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: 855
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.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top