conservation of energy in a system with non conservative forces.Ex 9C.Question 9.

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alevelmath1
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Q. A golf ball of mass 45.9 g is hit from a tee with speed 50 m/s .The ball rises to a height of 20 m, having travelled along a curved path of length 61.875 m. At the highest point of its path the ball is traveling at 40 m/s .

a) Find the magnitude of the average resistance force acting on the golf ball.

I have found it correctly as 51/275 N

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

Find the speed of the ball just before it lands on the green.
I also found this speed 35 m /s

c) The ball is travelling vertically when it lands on the green,where it is immediately brought to rest.
show that the energy absorbed by the green is 30.3 J.

I have a difficulty to understand the last part as what is the meaning of "energy absorbed "
as I know the the kinetic energy with mass 0.0459 kg is
(1/2)(0.0459)(35^2) = 28.11375 J.

My question is how I get 30.3 J. from this kinetic energy.
[ take g= 10( m/s)/s]
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RogerOxon
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(Original post by alevelmath1)
Q. A golf ball of mass 45.9 g is hit from a tee with speed 50 m/s .The ball rises to a height of 20 m, having travelled along a curved path of length 61.875 m. At the highest point of its path the ball is traveling at 40 m/s .

a) Find the magnitude of the average resistance force acting on the golf ball.

I have found it correctly as 51/275 N

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

Find the speed of the ball just before it lands on the green.
I also found this speed 35 m /s

c) The ball is travelling vertically when it lands on the green,where it is immediately brought to rest.
show that the energy absorbed by the green is 30.3 J.

I have a difficulty to understand the last part as what is the meaning of "energy absorbed "
as I know the the kinetic energy with mass 0.0459 kg is
(1/2)(0.0459)(35^2) = 28.11375 J.

My question is how I get 30.3 J. from this kinetic energy.
[ take g= 10( m/s)/s]
I've not done the calculations, but your approach is correct. The "energy absorbed" is purely the ball's kinetic energy upon landing. I don't understand how (I do why) they can assert that the ball lands vertically though. Had I have written the question, I'd have said that it lands at 90 degrees to the ground.
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