How Do Speeds and Angles Affect Post-Collision Ball Movement?

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Homework Help Overview

The discussion revolves around a collision problem involving two identical balls, where one ball strikes another at rest. The focus is on understanding the effects of speeds and angles on the post-collision movement of the balls, specifically analyzing momentum and velocity components.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to derive equations for the system, questioning the correct setup for momentum conservation in both x and y directions. There are attempts to resolve velocity vectors and establish relationships between the final speeds of the balls.

Discussion Status

Some participants have provided guidance on resolving the velocity components and forming equations based on momentum conservation. However, there is ongoing uncertainty regarding the correct relationships and calculations, particularly in determining the ratio of the final speeds.

Contextual Notes

Participants are working under the assumption that the two balls have equal mass and are questioning the implications of this on the momentum equations. There is also mention of a diagram that may influence the interpretation of angles and directions.

Naeem
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A ball of mass m = 2.1 kg moving with a speed of v0 = 15.2 m/s strikes an identical ball which was initially at rest. After the collision, the incoming ball (ball 1) goes off at q1 = 28° relative to its original direction and the struck ball (ball 2) moves off at q2 = 38° as shown in the diagram.

a) What is the magnitude of the initial momentum of the two ball system?

Ans, Used momentum m1v1 = m2v2 , Got the answer
b) What is the ratio of the balls' final speeds?
Since there is no vertical momentum before the collision, the net vertical momentum must be zero after the collision.
Since the vertical momentum must be zero after the collision, what does that tell us about the ratio of vertical momentum of the two balls?



c) What is the final speed of the incoming ball? v1

d)What is the final speed of the struck ball? v2

Need help with b , c , and d .
 

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If i read your question correct then c and d are to be solved before b ...

Ok have u derived any equation whatsoever for the system ?? If so post it , because note that the system has two unknowns v1 and v2 and u should have two equations for this.

-- AI
 
How do I do that ?

For the x direction would it be,

v1 cos theta1 and

for the y direction v2 sin (-theta2)

This is what I know, just a vauge idea, but I don't now if the eqns are correct .

Please guide.!
Appreciate ur help!111
 
resolve the velocity vectors v1 and v2 along x and y directions
find the momentum equation along x - direction
find the momentum equation along y - direction

these form the two equation in two unknowns...
solve these to find v1 and v2 ...

-- AI
 
There is only 1 vector pointing in the x and y directions

would it be

v1 cos theta1 + v1 sin theta2 = 0

and

v2 cos theta2 + v2sin theta2 = 0



How do i go about finding momentum... Please help
 
Yes i think u have the equations on the right track

though ur second question is at err,
it should be v1sintheta1 = v2sintheta2
(can u see why??)
Now solve these two equations ...
obtain v1 and v2 ..

-- AI
 
I have been trying this for a while now, but the ratio I am getting is incorrect.

Here is what I did:

Resolving components:

For the x- direction:

v1 cos theta1 + (-v1 cos(0)) = 0

Also,

-v2 cos (0) + v2 cos theta2 =0

For the y-direction:

v2 sin ( 90 - theta1) + (-v2 sin (90- theta2)) = 0

This is what I know so far,

Any insights would be appreciated!
 
Naeem said:
A ball of mass m...strikes an identical ball ... at rest.
... (ball 1) goes off at q1 = 28° ...(ball 2) moves off at q2 = 38° as shown in the diagram.
.

[if] m1 = m 2 → θ1 + θ2 [should be] = 90°.
28°+38° = 66°
 
Naeem said:
A ball of mass m = 2.1 ... After the collision, the incoming ball (ball 1) goes off at q1 = 28° relative to its original direction and the struck ball (ball 2) moves off at q2 = 38° as shown in the diagram.
.
if diagram is correct m2 should be ≈ 1
 

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