jinny1's Physics Questions Thread

[black]

quick question...2009 VCAA exam, Question 11

Question 11
The ball takes 9.0 s from the time Max hits it until it lands in the water. What is the height, h, of the cliff?

i used x = ut + 0.5at^2

where u = 0, t = 9 a = 10 x = ?

i got 405 metres, then minused the height the  ball rose initially, and ended up with a height of 360 metres

it wasn't the correct answer and i don't understand why doing it this way won't work?

[jinny1]

2. This is from TSSM 2010 exam,

(Image removed from quote.)

Question 14
Determine the spring coefficient of the mat required to bring the ball to rest at a height of
0.2 m. You must show your working.

I've attached the worked solutions below. It differs from what i did.

I did: 

GPE(f) + SPE = GPE(i) + KE(i)    and then solved for 'k' in the SPE     (i,f stand for inital and final respectively)

they seem to have forgotten that there was an initial GPE and just though of the final GPE.

or did i do something wrong???

The way i see that is the total energy at the top =



Now since the ball doesnt hit the bottom on it's way down, the height it has from the ground is it's new GPE which is



So the rest of the energy was converted to spring potential energy





Is this correct? :S

What about the GPE it loses as it drops from the top to 0.5m from the ground.??

[jinny1]

quick question...2009 VCAA exam, Question 11

Question 11
The ball takes 9.0 s from the time Max hits it until it lands in the water. What is the height, h, of the cliff?

i used x = ut + 0.5at^2

where u = 0, t = 9 a = 10 x = ?

i got 405 metres, then minused the height the  ball rose initially, and ended up with a height of 360 metres

it wasn't the correct answer and i don't understand why doing it this way won't work?

u shudve got a negative answer if you did it correctly, and that would be the height of the cliff..

[HCbigstick]

quick question...2009 VCAA exam, Question 11

Question 11
The ball takes 9.0 s from the time Max hits it until it lands in the water. What is the height, h, of the cliff?

i used x = ut + 0.5at^2

Working vertically, u = 30m/s  a = -10 m/s^2  t = 9s  x = ?
Use x = ut + .5at^2, you should end up with -135m relative to the top of the cliff. Therefore the height of the cliff is 135m.

where u = 0, t = 9 a = 10 x = ?

i got 405 metres, then minused the height the  ball rose initially, and ended up with a height of 360 metres

it wasn't the correct answer and i don't understand why doing it this way won't work?

[ttn]

2. This is from TSSM 2010 exam,

(Image removed from quote.)

Question 14
Determine the spring coefficient of the mat required to bring the ball to rest at a height of
0.2 m. You must show your working.

I've attached the worked solutions below. It differs from what i did.

I did: 

GPE(f) + SPE = GPE(i) + KE(i)    and then solved for 'k' in the SPE     (i,f stand for inital and final respectively)

they seem to have forgotten that there was an initial GPE and just though of the final GPE.

or did i do something wrong???

The way i see that is the total energy at the top =



Now since the ball doesnt hit the bottom on it's way down, the height it has from the ground is it's new GPE which is



So the rest of the energy was converted to spring potential energy





Is this correct? :S

What about the GPE it loses as it drops from the top to 0.5m from the ground.??

That was accounted for when he used h=0.2m for the mgh.

[jinny1]

2. This is from TSSM 2010 exam,

(Image removed from quote.)

Question 14
Determine the spring coefficient of the mat required to bring the ball to rest at a height of
0.2 m. You must show your working.

I've attached the worked solutions below. It differs from what i did.

I did:  

GPE(f) + SPE = GPE(i) + KE(i)    and then solved for 'k' in the SPE     (i,f stand for inital and final respectively)

they seem to have forgotten that there was an initial GPE and just though of the final GPE.

or did i do something wrong???

The way i see that is the total energy at the top =



Now since the ball doesnt hit the bottom on it's way down, the height it has from the ground is it's new GPE which is



So the rest of the energy was converted to spring potential energy





Is this correct? :S

What about the GPE it loses as it drops from the top to 0.5m from the ground.??

That was accounted for when he used h=0.2m for the mgh.

SHouldnt it be more than 0.2m??

Shouldnt it be the height at max height - 0.2m