Calculate the velocity of the eight ball after collision

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Homework Help Overview

The discussion revolves around a physics problem involving the collision of a cue ball and an eight ball in billiards, specifically focusing on the calculation of the eight ball's velocity after the collision. The problem is situated within the context of momentum conservation principles.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the application of momentum conservation, questioning the initial and final momentum equations presented by the original poster. There are discussions about the correct interpretation of the velocities involved and whether the balls move together post-collision.

Discussion Status

Several participants have provided feedback on the original poster's calculations, suggesting potential errors and clarifying the need to separate the velocities of the two balls. Some guidance has been offered regarding the correct setup of momentum equations, while others emphasize the importance of not providing complete solutions.

Contextual Notes

There is a noted concern regarding the assumption that the balls move together after the collision, as well as the potential misinterpretation of the velocities involved in the momentum calculations.

aarietta
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Hello, I was just wondering if I am doing this question correctly, if not, any suggestions please and thank you

A person takes aim at the eight bal in a game of billiards. Her shot gives the cue ball a velocity of 2.5m/s [forward]. The cue ball strikes the eight ball and continues with a velocity of 2.5 x 10^-1 m/s [forward]. If the mass of the cue ball equals the mass of the eight ball equals 400g, calculate the velocity of the eight ball after collision.

v1 of cueball = 2.5m/s [forward]
v2 of cueball = 2.5 x 10^1 m/s [forward]
m = 400g = .4kg

P1 = m1v1
P1 = 0.4(2.5)
P1 = 1 m/s

P2 = m2v2
P2 = 0.4(2.5 x 10^1)
P2 = 0.1 m/s

Ptotal = [(0.1)^2 + (1)^2]sqaure root
Ptotal = 1.0 m/s

Pafter = m1v1 + m2v2
1 = (0.4kg + 0.4kg)v
v = 1/0.8
v = 1.25 m/s



Thanks!
 
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Ptotal = [(0.1)^2 + (1)^2]sqaure root
Ptotal = 1.0 m/s
I don't get this. The total initial momentum of the system is just that of the cue ball before collision.

Pafter = m1v1 + m2v2
1 = (0.4kg + 0.4kg)v
v = 1/0.8
v = 1.25 m/s

The problem does not state the balls move together after the collision.
 
Hello,

I think you've made one small error on this question, in your equations it appears that you're looking for 1 velocity, when you're really looking for two. Your equation should look like this:

(0.4)(2.5) = (0.4)(0.25) + (0.4)V2prime

You already know one velocity, and you're searching for another (V2 prime) so you can't combine the two sections together (0.4+0.4)V2,
they must be separate.

I hope this was helpful.
 
aarietta said:
Hello, I was just wondering if I am doing this question correctly, if not, any suggestions please and thank you

A person takes aim at the eight bal in a game of billiards. Her shot gives the cue ball a velocity of 2.5m/s [forward]. The cue ball strikes the eight ball and continues with a velocity of 2.5 x 10^-1 m/s [forward]. If the mass of the cue ball equals the mass of the eight ball equals 400g, calculate the velocity of the eight ball after collision.

By conservation of total momentum,

Initial momentum = Final momentum (for any forms of collisions)
Initial P of cueball + Initial P of 8-ball = Final P of cueball + Final P of 8-ball
0.4(2.5) + 0 = 0.4(0.25) + 0.4v
v = 2.25ms
 
Please don't post full solutions .
You may suggest hints or point out where the OP is going wrong, but let him do it on his own .
 
"A person takes aim at the eight bal in a game of billiards. Her shot gives the cue ball a velocity of 2.5m/s [forward]. The cue ball strikes the eight ball and continues with a velocity of 2.5 x 10^-1 m/s [forward]. If the mass of the cue ball equals the mass of the eight ball equals 400g, calculate the velocity of the eight ball after collision"

the eight ball is supposed to move forward.
let "m" be the mass of each ball(the value is crap), which is given to be equal for both.
conserving momentum in the forward direction,

m*2.5 + m*0 = m*0.25 + m* v
* denotes multiplication and v to be found.

cancelling "m" on both sides,

v+0.25=2.5
therefore v=2.25 m/s... ANSWER.
 

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