For everyone asking about the ball question, here is what i did: For the first question to find the maximum height, the time was 6/20 because 20 was 1 second. And then use of equation s=ut+1/2at^2 because initial is zero as it falls from maximum height, all gpe no ke. You should get around 0.44m. Then for the second part, use the picture, if 0.44 was actual height, 6 cm was height on picture, that means total length(horizontal) was 9.3 cm. Convert both cm to meter. Use: If 6cm is 0.44m what is 9.3cm. You should get horizontal distance is 0.682m. Then use simple distance/time. Time for total should be 20/20 is 1 second and + 4/20 as it was 24 pictures and 20 was one second, so you need to add the extra 4, time should be 1.2. So velocity is 0.568m/s, atleast this is what i got, hopefully, it's correct.
I think you did most things right but then how come the horz. distance be greater than vertical fallen distance! ...0.6>0.44 ?? I got something like about 2 m..and the falling time is same as the time required to travel the horz.. As it was just a half projectile..so no required to do the time of flight..
exactly. we had to take those 6 pics in account between x and y not all
I'm quite sure they only asked about the horizontal velocity, not sure if i remember them asking specifically between x and y, that would seem a bit odd.
for that horiz velocity question, it said "use the photograph". so you had to measure the horiz distance between X and y and it came to be like 1.5cm. then use the scale from the prev question where 5.5 cm on pic was 0.44m 5.5 cm ----- 0.44m 1.5 cm ------0.12m
yeah you should get the mark.. I also said plot a graph of v against r^2.. then compare the equation for viscosity with y=mx+c.. then simply substitute m with eta( n )
what if we took weight of ball on y axis and 6pirv on x axes and get the gradient it was like wieght = n6pirv + upthrust y =m x + c
I'm quite sure they only asked about the horizontal velocity, not sure if i remember them asking specifically between x and y, that would seem a bit odd.
they hadnt mentioned that though. The question was ambiguous -_-
I'm quite sure they only asked about the horizontal velocity, not sure if i remember them asking specifically between x and y, that would seem a bit odd.
Yeah if thats what was asked to find then u r right..but i think it was a)ii) ques..so if i havent mistaken the ques. we were just asked to deal with the (x to y)..dont think they could have asked to do u anything more for a 4 mark ques we would alreday be doing a lot by scaling the distances from the image..not sure though u may be right only the ques papers can solve our ques. so will have to wait until **physics and maths tutor** post the model ans on their fb page..may b wait a couple of days 😊
but i made it on time. i started the paper from back xD
Yup fully agreed ..it was d most lenghtiest paper i did among all the ial papers..i was also seating for the jan 16 paper..the 18 ques papers are much better..
I think you did most things right but the how come the horz. distance be greater than vertical fallen distance! ...0.6>0.44 ?? I got something like about 2 m..and the falling time is same as the time required to travel the horz.. As it was just a half projectile..so no required to do the time of flight..
I did it for the entire, as in total velocity, because i dont remember them asking velocity specifically between X and Y, may be wrong, maybe both are correct
I did it for the entire, as in total velocity, because i dont remember them asking velocity specifically between X and Y, may be wrong, maybe both are correct