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    in this question, a ball strikes a smooth right cone (first part of the question it strikes it in the direction of the normal at the point of impact, second part it falls vertically onto the cone) sitting on a smooth horizontal table which causes the cone to slide horizontally across the table. since the cone is not free to move downwards because of the table (there are also no rotational effects), am I right in saying you can only do the conservation of linear momentum horizontally, parallel to the table since the cone can only slide that way?
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    Well, that's what I did (and got the right answer).
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    thing is, I do not get the right answer by taking momentum parallel to the normal (was trying to make the cancelling my unknown velocity and angle easier since restitution only considers the component of my unknown velocity parallel to the line of impulse ie the normal) or perpendicular to it (tried this later). the mark scheme once again mentions nothing on this.
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    From memory:

    You use Newton's law of restitution as normal (resolve into components parallel/normal to the cone, particle motion in parallel direction is conserved, motion normal to cone obeys N's law).

    Total horizontal momentum is conserved (since the floor is smooth).

    This gives enough to solve for the horizontal velocity of the cone.
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    I know all of that. I just want to check this bit of theory so if similar situations comes up again I won't go off in a wrong direction.
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    Momentum in the direction of the normal isn't conserved since there's a vertical impulse from the table that has a component in the direction of the normal.
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    So momentum is not conserved in all non-horizontal directions because an external force is acting on the system in all non-horizontal directions?
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    Yes.
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    'particle motion in parallel direction is conserved'

    But surely perpendicular to the impulse between particle and cone the component of the velocity of the particle in that direction changes due to the impulse between cone and table?
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    (Original post by HaouLelouch)
    'particle motion in parallel direction is conserved'

    But surely perpendicular to the impulse between particle and cone the component of the velocity of the particle in that direction changes due to the impulse between cone and table?
    As I recall, you assume that it doesn't. That is, the impulse between cone and table is exactly that required to keep the cone on the table surface (neither continuing down or bouncing up), and so the cone won't exert a new force on the particle.

    I should perhaps say that I wasn't convinced by this assumption either - it was definitely a "go on and see if you get the right answer - if so then the assumption must be OK".
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    (Original post by DFranklin)
    From memory:

    You use Newton's law of restitution as normal (resolve into components parallel/normal to the cone, particle motion in parallel direction is conserved, motion normal to cone obeys N's law).

    Total horizontal momentum is conserved (since the floor is smooth).

    This gives enough to solve for the horizontal velocity of the cone.
    Hey,

    in terms of content do STEP 1 mechanics questions require only m1 knowledge or other modules such as m2, m3 etc? I know the questions are very challenging but what would you say is hard about step in particular, Application or actual concepts?
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    (Original post by Extricated)
    Hey,

    in terms of content do STEP 1 mechanics questions require only m1 knowledge or other modules such as m2, m3 etc? I know the questions are very challenging but what would you say is hard about step in particular, Application or actual concepts?
    Step 1 will only require m1 at most because you're not expected to have done further maths for Step 1. And most schools go no further than M1 in regular maths
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    (Original post by hassi94)
    Step 1 will only require m1 at most because you're not expected to have done further maths for Step 1. And most schools go no further than M1 in regular maths
    That's not true. Many STEP I and STEP II questions require both M1 and M2 knowledge. Some of the earlier STEP I questions even need some M3 knowledge although the questions that require M3 nowadays will not come up due to a change in the syllabus around 2001 IIRC.
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    (Original post by Farhan.Hanif93)
    That's not true. Many STEP I and STEP II questions require both M1 and M2 knowledge. Some of the earlier STEP I questions even need some M3 knowledge although the questions that require M3 nowadays will not come up due to a change in the syllabus around 2001 IIRC.
    Oh well that's news to me! I thought they didn't go any further than m1 until Step II/III but I could very well be mistaken, I'm only in Year 12 so haven't had much experience of the STEP really.

    So do they not go to M3 on ANY of the STEP papers any more?
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    (Original post by hassi94)
    Oh well that's news to me! I thought they didn't go any further than m1 until Step II/III but I could very well be mistaken, I'm only in Year 12 so haven't had much experience of the STEP really.

    So do they not go to M3 on ANY of the STEP papers any more?
    M3, M4 and even M5 (On the specifications where it exists) knowledge is required to attempt some of the mechanics questions on STEP III. STEP I and II share the same syllabus so the questions are based on the same material.

    In order to have enough knowledge to attempt any applied STEP question on STEP I or II, you need to have covered material from M1, M2, S1, S2 and maybe the odd the chapter from S3 or M3 (as some things that appear on one examboard's S2 would appear on another board's S3, for example).
 
 
 
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