How Do You Determine the Velocity and Angle of a Billiard Ball After Collision?

Click For Summary

Homework Help Overview

The discussion revolves around determining the velocity and angle of a billiard ball after a collision, specifically focusing on the conservation of momentum in a two-dimensional context. The problem involves two billiard balls with given masses and initial velocities, and it requires calculating the post-collision speed and angle of one of the balls without assuming an elastic collision.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conservation of momentum equations and how to apply them to find the unknowns. There is an exploration of using trigonometric identities to assist in solving for angles. Some participants express uncertainty about the correctness of their calculations and whether to prioritize finding the angle or speed first.

Discussion Status

There is an ongoing examination of the calculations presented, with some participants questioning specific steps and suggesting corrections. Guidance has been offered regarding the use of trigonometric relationships to find the angle, indicating a productive direction in the discussion.

Contextual Notes

Participants are working under the constraint of not assuming an elastic collision, which influences the approach to solving the problem. There is also a noted confusion regarding the calculations, with some participants expressing a desire to reassess their work.

rvnt
Messages
14
Reaction score
0

Homework Statement


Billiard ball A of mass mA= 0.400kg moving with speed vA=1.80m/s strikes ball B initially at erst, of mass mB=0.500kg.As a result of the collision,ball A is deflected off at an angle of 30.0degrees with a speed v'A=1.10m/s. Taking the x-axis as the original direction of motion of ball A, solve for the speed v'B and angle Θ'B of ball B. Do not assume the collision is elastic.


Homework Equations


px conserved: mAvA=mAv'ACosΘ'A+mBv'BCosΘ'B
py conserved: mAvA=mAv'AsinΘ'A+mBv'BSinΘ'B



The Attempt at a Solution


0.72=0.38105+0.500v'BCosΘ'B
0.338948/0.500=v'BCosΘ'B
0.677897=v'BCosΘ'B
0.677897/CosΘ'B= v'B

0=mAv'AsinΘ'A+mBv'BSinΘ'B
0=0.22+0.5v'BSinΘ'B
-0.22/0.5=v'BSinΘ'B
-0.44=v'BSinΘ'B

-0.44=(0.677897/CosΘ'B)SinΘ'B
-1.5407=CosΘ'BSinΘ'B
 
Physics news on Phys.org
To solve for the angle, use a double angle trig identity:
sin(2θ) = 2 sinθ cosθ
 
Could you help me out with where I have went wrong please? or are my whole calculations incorrect and I should start over? Also, should I try figuring out the angle first or the speed?
 
rvnt said:
Could you help me out with where I have went wrong please? or are my whole calculations incorrect and I should start over? Also, should I try figuring out the angle first or the speed?
My bad. I didn't read your work carefully and just looked at the last line. You were on the right track until that point. That last line was an error.

rvnt said:

The Attempt at a Solution


0.72=0.38105+0.500v'BCosΘ'B
0.338948/0.500=v'BCosΘ'B
0.677897=v'BCosΘ'B
0.677897/CosΘ'B= v'B

0=mAv'AsinΘ'A+mBv'BSinΘ'B
0=0.22+0.5v'BSinΘ'B
-0.22/0.5=v'BSinΘ'B
-0.44=v'BSinΘ'B

-0.44=(0.677897/CosΘ'B)SinΘ'B
That looks OK.
-1.5407=CosΘ'BSinΘ'B
Not sure how you got this from the previous line. Note:
-0.44=(0.677897/CosΘ'B)SinΘ'B = (0.677897)(SinΘ'B/CosΘ'B) = (0.677897)(TanΘ'B)

That will give you the angle; then you can find the speed.
 
Oh right! k now I got it thank-you