Mechanics P4 question!

[alevel4a]

Could some1 explain the following questions?

41/O/N/09 Q 6 (ii) (iii)

Please give explanation as detailed as possible.

In part (ii)(a), it is said that P and Q move under gravity, but why in mark scheme, it use "a" instead of "g" ?????

In part (ii)(b), what is this formula in mark scheme, "2a", is it v = u + at , and how do I know the question is asked for final velocity or initial velocity?

Really have no ideas about this question, please help me~~

 

[alevel4a]

any 1?????

 

[alevel4a]

and 41/O/N/09 Q 7

 

[Zishi]

Try to post link to the questions, OR take screenshots of the questions by using the 'print screen' button on your keyboard, and press ctrl + v by opening paint, crop the question and attach it over here. That way many people will answer your questions. The time is not to be wasted these days, so we don't really waste time in opening the question papers, find questions and then answer them.

 

[XPFMember]

assalamoalaikum!!


yea true.. zishi

 

[DragonCub]

6. (ii) (a)
Yes they have gravity acting on themselves but that's not the only force. Tension in the string is also involved.
From (i), the acceleration should be calculated out to be 1 ms^(-2).
Use s = ut + 0.5 at^2 to solve (ii) (a). t = 2, a = 1, u = 0.
So s = 0.5 * 1 * 2^2 = 2m.
P is heavier so moves downwards. It height now is 5 - 2 = 3m
Q moves upwards, height is 5 + 2 = 7m

(ii) (b) v = u + at = 0 + 1*2 = 2ms^(-1)

(iii)
At the instant when the string breaks, height of P = 3m, of Q = 7m.
P's (initial) VELOCITY is -2m/s (it moves downwards), the displacement until it hit the ground should be -3m.
The equation is represented: -2t + 0.5 * (-10) * t^2 = -3
For Q, (initial) velocity = +2m/s, displacement = -7m
Then +2t + 0.5 * (-10) * t^2 = -7
Solve the two equations and get the two values of t, their difference should then be 0.8s.

 

[DragonCub]

7 (i)
Plug in t = 10s into v = 800 / (t^2) - 2, you get v = 800 / (10^2) - 2 = 6 m/s
That is velocity, then a = 6 / 10 = 0.6 m/(s^2)

7 (ii)
Differentiate the velocity-time function. You get dv/dt = -1600 / (t^3). This is acceleration's value during 10s and 20s.
-1600 / (t^3) = -0.6
t = 13.9s

7 (iii)
Calculate the displacement during 0s ---> 10s and 10s ---> 20s respectively and add them up.
0 to 10: 6 * 10 / 2 = 30m
10 to 20: integrate the function you get d = -400 / t -2t, then plug in t = 10s, t = 20s to get the values of d then find the difference.
d = (-400 / 20 - 2*20) - (-400 / 10 - 2*10) = (-40) - (-60) = 20m

Total displacement = 30 + 20 = 50m