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    Got this of Trinity Prep Paper for NatSci::: Help!

    A golfer driving from the tee swings his club so that the head completes a (virtually) complete
    circle in 0.5 s. Make a reasoned estimate of the maximum distance he could hit the ball.

    I've tried momentum, energy just can't get the speed of the ball.
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    (Original post by xaviars)
    Got this of Trinity Prep Paper for NatSci::: Help!

    A golfer driving from the tee swings his club so that the head completes a (virtually) complete
    circle in 0.5 s. Make a reasoned estimate of the maximum distance he could hit the ball.

    I've tried momentum, energy just can't get the speed of the ball.
    T=0.5 sec
    estimate r = 1

    v=rw (v of club)
    w=2pi/T

    v (club) = 1 * (2pi/0.5) = 12.6 ms^-1 (1dp)

    conservation of momentum (assuming club stops when it hits the ball lol :rolleyes: ):

    m1v1 = m2v2
    v2 = m1v1/m2

    estimate mass of club to be 1 kg, ball to be 0.1 kg
    m1 = 1
    m2 = 0.1

    v2 (of ball) = 1*12.6/0.1 = 126 ms^-1

    assume he hits it at say 45 degrees

    horizontal component (to get time in air) = 126 sin 45 = 88.9 ms^-1

    v = u + at
    t = (v - u)/a

    a = -9.8 ms^-2 (ie 'g')

    t = (0 - 88.9)/-9.8 = 9.1 s

    therefore distance travelled s = time in air * initial velocity (horizontal component) (assuming no air resistance)

    s = 9.1 * (12.6 cos 45) = 80.6 m ie ~80m ?


    kinda sounds about right... i dont know much about golf lol. as you need a maximum, id put in bigger mass for weight of club, larger height of club, smaller mass for ball, different angle etc.

    i think the method is ok(ish) apart from the cons. of momentum bit, which seems a bit iffy because it assumes all the momentum is transferred to the ball from the club.
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    Wouldn't you round to 81 m?
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    (Original post by edders)
    T=0.5 sec
    estimate r = 1

    v=rw (v of club)
    w=2pi/T

    v (club) = 1 * (2pi/0.5) = 12.6 ms^-1 (1dp)

    conservation of momentum (assuming club stops when it hits the ball lol :rolleyes: ):

    m1v1 = m2v2
    v2 = m1v1/m2

    estimate mass of club to be 1 kg, ball to be 0.1 kg
    m1 = 1
    m2 = 0.1

    v2 (of ball) = 1*12.6/0.1 = 126 ms^-1

    assume he hits it at say 45 degrees

    horizontal component (to get time in air) = 126 sin 45 = 88.9 ms^-1

    v = u + at
    t = (v - u)/a

    a = -9.8 ms^-2 (ie 'g')

    t = (0 - 88.9)/-9.8 = 9.1 s

    therefore distance travelled s = time in air * initial velocity (horizontal component) (assuming no air resistance)

    s = 9.1 * (12.6 cos 45) = 80.6 m ie ~80m ?


    kinda sounds about right... i dont know much about golf lol. as you need a maximum, id put in bigger mass for weight of club, larger height of club, smaller mass for ball, different angle etc.

    i think the method is ok(ish) apart from the cons. of momentum bit, which seems a bit iffy because it assumes all the momentum is transferred to the ball from the club.
    Ur suppose to be creative and make assumptions?
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    (Original post by Nylex)
    Wouldn't you round to 81 m?
    no, because its an estimate, so i gave it to the nearest 5 m
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    (Original post by 2776)
    Ur suppose to be creative and make assumptions?
    :rolleyes: ok lets see you do something clever
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    Thanx, I was using standard velocity equations, using the circumference of a circle, only met this recently in SHM,

    Thanks, preparing for interview on wednesday!! lol
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    (Original post by edders)
    :rolleyes: ok lets see you do something clever
    No, i was asking a question, are u suppose to like assume a club is like 1kg. What happens if u never played golf before?
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    (Original post by 2776)
    No, i was asking a question, are u suppose to like assume a club is like 1kg. What happens if u never played golf before?
    oh ok, your question was misleading. yeah, you have to assume values for the necessary variables, otherwise you wouldnt be able to do it
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    (Original post by xaviars)
    Thanx, I was using standard velocity equations, using the circumference of a circle, only met this recently in SHM,

