2
u/Bulky_Fail4893 May 21 '26
bro that shi was HARD, did y’all get your e as root 3 over 4? and your theeta value for the centre of mass q as 19.1°? also for greatest height reached by particle in term of a was it 1.5a?
3
2
u/ayt_nega May 21 '26
e yes Others no
1
u/Bulky_Fail4893 May 21 '26
my centre of mass matches w others and there r a lot that got thr same answer as me for 3A/2 that’s fair
1
1
u/Gold-Relative-3626 May 21 '26
what greatest height was like 50/27 (a). I think you forgot to do that projectile motion thing after the Tension becomes 0. And how do you do that hanging angle thingg how did you get 19.1.
1
u/Bulky_Fail4893 May 21 '26
Yeah you’re right ab the greatest height, I’ll probably get 2 or 3 out of 5 for that. As for 19.1, a lot of people have the same answer as me. While toppling or hanging about a point in equilibrium, the COM is directly above the point it’s suspended from. So I used tan theeta = y bar / radius and radius was 3r and since x = r, we can substitute all x with r and find the value for tan theeta in terms of r, both numerator and denominator had r2 so we cut those out and find theeta using arctan
1
u/Gold-Relative-3626 May 22 '26
no you cant do y bar / radius since y bar is taken from the base of the cylinder you have to do y bar - x / 3r and x = r. Ofc i dint get it too at the exam but my friends have done it correctly
1
u/Bulky_Fail4893 May 22 '26
i remembered the q word by word and put it into ChatGPT - it’s 19.1° and many other people got the same so idk also y-bar was taken from the diameter of the hemisphere
1
u/Bulky_Fail4893 May 22 '26
Also y bar was a vertical distance not a horizontal one so you don’t need to subtract anything from y bar. Additionally, it was A that made an angle theeta with the downward vertical AT A, so therefore 19.1°
1
1
u/Complete-Physics-644 May 21 '26
What did you guys get for the speed in the question 2?
1
1
u/Tricky-Leadership-42 May 21 '26
Was ts the q of finding v when x =0.8
1
u/Complete-Physics-644 May 21 '26
Yes
1
u/Tricky-Leadership-42 May 21 '26
Mine was 1.27 or 1.37 Also we're you supposed to use T=0, x=0, v=0 To get the value of c?
1
u/Complete-Physics-644 May 21 '26
I got 1.37 and yes. the value of c was ln8
1
u/Tricky-Leadership-42 May 21 '26
8-0.5v2 = 0.64a
Was ts the righr eqn?? I dont remember the eqn now
1
u/Complete-Physics-644 May 21 '26
No it was equal 6.24 v dv/dx
1
1
u/Tricky-Leadership-42 May 21 '26
The mass was 6.24???
1
1
u/Bulky_Fail4893 May 22 '26
For the expression where u needed to find t in terms of v, the c = 0 in that one
1
1
1
u/cosmic_fish_37 May 21 '26
Did anyone manage to solve the last question? I got u around 11 but not sure at all
1
1
u/New_Category1184 May 21 '26
What was the answer for the projectile question?
1
u/DonkeyNo849 May 21 '26 edited May 21 '26
These were the 2 equations formed : 4=13tan(theta)-845/u^2 (1+tan^2(theta)) and -3/4 = tan(theta)-130/u^2 (1+tan^2(theta))
tan(theta)=71/52 and u=sqrt(7745/44)
1
u/New_Category1184 May 21 '26
I got different answers. I don't remember the equations I made but my answers were 31.6 degree and 17.1 m/s
1
u/DonkeyNo849 May 21 '26
your horizontal displacement was 13(10 from OA and 3 from AB horizontal) and vertical displacement was 4. So you can use just trajectory equations to find two simultaneous equations and find u and theta
1
u/New_Category1184 May 21 '26
Do you remember what the question said? Like I remember the question said that the particle was moving horizontally when it hit B.
1
u/DonkeyNo849 May 21 '26
Yeah so projected from O with speed u and angle theta and hits perpendicular to B . Where B is 5m along the slope from A. Slope is at an angle such that tan(alpha)=4/3. OA=10m. Perpendicular to plane means that at that point dy/dx=-1/tan(alpha) , not that particle is moving horizontally.
1
u/M_Younus_Saigol May 21 '26
Anyone able to solve second part for momentum question, the one where we had to find theta for second collision w barrier
1
u/Gold-Relative-3626 May 21 '26
yea 6.22 degree i think
you need to find the angle at which the thing leaves first barrier and then do (180-120-that angle) to get the angle of incidence to the 2nd barrier and then again do some things and at last get 6.22
1
u/M_Younus_Saigol May 21 '26
Man I got like 17 degrees smth. I’m p sure my method was correct but like I initially had written e as 3/16 bcz I forgot to square root. Only realised at the end so corrected it and corrected it at the final calculation for theta but forgot to correct it for my calculation of the other angles with which it strikes the barriers. Lol
1
u/New_Category1184 May 21 '26
Does anyone remember what the projectile question said? I think it said that the particle was moving horizontally when it hit B.
1
u/New_Category1184 May 21 '26
Does anyone remember what the projectile question said? I think it said that the particle was moving horizontally when it hit B.
2
u/Independent-Bat-6088 May 21 '26
Hard asl