Physics: Post your doubts here!

[princess Anu]

If there is any region where the velocity is negative you would need to treat each of those sections as yet more pieces. In those pieces the DISTANCE is positive and is added on prior to calculating average speed.[/QUOTE]

Yes the answer is 0.15m
Btw If we have to find average velocity, the displacement will be -ve in those regions of the section where velocity is below zero, right?

 

[Thought blocker]

If there is any region where the velocity is negative you would need to treat each of those sections as yet more pieces. In those pieces the DISTANCE is positive and is added on prior to calculating average speed.

Yes the answer is 0.15m
Btw If we have to find average velocity, the displacement will be -ve in those regions of the section where velocity is below zero, right?[/QUOTE]
Find the time with, t = s/v = 5.5/32 = 0.171875
Then use s = 1/2at^2 = 5*0.171875^2 = 0.15m

The DISPLACEMENT is negative and is subtracted before calculating average velocity.

 

[princess Anu]

why can't we directly use v^2= u^2+2as ?

 

[Thought blocker]

why can't we directly use v^2= u^2+2as ?

I bet you cannot get answer with that equation.
We need time to calculate that distance asked.
SO we used s/v
Use everything given in the question and go according to the format.

 

[princess Anu]

Yea i know But WHYYY
We've got everything except displacement why can't we use this eq

 

[princess Anu]

Yes the answer is 0.15m
Btw If we have to find average velocity, the displacement will be -ve in those regions of the section where velocity is below zero, right?

Find the time with, t = s/v = 5.5/32 = 0.171875
Then use s = 1/2at^2 = 5*0.171875^2 = 0.15m

The DISPLACEMENT is negative and is subtracted before calculating average velocity.[/QUOTE]
The eq u used isn't it incomplete ? where is u*t? initial velocity is not even O in this case?

 

[Thought blocker]

Yea i know But WHYYY
We've got everything except displacement why can't we use this eq

Physicist I am unable to help her. I am not much of any help. Help her.

 

[Thought blocker]

Find the time with, t = s/v = 5.5/32 = 0.171875
Then use s = 1/2at^2 = 5*0.171875^2 = 0.15m

The DISPLACEMENT is negative and is subtracted before calculating average velocity.

The eq u used isn't it incomplete ? where is u*t? initial velocity is not even O in this case?[/QUOTE]
It is 0

Physicist help her.

 

[Awesome12]

Yea i know But WHYYY
We've got everything except displacement why can't we use this eq

See in such questions related to projectile motion, we need to deal with the horizantal and vertical components separately.
The ball is hit HORIZANTALLY, with a speed of 32 m/s. The HORIZANTAL distance is 5.5 m.


Since the ball is hit horizantally there is no VERTICAL omponent of speed, The acceleration is 10m/s. Acceleration due to gravity is itself a VERTICAL component and not a horizantal. We need to find the distance the ball has dropped. This is the vertical component of distance.

You cant use the equation you stated because:
if we were to use the equation for the horizantal component, we wouldnt be able to get an answer as the "a" in the formula will have no value. LIKE I SAID BEFORE ACCELERATION DUE TO GRAVITY IS PART OF THE VERTICAL COMPONENT ONLY.

 

[SIstudy]

Can somebody plis help me with the P3 Questions
like in the Q1 after graph he gives some formula and we have to substitute
I cant just solve these questions
for eg Q1 part (G) of this paper http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w03_qp_3.pdf
(for part (F) the gradient was 5.95x10-3)
and y-intercept qas 5.84
Q1 (G) of this ppr http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s08_qp_31.pdf

 

[janasalem]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w04_qp_2.pdf
question 3 is the maximum height the top of the loop or the top of h in the diagram ?

 

[SIstudy]

explanation are available at
http://physics-ref.blogspot.com/2014/05/physics-9702-notes-worked-solutions-for.html

(at the bottom)

hope it helps

There's no P3 solved on this site :/

 

[adarsha]

did not have the answer to this one....
A ball having a momentum p hits a bat and its momentum becomes -P. what is the change in momentum of the ball?

 

[snowbrood]

did not have the answer to this one....
A ball having a momentum p hits a bat and its momentum becomes -P. what is the change in momentum of the ball?

just calculate the change p-(-p)=2p

 

[Thought blocker]

just calculate the change p-(-p)=2p

Change in momentum = Final momentum - Initial momentum
Hence in shouldn't be -P - P = -2P?

 

[adarsha]

i did this and just was wondering if i got it correct

changed momentum = initial momentum - negative momentum
because in momentum energy is not lost but transferred.

where negetive momentum is the momentum of the bat...

still correct?

thanks again guys

 

[Thought blocker]

-->

Line 1: The change in momentum equals the mass times the change in velocity.

Line2: The change in velocity is the final velocity minus the original velocity.

Line 3: Distribute the m.

Line 4: The change in momentum equals the final momentum minus the original momentum. Not initial minus final.

 

[marduk]

can someone please provide me with As level PRACTICAL tips and tricks. Any help will be highly appreciated

 

[Thought blocker]

https://www.xtremepapers.com/community/threads/physics-practical-tips.6306/

can someone please provide me with As level PRACTICAL tips and tricks. Any help will be highly appreciated

 

[janasalem]

On a particular railway, a train driver applies the brake of the train at a yellow signal, a distance of
1.0km from a red signal, where it stops.
The maximum deceleration of the train is 0.2ms–2
.
Assuming uniform deceleration, what is the maximum safe speed of the train at the yellow signal?
A 20ms–1 B 40ms–1 C 200ms–1 D 400ms–1