# Physics Again..

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.
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 ):

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.
Wouldn't you round to 81 m?
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.

Ur suppose to be creative and make assumptions?
Nylex
Wouldn't you round to 81 m?

no, because its an estimate, so i gave it to the nearest 5 m
2776
Ur suppose to be creative and make assumptions?

ok lets see you do something clever
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
edders
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?
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
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
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?
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.

wow,that looks...interesting...
BTW any tips for Nat Sci (Physics) at Trinity Cam!??!?
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.

hmmmmmm That isn't very realistic - gold balls are hit much farther than that! I'm not saying you're wrong....
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.

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.