    Thanks, preparing for interview on wednesday!! lol
    no probs, it was good practice for me too!! ive got an interview at exeter college, oxford, for physics&philosophy next sunday. ive done sod all revision :eek:
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    (Original post by edders)
    oh ok, your question was misleading. yeah, you have to assume values for the necessary variables, otherwise you wouldnt be able to do it
    so what do gonna do in uni?
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    (Original post by edders)
    T=0.5 sec
    estimate r = 1

    v=rw (v of club)
    w=2pi/T

    v (club) = 1 * (2pi/0.5) = 12.6 ms^-1 (1dp)

    conservation of momentum (assuming club stops when it hits the ball lol :rolleyes: ):

    m1v1 = m2v2
    v2 = m1v1/m2

    estimate mass of club to be 1 kg, ball to be 0.1 kg
    m1 = 1
    m2 = 0.1

    v2 (of ball) = 1*12.6/0.1 = 126 ms^-1

    assume he hits it at say 45 degrees

    horizontal component (to get time in air) = 126 sin 45 = 88.9 ms^-1

    v = u + at
    t = (v - u)/a

    a = -9.8 ms^-2 (ie 'g')

    t = (0 - 88.9)/-9.8 = 9.1 s

    therefore distance travelled s = time in air * initial velocity (horizontal component) (assuming no air resistance)

    s = 9.1 * (12.6 cos 45) = 80.6 m ie ~80m ?


    kinda sounds about right... i dont know much about golf lol. as you need a maximum, id put in bigger mass for weight of club, larger height of club, smaller mass for ball, different angle etc.

    i think the method is ok(ish) apart from the cons. of momentum bit, which seems a bit iffy because it assumes all the momentum is transferred to the ball from the club.
    wow,that looks...interesting...
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    BTW any tips for Nat Sci (Physics) at Trinity Cam!??!?
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    (Original post by edders)
    T=0.5 sec
    estimate r = 1

    v=rw (v of club)
    w=2pi/T

    v (club) = 1 * (2pi/0.5) = 12.6 ms^-1 (1dp)

    conservation of momentum (assuming club stops when it hits the ball lol :rolleyes: ):

    m1v1 = m2v2
    v2 = m1v1/m2

    estimate mass of club to be 1 kg, ball to be 0.1 kg
    m1 = 1
    m2 = 0.1

    v2 (of ball) = 1*12.6/0.1 = 126 ms^-1

    assume he hits it at say 45 degrees

    horizontal component (to get time in air) = 126 sin 45 = 88.9 ms^-1

    v = u + at
    t = (v - u)/a

    a = -9.8 ms^-2 (ie 'g')

    t = (0 - 88.9)/-9.8 = 9.1 s

    therefore distance travelled s = time in air * initial velocity (horizontal component) (assuming no air resistance)

    s = 9.1 * (12.6 cos 45) = 80.6 m ie ~80m ?


    kinda sounds about right... i dont know much about golf lol. as you need a maximum, id put in bigger mass for weight of club, larger height of club, smaller mass for ball, different angle etc.

    i think the method is ok(ish) apart from the cons. of momentum bit, which seems a bit iffy because it assumes all the momentum is transferred to the ball from the club.

    hmmmmmm That isn't very realistic - gold balls are hit much farther than that! I'm not saying you're wrong....
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    (Original post by edders)
    T=0.5 sec
    estimate r = 1

    v=rw (v of club)
    w=2pi/T

    v (club) = 1 * (2pi/0.5) = 12.6 ms^-1 (1dp)

    conservation of momentum (assuming club stops when it hits the ball lol :rolleyes: ):

    m1v1 = m2v2
    v2 = m1v1/m2

    estimate mass of club to be 1 kg, ball to be 0.1 kg
    m1 = 1
    m2 = 0.1

    v2 (of ball) = 1*12.6/0.1 = 126 ms^-1

    assume he hits it at say 45 degrees

    horizontal component (to get time in air) = 126 sin 45 = 88.9 ms^-1

    v = u + at
    t = (v - u)/a

    a = -9.8 ms^-2 (ie 'g')

    t = (0 - 88.9)/-9.8 = 9.1 s

    therefore distance travelled s = time in air * initial velocity (horizontal component) (assuming no air resistance)

    s = 9.1 * (12.6 cos 45) = 80.6 m ie ~80m ?


    kinda sounds about right... i dont know much about golf lol. as you need a maximum, id put in bigger mass for weight of club, larger height of club, smaller mass for ball, different angle etc.

    i think the method is ok(ish) apart from the cons. of momentum bit, which seems a bit iffy because it assumes all the momentum is transferred to the ball from the club.
    With this amount of uncertainty it would almost be just as good to say "Well, I know that a golfer can hit a ball somewhere around 80m."
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    (Original post by Jonatan)
    With this amount of uncertainty it would almost be just as good to say "Well, I know that a golfer can hit a ball somewhere around 80m."
    Well, that's not the point - the point is to see if you can make an EDUCATED guess - they want to see how you would calculate it - i.e. how your brain works.
 
 
 

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