help with math mechanics paper4 m/j 03 question!!!!!!please!

[alevel4a]

can somebody help me with mechanics paper4 m/j 03 question 3(i). totally have no idea of drawing a graph.
if you can upload an attachment and explain to me how to do these kind of question, it would be better.
for example, m/j 02 question7(iii)
i really appreciate your help.

 

[alevel4a]

Re: help with math mechanics paper4 m/j 03 question!!!!!!ple

and also paper4 m/j 03 question 5(ii)

 

[alevel4a]

Re: help with math mechanics paper4 m/j 03 question!!!!!!ple

and paper4 m/j 03 question 6 (iv) (v).
thank you so much.

 

[alevel4a]

Re: help with math mechanics paper4 m/j 03 question!!!!!!ple

one more question:
o/n 03 question 7(iii)

 

[ks136]

Re: help with math mechanics paper4 m/j 03 question!!!!!!ple

can somebody help me with mechanics paper4 m/j 03 question 3(i). totally have no idea of drawing a graph.
if you can upload an attachment and explain to me how to do these kind of question, it would be better.
for example, m/j 02 question7(iii)
i really appreciate your help.

According 2 me...it would golike this
* As the area under the line represents distance....therefore...u would draw a line from T upto the line P and shade the smaller triangle

TELL ME IF I AM WRONG

 

[DragonCub]

(Re) Solution to paper4 m/j 03 Q5(ii)

As we can see, A's weight (Wa) = B's weight (Wb) = 2N. So total weight (W) = 4N
When S1 cut, the total resistance = 0.2N + 0.4N = 0.6N
So resultant force = 4N - 0.6N = 3.4N
Use F = ma, m = total mass (string mass negligible) = 0.2kg + 0.2kg = 0.4kg
a = F/m = 3.4N / 0.4kg = 8.5 ms^(-2) <-----Acceleration

For the tension (T), it should be clear that: resultant force on A (Fa) = Wa + T - (resistance on A)
Use F = ma again. Fa = A's mass * a = 0.2kg * 8.5ms^(-2) = 1.7N
So Wa + T = 1.7N + resistance on A = 1.7N + 0.4N = 2.1N
Hence T = 2.1N - Wa = 2.1N -2N = 0.1N <----- Tension in the string

(If you find anything incorrect in my solution, please notify me. Thanks. )

 

[DragonCub]

(Re) Solution to paper4 m/j 03 Q6(iv)

As figured out in (iii), AB = 20m and acceleration (a) is still -0.25ms^(-2).
Here the equation v^2 = u^2 + 2as is required.
v = 3.5 m/s, a = -0.25ms^(-2) and s = AB = 20m
So (3.5)^2 = u^2 + 2 * (-0.25) * 20
12.25 = u^2 - 10
u^2 = 22.25
u ≈ 4.72 m/s

Hope I didn't make mistakes during calculation...