Im sorry but are u solid for q6 part 2 answer im not familiar with your method but im pretty sure that as p rise up it will be ma= Tension- mg for particle P and for Q it will be ma = (1-m)g – T for Q putting the value of a as 4 from the graph’s gradient will give u for p 4m= Tension – 10m which is 14m=T or 14m-T=0 and for q its 4m= 10 – 10m – T giving u 14m = 10 – T or 14 m + T = 10 solving these two equations simaltaneously would give Tensiom as 5 N and m as 5/14 please review what i said and tell me if im right or no

In Question 7 part iii) he said that he attached a mass m kg to ring R which is already 0.2 kg which means 10(m+0.2) so why u found m without inluding 2 N ?

Im sorry but are u solid for q6 part 2 answer im not familiar with your method but im pretty sure that as p rise up it will be ma= Tension- mg for particle P and for Q it will be ma = (1-m)g – T for Q putting the value of a as 4 from the graph’s gradient will give u for p 4m= Tension – 10m which is 14m=T or 14m-T=0 and for q its 4m= 10 – 10m – T giving u 14m = 10 – T or 14 m + T = 10 solving these two equations simaltaneously would give Tensiom as 5 N and m as 5/14 please review what i said and tell me if im right or no

for Q it will be (1-m)a = (1-m)g – T

Oh yea youre right damn forgot the 1-m thanks man 🙂 hope i dont lose more than 2 morks for that 🙁

What do u think the marks that i might lose for that mistake?

In Question 7 part iii) he said that he attached a mass m kg to ring R which is already 0.2 kg which means 10(m+0.2) so why u found m without inluding 2 N ?

i didn’t read the question