Determining Velocity and Direction of Second Ball After Collision

  • Thread starter konceptz
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In summary, the problem involves a steel ball of mass 0.50 kg colliding with a second ball of mass 0.30 kg, initially at rest. The collision is glancing and the first ball has a velocity of 1.5 m/s [30 degrees N of E] after the collision. The task is to determine the velocity of the second ball after the collision. Using conservation of momentum and considering the horizontal and vertical components separately, the velocity of the second ball was found to be 1.7 m/s [E 47 degrees S].
  • #1
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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  • #2
Since momentum is a vector you can get two conservation equations by considering horizontal (E) and vertical (N) components separately.
 
  • #3
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.
 
  • #4
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!)
 
  • #5
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.
 
  • #6
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?)
 
  • #7
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.)
 
  • #8
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.
 

1. What is the difference between a collision and momentum?

Collision refers to the event when two or more objects come into contact with each other. Momentum, on the other hand, is a property of an object that describes its motion and is defined as the product of its mass and velocity.

2. How is momentum conserved in a collision?

In a closed system, the total momentum of the objects before a collision is equal to the total momentum after the collision. This means that momentum is conserved in a collision, and it can only be transferred between objects rather than created or destroyed.

3. What is the formula for calculating momentum?

The formula for calculating momentum is p = m * v, where p is momentum, m is mass, and v is velocity. Momentum is measured in units of kg*m/s.

4. How does the mass and velocity of an object affect its momentum?

The momentum of an object is directly proportional to its mass and velocity. This means that an object with a larger mass or a higher velocity will have a greater momentum compared to an object with a smaller mass or a lower velocity.

5. Can momentum be negative?

Yes, momentum can be negative. This can occur when an object is moving in the opposite direction of its defined positive direction, or when it experiences a negative change in velocity. It is important to keep in mind that the negative sign only indicates the direction of momentum, not its magnitude.

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