Velocity of Striking Ball After Collision: Billiard Ball Physics Problem

[ken62310]

PHYSIC momentum QUESTION~

Homework Statement

A billiard ball moving with velocity 4i+3j m/s strikes a stationary ball of the same mass. After the collision the struck ball moves with a speed of 4m/s at an angle of 30 degree with respect to the line of original motion of the striking ball. Find velocity of the striking ball after collision.

Homework Equations

Pi=Pf
P=mv

The Attempt at a Solution

Pxi=mv1=4
Pxf=mv2cos(theta)+m(4)cos30

Pyi=mv1=3
Pyf=mv2sin(theta)+m(4)sin30


am i correct with that?? i am really confuse

 

[Mindscrape]

You have the right idea, however the problem says that the stricken ball goes at a 30 degree angle with respect to the stricker's original line of motion, so you will want to add the two angles.

 

[ken62310]

You have the right idea, however the problem says that the stricken ball goes at a 30 degree angle with respect to the stricker's original line of motion, so you will want to add the two angles.

i am not quite understand what you mean.
how do i add the two angles? could you be more detailed?

 

[Mindscrape]

Well, let me preface this by saying that what I am telling you is only one approach. I can think at least one other reasonable way to do this problem, however, what I am telling you is pretty straightforward, or at least hopefully it will seem that way.

So if I am walking along some path at 50 degrees and then the path forks off so that one trail goes at 30 degrees with respect to the original direction I am walking, what is the total angle that I have changed by? It would be 80 degrees, right? The same is true for your ball. The first ball has some trajectory with an angle (that I'll let you find), and then it deviates from the angle by an additional 30 degrees.

 

[ken62310]

Well, let me preface this by saying that what I am telling you is only one approach. I can think at least one other reasonable way to do this problem, however, what I am telling you is pretty straightforward, or at least hopefully it will seem that way.

So if I am walking along some path at 50 degrees and then the path forks off so that one trail goes at 30 degrees with respect to the original direction I am walking, what is the total angle that I have changed by? It would be 80 degrees, right? The same is true for your ball. The first ball has some trajectory with an angle (that I'll let you find), and then it deviates from the angle by an additional 30 degrees.

so that means i need to find the angle that formed by the 4i+3j right?
now i have another question.
I found that the angle of the 4i+3j would be 37 degree.
and for x i will add that to the 30 degree right. then what about for the y?
do i need to add the different angle? should it be respect to the y-axis?

 

[ken62310]

for x: mv1=mv2 cos67+ mv3 cos theta.
for y: mv1=mv2 sin30+ mv3(37-theta).
...
please tell me if i am correct..

 

[katchum]

Why mv1? If you projected it to the x-axis, it should have a cosinus in it. Am I missing something?

 

[ken62310]

Why mv1? If you projected it to the x-axis, it should have a cosinus in it. Am I missing something?

oh...so for x: mv1 cos 37 and for y: mv1 cos 53 ? am i right?

 

[rl.bhat]

Initial velocity of the striking ball is 5 m/s. If the collision is perfectly elastic KE is conserved. Since masses of the balls are same V^2 = V1^2 + V2^2. Therefore the velocity of the strking ball after collision will be 3 m/s. And the two balls will move perpendicular to each other.

 

[ken62310]

Initial velocity of the striking ball is 5 m/s. If the collision is perfectly elastic KE is conserved. Since masses of the balls are same V^2 = V1^2 + V2^2. Therefore the velocity of the strking ball after collision will be 3 m/s. And the two balls will move perpendicular to each other.

so all the angle stuff are useless in this case??

 

[Mindscrape]

It depends on what you have learned so far in physics. If you don't know about kinetic energy and perfectly elastic collisions then conservation of momentum is the only choice you have. Anyway, for the momentum case you should get something to the effect of

v_{xi1} = 4 = v_{xf1} + 4cos(arctan(3/4) + 30)

and

v_{yi1} = 3 = v_{yf1} + 4sin(arctan(3/4) + 30)

so

||\vec{v}|| = (v_{yf1}^2 + v_{xf1}^2)^{(1/2)}

 

[katchum]

I have a question of my own.

How do you find the angle of the first ball after the collision?

Here is my attempt:

x: m.vi1.cos37 = 4 = m.vf1.cosx + m.vf2.cos7 and vf2 = 4
y: m.vi1.sin37 = 3 = m.vf1.sinx + m.vf2.sin7

Two parameters vf1 and x.

vf1 is the final velocity. (absolute)
x is the angle between the first ball after collision and the x axis.


I find it odd that I didn't use the equation of kinetic energy, there has to be something wrong...

 

[rl.bhat]

Momentum

If you take the direction of the initial velocity as axis of reference, the problem becomes easy. If you shift the co-ordinate of the initial velocity to the point of impact without any rotation, the components of final velocities of two balls involve several angles. Angle of V1 with new X-axis will be 37 +30 degree. But the angle of V2 with new X-axis is unknown.. If you use coservation of energy we can avoid all these complication.
I didn't understood mindscrape's equations. vxf1 is not in the direction of x-component of the initial velocity and y-component should be the difference of two components of v1 and v2 along y axis.