1+m2*0 = m1*v1' * cos(30) + m2v2' =>
0.4 * v1 + 0 = 0.4 * 1.2 * 0.866 + 0.3 * 1.5 => v1 = 2.164

correction: originally I used sin(30). That's wrong. It should be cos(30).

Upon further thought, the answer below is the right one. You can generalize the problem in two dimensions. My bad.

Answer 5

m1v1+m2v2 = m1v´1 +m2v´2( dont forget that the velocitys are vectors so you have to take their proyections on two axis
But this is not enough.As there is no information about friction I´ll suppose the process frictionless.
So There must be conservation of quinetic energy
1/2m1(v1^2 )+1/2m(v2)^2= 1/2m1(v´1)^2
+1/2m2(v´2)^2
Here you should take the modulus of velocities
Both equation will solve you the problem
The conservation of momentum is NOT enough

Answer 6

use m1v1 +m2v2 = m1v'1 + m2v'2

0.4*v1 + 0 = 0.4*1.2 + 0.3*1.5

0.4v1 = 0.93

v1 = 2.33m/s