Momentum Problem and final velocities

  • Thread starter Thread starter Foon
  • Start date Start date
  • Tags Tags
    Final Momentum
Click For Summary
Momentum conservation is key to solving the collision problem involving two balls of equal mass. The initial momentum of the moving ball, with a speed of 2.25 m/s, can be expressed in terms of the final velocities of both balls after the collision. By breaking down the velocities into x and y components using trigonometry, the equations can be simplified to isolate the final velocities. The relationship established allows for the calculation of one ball's final velocity in terms of the other, facilitating the solution. Understanding and applying these principles will lead to determining the final speeds of both balls post-collision.
Foon
Messages
5
Reaction score
0
I've been hacking away at this question for a while but I'm really getting nowhere. All I know is that its supposed to be broken down into components or something. Any helps would be appreciated.

Question:

Two balls of equal mass (m) undergo a collision. One ball is initially stationary. after the collision, the velocities of the balls make angles of 31.1 degrees and 48.9 degrees relative to the original direction of motion of the moving ball. (momentum is conserved)

b) If the initial speed of the moving ball is 2.25 m/s what are the speeds of the balls after the collision?

I'm pretty sure that we're searching for the final velocties. Again, any help would be great. :smile:

Thanks!
 
Physics news on Phys.org
You said it yourself, momentum is conserved. So write the conservation of momentum equations for each component. Hint: choose your coordinate system so that one axis lies along the initial direction of the moving mass.
 


First of all, don't worry if you're struggling with this problem. Momentum problems can be tricky, but with some practice and understanding of the concepts, you'll be able to solve them easily.

To solve this problem, you'll need to use the law of conservation of momentum, which states that in a closed system, the total momentum before a collision is equal to the total momentum after the collision. In this case, the two balls are the only objects involved in the collision, so we can apply this law to them.

To break down the problem into components, we'll use trigonometry. The initial velocity of the moving ball can be represented as v1, and the final velocities of the two balls can be represented as v2 and v3. We can also break down the velocities into their x and y components, using the angles given in the question.

Using the law of conservation of momentum, we can write the following equation:

m*v1 = m*v2*cos(31.1) + m*v3*cos(48.9)

Since the masses of the two balls are equal, we can cancel them out from both sides of the equation. This leaves us with:

v1 = v2*cos(31.1) + v3*cos(48.9)

We also know that the initial velocity of the moving ball is 2.25 m/s, so we can substitute this value into the equation. This gives us:

2.25 = v2*cos(31.1) + v3*cos(48.9)

Now, we can solve for the final velocities by using basic algebra. First, we'll isolate v2 by subtracting v3*cos(48.9) from both sides of the equation. This gives us:

2.25 - v3*cos(48.9) = v2*cos(31.1)

Next, we'll divide both sides of the equation by cos(31.1) to isolate v2. This gives us:

(2.25 - v3*cos(48.9)) / cos(31.1) = v2

Finally, we can plug in the values for the given angles and solve for v2:

v2 = (2.25 - v3*cos(48.9)) / cos(31.1)

Now, we can use the same process to solve for v3. We'll
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 13 ·
Replies
13
Views
1K
Replies
10
Views
1K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
20
Views
2K
Replies
9
Views
4K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
10
Views
3K