Momentum of ball on a pool table

[ih8pa1n]

Homework Statement

Picture looks like this:
----------------------- ○ (empty space) ●
(Cue) (White ball) (Black Ball)

Mass: Cue (595g), White Ball (170g), Black Ball (155g)

Question: On the given pool table, the white ball bounces off of the black ball and moves in the direction opposite to its original direction. If the speed of the white ball immediately after the collision is 3.13m/s, then the speed of the black ball immediately after the collision is _______m/s. (Record your three-digit answer)

Homework Equations

Conservation of Momentum
Pi=Pf

The Attempt at a Solution

0=(-3.13m/s)×(170g) + (155g)×V
532.1kgm/s = 155g×V
V= 3.43m/s
Apparently this is not the answer... What am I doing wrong??

4. Keybook Solution
Pi=Pf
MV (1) + MV (2) = MV'(1) + MV'(2)
170g×(3.13m/s) + 0 = 170g×(-0.147m/s) + (155g)V '(2)
V'(2) = 3.59m/s

How do you get (-0.147m/s)?

 

[FactChecker]

The problem statement seems incomplete. What is the speed of the white ball before the collision?

 

[Jamison Lahman]

If the speed of the white ball immediately after the collision is 3.13m/s,

Are you sure it isn't supposed to be the speed immediately before the collision?

 

[jbriggs444]

If the black ball is stationary and the white ball is more massive than the black then there is no way for an elastic collision to end with the white ball reversing course. [Conservation of momentum would require the black ball to move away rapidly. Its resulting kinetic energy would exceed that of the impacting white ball]

If one assumes an elastic collision and a black ball more massive than the white then the problem becomes solvable.

 

[haruspex]

0=(-3.13m/s)×(170g) + (155g)×V

That equation says the system has no momentum, but the white ball must have had some before the collision.
As jjbriggs says, the masses must be wrong. Are you sure you quoted them correctly? Try swapping them around.

 

[Miraclo]

I actually contacted the company who published this question in their book and it has been reviewed. They sent me the correct information for the question as it was an error in printing and will be corrected in future printed copies. The question should read as follows.

"On the given pool table, the white ball traveling with a speed of 3.13 m/s strikes the black ball and then rolls in the direction opposite to its original motion.If the speed of the white ball immediately after the collision is 0.147 m/s, then the speed of the black ball immediately after the collision is ____ m/s. (Record your three-digit answer.)"

This explains where the random -0.147m/s comes from. A few years late, but I hope this helps anyone else who finds this question equally as confusing.

Below is the corrected question PDF I received from the publishing company.

 

[haruspex]

I actually contacted the company who published this question in their book and it has been reviewed. They sent me the correct information for the question as it was an error in printing and will be corrected in future printed copies. The question should read as follows.

"On the given pool table, the white ball traveling with a speed of 3.13 m/s strikes the black ball and then rolls in the direction opposite to its original motion.If the speed of the white ball immediately after the collision is 0.147 m/s, then the speed of the black ball immediately after the collision is ____ m/s. (Record your three-digit answer.)"

This explains where the random -0.147m/s comes from. A few years late, but I hope this helps anyone else who finds this question equally as confusing.

Below is the corrected question PDF I received from the publishing company.

Well done in following up.

 

[jbriggs444]

I actually contacted the company who published this question in their book and it has been reviewed.

Below is the corrected question PDF I received from the publishing company.

Looking at the PDF, the problem is still broken.

Since the masses of the balls have not been updated, we must conclude that there is a tiny blasting cap located on the stationary black ball at the point of the collision. The collision is neither inelastic (losing kinetic energy) nor perfectly elastic (preserving kinetic energy). Instead is is super elastic, increasing kinetic energy.

If we ignore energy considerations, the rest of the problem is nothing but a mundane balancing of momentum before and after.

Edit: Like @haruspex, I am surprised and pleased by your determination in following up.

 

[Miraclo]

Looking at the PDF, the problem is still broken.

Since the masses of the balls have not been updated, we must conclude that there is a tiny blasting cap located on the stationary black ball at the point of the collision. The collision is neither inelastic (losing kinetic energy) nor perfectly elastic (preserving kinetic energy). Instead is is super elastic, increasing kinetic energy.

If we ignore energy considerations, the rest of the problem is nothing but a mundane balancing of momentum before and after.

Edit: Like @haruspex, I am surprised and pleased by your determination in following up.

The question in the book is in the chapter on Conservation of Momentum and elastic/inelastic collision equations are later within the book. I believe they are trying to make it as simple as possible without overcomplicating the question as more complex equations involving the conversation of momentum including energy are brought up later.

Either way, I do hope that this PDF that they provided helps to clear up a better understanding.The Answer in the book in case anyone is wondering how to solve this is:m1v1+m2v2=m1v1'+m2v2'

170g(3.13m/s)+(0)= 170g(-0.147 m/s) + (155g)v2'

v2' = 3.59 m/s

 

[jbriggs444]

The question in the book is in the chapter on Conservation of Momentum and elastic/inelastic collision equations are later within the book. I believe they are trying to make it as simple as possible without overcomplicating the question as more complex equations involving the conversation of momentum including energy are brought up later.

Yes, I expect that you are correct. However, a different choice of values could make for an equally simple problem that invites the same momentum conservation approach without requiring the injection of unexplained kinetic energy.

Edit: Today I learned that in coin operated billiard tables, the cue ball is intentionally more massive than the target balls, possibly having a magnetic core. This allows for automated ball sorting. So if one wants to make the problem realistic in ball mass ratio and to retain the feature of a cue ball that recoils from the collision, one is forced into a scenario where energy is added.

 

[Herman Trivilino]

Then to make it realistic the cue ball needs to continue to move in the same direction.

 

[hutchphd]

The question in the book is in the chapter on Conservation of Momentum and elastic/inelastic collision equations are later within the book. I believe they are trying to make it as simple as possible without overcomplicating the question as more complex equations involving the conversation of momentum including energy are brought up later.

Either way, I do hope that this PDF that they provided helps to clear up a better understanding.The Answer in the book in case anyone is wondering how to solve this is:m1v1+m2v2=m1v1'+m2v2'

170g(3.13m/s)+(0)= 170g(-0.147 m/s) + (155g)v2'

v2' = 3.59 m/s

That is physically impossible answer. You said v1' was +0.147 in which case energy will be conserved.

 

[Miraclo]

That is physically impossible answer. You said v1' was +0.147 in which case energy will be conserved.

That is the answer written as is In the book, which I hope will be corrected in their future prints.