Elastic Collision: Solving Components of the Problem

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The discussion focuses on solving an elastic collision problem using conservation of momentum and kinetic energy equations. Participants emphasize the need to break down the vector equations into scalar components, suggesting directions like West and North for clarity. One user mentions a friend's advice to simplify the problem by using a triangle instead of components, leading to the formulation of two key equations. The first equation relates the velocities post-collision, while the second equation ensures kinetic energy conservation. The user expresses gratitude for the assistance and indicates they will continue working on the problem.
harisf
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1. Homework Statement
The problem is in the picture(attached)


2. Homework Equations
I know we have to use
m1v1 + m2v2 = m1v1' + m2v2'
(1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)m1v1'^2 + (1/2)m2v2'^2

3. The Attempt at a Solution
All I know is that it needs to be split up into components.
Please help me out.
 

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Welcome to PF!

Welcome to PF! :smile:

(please type the question in future … it's much easier to read than having to look at a picture in a separate window)

(and try using the X2 and X2 tags just above the Reply box :wink:)
harisf said:
I know we have to use
m1v1 + m2v2 = m1v1' + m2v2'
(1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)m1v1'^2 + (1/2)m2v2'^2

All I know is that it needs to be split up into components.

Yes, the first equation is a vector equations, and needs to be split up into two scalar equations.

(You could choose West and North for the component directions, for example.)

So … show us what you get. :smile:
 
Health Warning::smile:

This problem can get very messy unless you translate it into
the reference frame in which the c of m is stationary.

David
 
davieddy said:
Health Warning::smile:

This problem can get very messy unless you translate it into
the reference frame in which the c of m is stationary.

David

No, you should be able to do it in the reference frame given.
 
Thanks guys, and yes I will try this question out. I got some help today from a friend, he said that there was no need for NS/EW components. He told me to make a triangle. So therefore the resulting velocity after the collision of the 2amu mass would be v1'cos10.

2(2km/s) + 1(0m/s) = 2(v1'cos10) + v2'
4 - 2v1'cos10 = v2' (this will be equation one)

We know kinetic energy is conserved:
2(2km/s)2 + 1(0m/s)2 = 2(v1'cos10)2 + v2'2
8 - 2(v1'cos10)2 = v2'2 (our second equation)

Then sub in equations into each other for v2'.
I have to go for now but I will get back to you guys soon on the problem. Please keep in mind that I am only in grade 12:confused:. Thanks again.
 
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?

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