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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.

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.

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.

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 ):

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.

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 ):

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.

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 ):

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?

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

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

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 ):

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.

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 ):

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...

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 ):

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.

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 ):

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....

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 ):

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."

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.

- Changing degree in 2nd year?
- Is studying to become an electrician not good enough?
- 4 A-levels with Physics for Maths degree?
- Isaac Physics SPC Summer School
- As topics structure (maths/physics)
- What A-Levels for Neurosurgery
- Maths and physics super curriculars
- A-Levels: Bio, Chem, Maths, English Lit for Medicine/Dentistry? is this a good combo?
- Natural Sciences Vs Chemistry
- Physics a level without maths a level
- Physics at Nottingham
- revision
- How do you tell when you don't want to do your course any more?
- switching degrees before first year
- I got E in as level
- Physics or English A level
- GYG
- Complicated situation with bf. Back together again
- Where can I study general relativity?
- Physics and biology A-level come back gyg

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