Determining Velocity and Direction of Second Ball After Collision

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Homework Help Overview

The problem involves a collision between two steel balls of different masses, where one ball is initially at rest. The scenario requires determining the velocity and direction of the second ball after a glancing collision, given the initial conditions and the final velocity of the first ball.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using conservation of momentum in both horizontal and vertical components to solve for the unknowns. There are attempts to isolate variables related to the second ball's speed and direction, with some questioning the accuracy of their calculations.

Discussion Status

Some participants have expressed confusion regarding their results, noting discrepancies between their calculations and the expected answers. There is acknowledgment of the need to ensure consistency with conservation laws, and guidance has been offered to check the equations used. Multiple interpretations of the problem setup are being explored.

Contextual Notes

There is a lack of clarity regarding whether the collision is elastic, which has influenced the approaches taken. Participants have also noted potential typographical errors in the problem statement regarding angles.

konceptz
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Homework Statement


A steel ball of mass 0.50 kg, moving with a velocity of 2.0 m/s [E], strikes a second ball of mass 0.30 kg, initially at rest. The collision is a glancing one, causing the moving ball to have a velocity of 1.5m/s [30 degrees N of E] after the collision. Determine the velocity of the second ball after the collision.


Homework Equations


m1v1 + m2v2 = m1v1' + m2v2'


The Attempt at a Solution


I've tried using the y-component to solve for the speed on v2y' but there's no direction given so there are two variables to solve for. The final speed of the second ball and the direction of the second ball after the collision. Using the y-component i had the equation:

0 = (0.50)(1.5sin33) + (0.30)(v2')(siny)

How can I solve for both the speed and the direction? I've tried subbing in v2 in terms of siny but I just can't seem to get the right answer. Can someone solve this problem and show me how you solved it? Thanks.
 
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Since momentum is a vector you can get two conservation equations by considering horizontal (E) and vertical (N) components separately.
 
Yes, I've noticed that. I've tried separating the x and y components and subbing in for v2'. I've reached an answer however, the answer does not match the correct solution for the problem. It's either I'm doing something wrong or the book had a missprint. My answer was 18m/s [E 48 degrees S] while the answer in the book was 17m/s [E 47 degrees S] I've also carried at least four sig figs to ensure my accuracy.
 
Please show how you got your answer. Those answers seem rather unlikely, as the final KE would be much greater than the initial KE. (Quite interesting that your answer is close to that of the book, though!)
 
As the question does not mention the collision as an elastic one, it would be wrong to assume that it is. Therefore, I excluded the use of the kinetic energy theorem.
 
Hi konceptz,

Your result for the speed seems too large, which is what I believe Doc Al was referring to, and why he asked to see your work. Notice that if you plug your speed and angle into the y-equation you have in your original post, they don't work, so something seems to have gone wrong.

(Also you have an angle of 30 degrees in the problem statement, but an angle of 33 degrees in that equation. Which one is right?)
 
konceptz said:
As the question does not mention the collision as an elastic one, it would be wrong to assume that it is. Therefore, I excluded the use of the kinetic energy theorem.
As alphysicist already explained, your answers are not consistent with conservation of momentum, which is all you need to solve this problem--so you must have made an error somewhere. (I did not mean to imply that you had to use conservation of KE to solve this problem--you don't.)
 
Sorry, my speed was 1.7 not 17, forgot to type in the decimal. As for the degrees the correct one was 30. I've solved the problem now, thank you reminding me to include the x and y components.
 

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