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    Particle A of mass 8kg and Particle B of mass 5kg are moving towards each other at speeds of 2ms^-1 and 1ms^-1 respectively. Given B rebounds with speed 3ms^-1 in the opposite direction to its initial velocity, find the velocity of A after the collision.

    i drew my diagram and did these workings

    m1u1+m2u2=m1v1+m2v2

    (8x2)+(5x-1)=(8xV)+(5x3)
    16-5=8V+15
    11=8V+15
    8V=-4
    V=-2ms^-1

    Aside from the -2ms^-1
    where else did i go wrong?
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    (Original post by thefatone)
    Particle A of mass 8kg and Particle B of mass 6kg are moving towards each other at speeds of 2ms^-1 and 1ms^-1 respectively. Given B rebounds with speed 3ms^-1 in the opposite direction to its initial velocity, find the velocity of A after the collision.

    i drew my diagram and did these workings

    m1u1+m2u2=m1v1+m2v2

    (8x2)+(5x-1)=(8xV)+(5x3)
    16-5=8V+15
    11=8V+15
    8V=-4
    V=-2ms^-1

    Aside from the -2ms^-1
    where else did i go wrong?
    Nothing wrong with getting a negative velocity but check your masses in your calculation.
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    (Original post by SeanFM)
    Nothing wrong with getting a negative velocity but check your masses in your calculation.
    sorry i typed the question slightly wrong xD i edited it now
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    (Original post by thefatone)
    sorry i typed the question slightly wrong xD i edited it now
    Now check the last two lines of your calculation
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    (Original post by SeanFM)
    Now check the last two lines of your calculation
    11=8V+15 -15 from both sides gives me
    -4=8V divide by 8 on both sides gives me
    V=-0.5ms^-1
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    (Original post by thefatone)
    11=8V+15 -15 from both sides gives me
    -4=8V divide by 8 on both sides gives me
    V=-0.5ms^-1
    Voila
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    (Original post by thefatone)
    11=8V+15 -15 from both sides gives me
    -4=8V divide by 8 on both sides gives me
    V=-0.5ms^-1
    If you're with Edexcel, -0.5 won't get you the final mark. Your choice of a positive direction was up to you. You need to say "in the opposite direction to its velocity before the collision" or "away from B".
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    (Original post by tiny hobbit)
    If you're with Edexcel, -0.5 won't get you the final mark. Your choice of a positive direction was up to you. You need to say "in the opposite direction to its velocity before the collision" or "away from B".
    i know you're very critical and like to be very pedantic and picky about details(just like me) but do i need to state that? On my diagram i drew an arrow stating which way Vms^-1 was going so should i reverse the arrow and change my answer or just state that it's going in the opposite direction?
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    (Original post by thefatone)
    i know you're very critical and like to be very pedantic and picky about details(just like me) but do i need to state that? On my diagram i drew an arrow stating which way Vms^-1 was going so should i reverse the arrow and change my answer or just state that it's going in the opposite direction?
    I'm not being "very critical", I'm giving you advice.

    From an Edexcel M1 mark scheme:
    "Second A1 for 1.2u oe due E (or ‘reversed’ or ‘original direction of P)
    But A0 if just ‘changed’ or ‘to the right’ or ‘in positive direction’"

    Here's another one:
    "B1 for (original) direction of B or opposite to original direction (of A) oe.
    (B0 for ‘left’ or direction changed)."
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    (Original post by tiny hobbit)
    I'm not being "very critical", I'm giving you advice.

    From an Edexcel M1 mark scheme:
    "Second A1 for 1.2u oe due E (or ‘reversed’ or ‘original direction of P)
    But A0 if just ‘changed’ or ‘to the right’ or ‘in positive direction’"

    Here's another one:
    "B1 for (original) direction of B or opposite to original direction (of A) oe.
    (B0 for ‘left’ or direction changed)."
    oh okay thanks i'll make sure i put those next time
 
 
 
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