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    I usually get pissed off with questions like this, and mostly I do them wrongly. The two ways to solve which I've heard are either using law of conservation of momentum OR speed of approach = speed of separation(though concept of co-efficient of restitution is not in the A level syllabus). I'll be grateful if anyone solves and explains it me. . .
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    Woah dude ever thought of creating one thread and posting all of them in there i've just seen another one of yours I believe only a couple of seconds ago...

    as for the question soz im stuck on linear expressions and correlations
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    (Original post by Smilingsam)
    Woah dude ever thought of creating one thread and posting all of them in there i've just seen another one of yours I believe only a couple of seconds ago...

    as for the question soz im stuck on linear expressions and correlations
    That thread was created 11 hours ago. And I've already asked from a mod about this, so it's okay to make new threads after time intervals.

    Let's see if anybody can help us.
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    For a perfectly elastic collision, speed of approach = speed of separation. The speed of approach is u1+u2 as the spheres are moving towards each other so their velocities relative to each other are the sum of the velocities. After the collision the spheres move in the same direction. We can assume v2>v1 as otherwise the sphere would immediately collide again. Therefore the velocity of the second sphere relative to the first is v2-v1 (as the first sphere is moving in the same direction). So D is the correct answer.
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    (Original post by Hudzy)
    For a perfectly elastic collision, speed of approach = speed of separation. The speed of approach is u1+u2 as the spheres are moving towards each other so their velocities relative to each other are the sum of the velocities. After the collision the spheres move in the same direction. We can assume v2>v1 as otherwise the sphere would immediately collide again. Therefore the velocity of the second sphere relative to the first is v2-v1 (as the first sphere is moving in the same direction). So D is the correct answer.
    Hmm, thank you!
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    (Original post by Zishi)
    That thread was created 11 hours ago. And I've already asked from a mod about this, so it's okay to make new threads after time intervals.

    Let's see if anybody can help us.
    Sorry if i sounded a bit negative didnt have any sleep glad you managed to get your answer and yeah i might post mine later.. just hope someone can help me i have no clue how to do it :mad:
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    (Original post by Smilingsam)
    just hope someone can help me i have no clue how to do it :mad:
    There are two ways of solving this question (I mean I know two.)

    For elastic collisions, total energy of the system is conserved. Momentum of the system is conserved in any collision. Can you write down the laws of conservation of momentum and energy for these to balls in the form of equations?

    Another, quicker approach would be to notice what Hudzy suggests. He says that "speed of approach = speed of separation", or to put it in other words, relative velocity of the two balls is conserved*. This fact follows from the laws of conservation (you should be able to prove it yourself), but is so useful that it is worth remembering that instead of deriving it from a system of equations each time.


    *Of course this is only true for perfectly elastic collisions.
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    (Original post by jaroc)
    There are two ways of solving this question (I mean I know two.)

    For elastic collisions, total energy of the system is conserved. Momentum of the system is conserved in any collision. Can you write down the laws of conservation of momentum and energy for these to balls in the form of equations?

    Another, quicker approach would be to notice what Hudzy suggests. He says that "speed of approach = speed of separation", or to put it in other words, relative velocity of the two balls is conserved*. This fact follows from the laws of conservation (you should be able to prove it yourself), but is so useful that it is worth remembering that instead of deriving it from a system of equations each time.


    *Of course this is only true for perfectly elastic collisions.
    Sorry to put you through writing that answer out however I was on about the A level maths questions i am being forced to do for my post grad they are a lot easier than this but, last time I did maths was 4 years ago and never touched on any in my topic including (linear expressions, correlations and standard definitions are just a few) .
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    (Original post by Smilingsam)
    Sorry to put you through writing that answer out however I was on about the A level maths questions i am being forced to do for my post grad they are a lot easier than this but, last time I did maths was 4 years ago and never touched on any in my topic including (linear expressions, correlations and standard definitions are just a few) .
    I'm not sure if I understand you properly now but I guess I have misinterpreted your post, right? Never mind
 
 
 
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