What Are the Final Velocities of Ball A and Ball B After Collision?

AI Thread Summary
The discussion revolves around calculating the final velocities of two colliding balls, A and B, with given masses and initial conditions. Ball A, with a mass of 0.030 kg, moves at +5.5 m/s, while Ball B, with a mass of 0.055 kg, is initially at rest. The collision is not head-on, and the angles of deflection for both balls are provided. Participants emphasize the need to apply conservation of momentum equations in both the x and y directions to solve for the final velocities. The challenge lies in determining the two unknown velocities without additional information, highlighting the importance of using the angles effectively in the calculations.
Meco
Messages
8
Reaction score
0

Homework Statement


The drawing shows a collision between two balls. Ball A has a mass of .030kg and is moving along the x-axis with a velocity of +5.5 m/s. It makes a collision with ball B, which has a mass of .055kg and is initially at rest. The collision is not head-on. After the collision, the two balls fly apart with the angles shown in the drawing below
What is the final velocity of ball A and B?

... ... | Ball A
... ... | 65°
. A ---B---
... ... | 37°
... ... | Ball B

Can't get above picture to look right, going to take a picture of it and save it in the post

A and B show the start position and Ball A and Ball B shows where they finish

Homework Equations


M1V1 + M2V2 = M1V1'+ M2V2'
Using inverse trig?

The Attempt at a Solution


I saw basically the same problem in Khan's academy and in there he had one of the two ball's final velocity given. I have no real idea how to solve for two variables but here was my try:
Momentum in X initial= 5.5 * .03 = .165
Momentum in Y initial= 0
m1v1+ m2v2= m1v1'+ m2v2'
.030*5.5 + 0 = .03v1' +.055v2'
 

Attachments

  • chapt7 quest 2.jpg
    chapt7 quest 2.jpg
    12.7 KB · Views: 860
Last edited:
Physics news on Phys.org
You should have three equations,

Energy before = energy after

Momentum in x direction before = momentum in x direction after

Momentum in y direction before = momentum in y direction after
 
Ok, well the energy before and after equation is:
(.03*5.5)+ (0.055*0)= (.03*V1')+(.055*V2')
Momentum in X before:
.030*5.5
Momentum in the Y before:
0
Momentum final for X must equal .165
Momentum final for Y must equal 0
I do not really understand how I can use this to find two velocities without knowing atleast one of them. How do I use the angles given to help me figure this out?
 
Write it out, if you see it written out it might click, if not it did not take too much time. I don't think you need the energy equation. You have two unknowns, the magnitudes of the velocity of each mass after collision so you only need two equations in the unknowns.
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top