What is the initial velocity of the second ball in a 2-dimensional collision?

AI Thread Summary
In a discussion about a 2-dimensional collision between two golf balls of equal mass, one ball has an initial velocity of 2.7 m/s to the east. After the collision, the balls move at 2.49 m/s at an angle of 62.8 degrees north of west and 2.37 m/s at 69.2 degrees south of east. Participants emphasize the importance of using conservation of momentum and vector diagrams to analyze the collision. They suggest resolving the velocities into x and y components to apply momentum conservation in both directions. The conversation highlights the complexity of 2D collisions compared to 1D scenarios, with a focus on correctly calculating the initial velocity of the second ball.
pinkyjoshi65
Messages
260
Reaction score
0
Two rolling golf balls of the same mass collide. The velocity of one ball is initially 2.7m/s (E). After the collision, the velocities of the balls are 2.49m/s(62.8 N of W) and 2.37m/s (69.2 S of E). What is the initial velocity of the second ball?

So i know that the masses are the same, and the initial velociy of the 1st ball.
Hence by using the conservation of momentum formula, I get
v_2= (V_1+V_2)v1
(here, the capital V's r the final velocities and the small v's are the initial velocities)
So now if it was one dimension, it straight forward. But since this is 2-dimensional, its confusing.

I know we have to draw somthing like a vector diagram. So, i tried doing that.
I tried attaching the file but I can't seem to do it. So, i'll try to explain the my diagram in words!

So one ball is moving in the north west direction, the other in the south east. I drew a line frm the centre of the 1st ball going NW. Thats V_1. With that line as the hypotunese, i drew V_1x and V_1y. And the angle betwee V_1y and V_1 is 62.8 degrees. SO i used cos and sine to figure out V_1x and V_1y. I did the same thing for the 2nd ball..This is all i could do...Any help..
 
Physics news on Phys.org
Define an x-axis, say the West-East line, then Nort-South is your y-axis. The resolve everything along those two axes, and work with conservation of momentum in those directions.
 
i still don't get it..ok..so i know the final velocities of both the golf balls. Hence i found the components of each final velocity.
V1x= Sin(62.5)V1, V1y= Cos(62.5)V1
V2x= Cos(69.2)V2, V2y= Sin(69.2)V2
And by using the conservation of momentum formula, i got v2= (V1=+V2)/v1
 
so now if i take the y components of each, i'll get
0=(V1y+V2y)/v1y
But v1y should be zero too yes..?..This is what i could do...
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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...
Back
Top