AQA A-level Further Mathematics Paper 3 Mechanics (7367/3M) - 14 June 2023

OP

Original post by ed729781

messed up circular motion (question 6) and i honestly don’t know about question 8? everyone i’ve spoken to seems to have gotten a different highest velocity

i think for the most part the paper was really good though considering i struggle w mechanics generally

I think I got around 5.3 for max velocity of the block on q8 by considering: Work done by pulling force - Work done against friction - Elastic potential = Kinetic.

For the tension question I got T=mgsec(theta) then differentiated and showed max tension is at theta = 0 and I also got max velocity of the pendulum as something around 2 by doing mg(change in h)=1/2mv^2

Reply 7

Fort21

6

Original post by raducorha

I think I got around 5.3 for max velocity of the block on q8 by considering: Work done by pulling force - Work done against friction - Elastic potential = Kinetic.

For the tension question I got T=mgsec(theta) then differentiated and showed max tension is at theta = 0 and I also got max velocity of the pendulum as something around 2 by doing mg(change in h)=1/2mv^2

I got 5.3 and 2 but the 5.3 is wrong apparently?

Reply 8

bluewhale87

12

i got 3.4? i assumed it was at the point when the string became taught

Reply 9

Fort21

6

Original post by bluewhale87

i got 3.4? i assumed it was at the point when the string became taught

I think it’s at the equilibrium point

Reply 10

bluewhale87

12

Original post by Fort21

I think it’s at the equilibrium point

but the tension in the elastic string would decelerate the box… at the equilibrium point it’s velocity is 0 is it not ?

Reply 11

OP

Original post by bluewhale87

but the tension in the elastic string would decelerate the box… at the equilibrium point it’s velocity is 0 is it not ?

I dont think so cause there is a constant pulling force. There’s always a resultant force to the right before the equilibrium, and at equilibrium the acceleration is 0 due to F=ma, not velocity.

Reply 12

Fort21

6

Original post by bluewhale87

but the tension in the elastic string would decelerate the box… at the equilibrium point it’s velocity is 0 is it not ?

Acceleration is 0 at the equilibrium point, so v is a maximum.

Reply 13

Original post by bluewhale87

i got 3.4? i assumed it was at the point when the string became taught

do you mean when it became slack? so when T=0

i did that and got 3.4 too

Reply 14

how did you guys find a distance it moves to put into your energy equation if you didn’t use the tension from the string to the left of the block is 0

Reply 15

OP

Original post by ed729781

how did you guys find a distance it moves to put into your energy equation if you didn’t use the tension from the string to the left of the block is 0

I found the extension at equilibrium and added it to the distance it moved before equilibrium which was 1m.

Reply 16

nhbvugctx

1

Original post by raducorha

I found the extension at equilibrium and added it to the distance it moved before equilibrium which was 1m.

Same

Reply 17

Original post by raducorha

I found the extension at equilibrium and added it to the distance it moved before equilibrium which was 1m.

i mean for the second part where you’re trying to find v not the show that it’s 3.9 meters

Reply 18

Original post by ed729781

i mean for the second part where you’re trying to find v not the show that it’s 3.9 meters

oh did you use the extension that you got when finding 3.9m (2.9) when finding v for the next question?

(edited 10 months ago)

Reply 19

OP