I got something like 0.112J I don't remember exactly. What was the answer for 2nd mcq the one which said tensile force 60N is applied on 9cm wire and when 100N is applied it is 11 cm

Original post by Kamel1023

How much did you get for transfer of gravitational potential energy

Reply 67

usman200186

My answer was 0.26mm and in powers it was 10^-4. So I wrote it is not safe

Original post by 686868t

Do u remember the answerr for the YM extension question? was the final answer to the power of -4 or -5

Reply 68

686868t

Original post by usman200186

I got something like 0.112J I don't remember exactly. What was the answer for 2nd mcq the one which said tensile force 60N is applied on 9cm wire and when 100N is applied it is 11 cm

I have no idea i guessed it as 6cm. anyone?

Reply 69

usman200186

I did it by subtracting 90-70=20 which was angle with horizontal. Then I found weight which was 10.8N. Then we had to find tension for both wires, I used 2Tsin20=10.8 and this question was a repeated one and this was the same exact answer to that one too

Original post by ree.x09

Also what was the value for force? The rope was 70 degree to the vertical. Did i have to calculate the Tension or did i have to use the weight only? I might lose 1/2 marks in this question.

Reply 70

usman200186

Because law of energy conservation states that it cannot be created or destroyed but transformed to other forms in an isolated system. So some of it transferred to heat and sound energy.

Original post by 686868t

what about the energy question when you had to explain why the gpe is much less than the total energy thing question

Reply 71

usman200186

Original post by 686868t

I have no idea i guessed it as 6cm. anyone?

Yeah, 6cm is correct because we had to find the ratio. For every 40N the wire is imcreased by 2cm. So if we subtract 60 from 40 it will -2cm so 7 cm. Then 20N is left and if we substract it to we will lose 1 cm only because its half. So 7cm-1cm= 6cm

Reply 72

usman200186

Only one question that was the most confusing one was this.

How do we had to find the freaking time???

Reply 73

usman200186

It was easy.
For horizontal velocity, use s=ut+1/2at^2 with u = 0 and s= vertical height to find time. Then use v=s/t to find the velocity. My answer was 27.smthing
For vertical velocity, use v^2=u^2+ 2as with u=0. My answer was 3.2.
For magnitude, use pythagoras theorem using vertical and initial velocity. My answer was 31.2 something. Then find the angle using tan^-1, my angle was 6.7.

Original post by 686868t

there were 2 questions that were ‘show’ pretty straight forward. i think i used 1/2at^2 for the first part and got 27.6 m/s
Second part was v^2=u^2+2as and i forgot the value i got but around 2. smthn