HOW WAS PHYSICS PAPER 2 AS!!!!!!!!!!!

[Saad Sarfraz]

OUTPUT POWER is the effective power once you subtract the work done against friction. Have you ever calculated efficiency of a system? If you have then you might remember that output power is the effective power.

Same here i guess most people wrote NO but i discussed this with my sir he said that both answers could be right since he was asking your opinion i went with yes caz output power is always usefull work done. OLEVEL CONCEPT

 

[Ahmed Tariq]

Oh, and also, for the very last part of question two where you had to state and explain the over all change in energy, what did you write?
only about the conversion of GE to KE at the plate and then again about the conversion of KE to GE at B?

At A, the ball has both kinetic energy and potential energy which is converted to the internal energy of the atoms of the ball and ground when it hits the ground, which is again converted to kinetic and potential energy at B.

 

[Hertz]

is it 500? or is it 1000?

500

 

[rabiafarooq101]

you don't understand the ques dudette

No sweetheart I do. Have you ever included the heat loss in the power output when you calculate the efficiency? No right. i think you just need to clarify you concepts before you even speak. DUDE.

 

[oldfashionedgirl]

in phy paper i wrote that micro meter screw guage will be used to calculate cross-sectional are rather than dia will be measured by micro meter and then area???? will i be penalised???

it depends on the marks that question carried which were not that much....so u might not be penalized if u wrote a direct answer although it would have been better to describe it just to be on the safe side

 

[rabiafarooq101]

total output power is just work done/time taken....not useful work done/time taken.............i say my car has a 20Kilo Watt powered engine, in order to calculate top speed i have to subtract power against resistence first.......it isnt included as you assume...any how..gain in pe was less as the motor does work against resistive forces too.

Now you're just confusing the total input power and output power. The TOTAL output power is always the effective work done and effective means subtracting the frictional force. Cheers!

 

[Unicorn]

hahaa dw same as my reaction i think dis variant 2

most likely

 

[JD REBORN]

PE was 2.02J.
and you had to use the difference in height between A and B

Explain how u did it

 

[JD REBORN]

Can someone explain how they calculated KE change and PE change?

 

[good man12]

Can someone explain how they calculated KE change and PE change?

my P.E was mgh= (0.050)(9.81)(5.0-0.9)=2.011= which i wrote as 2.0 as the data had a min of 2 sig fig..

 

[mhalvi]

my P.E was mgh= (0.050)(9.81)(5.0-0.9)=2.011= which i wrote as 2.0 as the data had a min of 2 sig fig..

Exactly!!!btw how many marks were there for this KE and PE question?

 

[good man12]

Exactly!!!btw how many marks were there for this KE and PE question?

i think 2 for kinetic en, and 3 for mgh as we had to calc final height also

 

[AppleGreen]

Now you're just confusing the total input power and output power. The TOTAL output power is always the effective work done and effective means subtracting the frictional force. Cheers!

Or should I say dudette. You really need to study. A levels is ending. I'm sure you'll be able to complete your A levels in the next five years. Cheers

marking scheme will prove that who is the confused 1. LOL if i gotta complete in next five years, see you graduating from pre-u in next 10 years
http://www.a-levelphysicstutor.com/m-kinetics-power-efficiency.php
go and study the question

 

[littlecloud11]

Explain how u did it

well the height of A was 5m
and when you use v^2 =u^2 +2as ( v=0, u= 4.2) after the ball rebounds you find that the height of B above the ground is .9m
so GE= .05 * 9.8 * (5-.9)

 

[fathimaash]

wat is da K.E?

 

[rogue94]

wat is da K.E?

KE was 1.8 J and GPE was 2.0 J

 

[fathimaash]

really i got KE as 3.76
0.5*0.05*(13^2 - 4.3^2) = 3.76j
it hits wit speed 13ms-1 nd leaves wit speed be 4.3ms-1. so change in KE is as shown.

 

[AppleGreen]

really i got KE as 3.76
0.5*0.05*(13^2 - 4.3^2) = 3.76j
it hits wit speed 13ms-1 nd leaves wit speed be 4.3ms-1. so change in KE is as shown.

initial u= 8.2 and v= 0 so KE= 1.76

 

[AdeelRox]

initial u= 8.2 and v= 0 so KE= 1.76

initial was 8.4 and not 8.2 but the anser is crect

 

[WhereAmazingHappens]

really i got KE as 3.76
0.5*0.05*(13^2 - 4.3^2) = 3.76j
it hits wit speed 13ms-1 nd leaves wit speed be 4.3ms-1. so change in KE is as shown.

they wanted k.e from A to B
initial speed is 8.4 or something and final speed is 0

cheers