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Mechanics- work energy principle and power

A golf ball of mass 45.9g is hit from a tee with speed 180km/h . The ball rises to a height of 20m, having traveled along a curved path of length 61.875m. At the highest point of its path the ball is travelling at 144km/h.
a) Find the magnitude of the avg resistance acting on the ball
I calculated this and got resistance = 0.2N

The ball travels a further 105.8m along a curved path to land on the green. The green is 4m lower than the tee. The average resistance remains unchanged.

b) find the speed of the ball just before it lands on the green
Iam getting the ans v= 36m/ s but the textbook ans is 35m/s
m= 0.0459kg, u= 40m/s, s= 105.8- 61.875=43.925m, v=?
Increase in KE = 0.02295v^2-36.72J
Increase in GPE= mgh =-1.836J
Work done against resistance =-8.785J
When I do total mechanical energy = total work done, I get v= 36m/s

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Reply 1
Original post by Shas72
A golf ball of mass 45.9g is hit from a tee with speed 180km/h . The ball rises to a height of 20m, having traveled along a curved path of length 61.875m. At the highest point of its path the ball is travelling at 144km/h.
a) Find the magnitude of the avg resistance acting on the ball
I calculated this and got resistance = 0.2N

The ball travels a further 105.8m along a curved path to land on the green. The green is 4m lower than the tee. The average resistance remains unchanged.

b) find the speed of the ball just before it lands on the green
Iam getting the ans v= 36m/ s but the textbook ans is 35m/s
m= 0.0459kg, u= 40m/s, s= 105.8- 61.875=43.925m, v=?
Increase in KE = 0.02295v^2-36.72J
Increase in GPE= mgh =-1.836J
Work done against resistance =-8.785J
When I do total mechanical energy = total work done, I get v= 36m/s

It would help to see the full working, but for a) I get 0.188N which may account for the later "rounding error"?
Reply 2
Original post by mqb2766
It would help to see the full working, but for a) I get 0.188N which may account for the later "rounding error"?

The rounding to one decimal place is the ans in the textbook. That's the reason I wrote the resistance as 0.2N
Reply 3
Original post by mqb2766
It would help to see the full working, but for a) I get 0.188N which may account for the later "rounding error"?

I have posted the working for b
Reply 4
Original post by Shas72
The rounding to one decimal place is the ans in the textbook. That's the reason I wrote the resistance as 0.2N

Still help to see your working though. If you're rounding to 1 sig fig, its hard to worry about later answers which are out by 2 sig figs. Have you tried the more accuate resistance?
(edited 2 years ago)
Reply 5
ok sure I will
Reply 6
Note I don't understand the s=43? The ball travels a further 105m, so if you're taking ti from the peak, s=105?
But I don't see how you've defined terms (full working).
Reply 7
so m= 0.0459 kg, u= 50m/s, h=20m, s=61.875m, v=144m/s
increase in KE= -20.655J
increase in GPE= mgh = 9.18J
workdone against resistance = -FΓ—S= -61.875F
total mechanical energy = total work done
-20.655+9.18= -61.875F
F=0.185N
Reply 8
sorry v=40m/s
Reply 9
Guessing you;re taking g=10 rather than my 9.8? Pls upload the full working in one post preferably if you want it checked.
(edited 2 years ago)
Reply 10
Original post by mqb2766
Guessing you;re taking g=10 rather than my 9.8? Pls upload the full working in one post preferably if you want it checked.

For q(b)
m=0.0459kg, u= 50m/s, v=?, resistance = 0.2N, s= 105.8m, h=20-4=16m
Increase in KE= 1/2mv^2-1/2mu^2=0.02295v^2-57.375 J
Increase in GPE= mgh= 7.344J
Work done against resistance = -FΓ—S=-21.16J
Total mechanical energy = total work done
I get v= 35 m/s
Reply 11
Original post by mqb2766
Guessing you;re taking g=10 rather than my 9.8? Pls upload the full working in one post preferably if you want it checked.

Iam stuck with the next question (c)
The ball is travelling vertically when it lands on the green where it is immediately brought to rest.
c) show that the energy absorbed by the green is 30.3J
Can you pls help me with the solution and how to calculate
Original post by Shas72
Iam stuck with the next question (c)
The ball is travelling vertically when it lands on the green where it is immediately brought to rest.
c) show that the energy absorbed by the green is 30.3J
Can you pls help me with the solution and how to calculate

So when the ball comes to rest, what energy is lost by the ball?
Reply 13
mechanical energy
Original post by Shas72
mechanical energy

Sort of, but I don't know a formula for that. Can you be more specific?
Reply 15
Original post by mqb2766
So when the ball comes to rest, what energy is lost by the ball?

KE + GPE is lost by the ball
Original post by Shas72
KE + GPE is lost by the ball

Again sort of. You might want to sketch the scenario of the ball impacting with the green and think about how those terms are represented for an impact.

For part b), a sketch would come in handy (pls upload) to make sure the distances, height, signs are correct.
Reply 17
Iam not getting the ans
I took the KE just before it falls on the ground= -28.11J
final GPE before it falls= -1.836 J taking h=4m
work done against resistance = -0.2Γ—4=-0.8J
so total comes to 30.7J lost by the ball gained by the green.
can you pls help if you can? I have no clue how to get the text book ans 30.3J. pls help
Original post by Shas72
Iam not getting the ans
I took the KE just before it falls on the ground= -28.11J
final GPE before it falls= -1.836 J taking h=4m
work done against resistance = -0.2Γ—4=-0.8J
so total comes to 30.7J lost by the ball gained by the green.
can you pls help if you can? I have no clue how to get the text book ans 30.3J. pls help

As per previous post, can you upload your sketch with the values for the tee, highest point, green etc with the corresponding working You seek to be getting confused over which distance is which and how they're used. I simply don't undersstand your working enough.
(edited 2 years ago)
Reply 19
I have understood the first and the second question. I don't know how to calculate the energy absorbed by the green . i haven't drawn diagram as I have taken upwards as gain in gpe and loss of KE, downwards there is gain in KE and loss of GPE, height upwards is positive and downwards is negative.

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