Mechanics Help

I am revising past papers for my mechanics exam and I am stuck on the last question.

Two particles P and Q, of mass 4kg and 6kg respectively, are joined by a light inextensible string. Initially the particles are at rest on a rough horizontal plane with the string taut. The coefficient of friction between each particle and the plane is 2/7. A constant force of magnitude 40 N is then applied to Q in the direction PQ.

(a) Show that the acceleration of Q is 1.2ms-².
(b) Calculate the tenson in the string when the system is moving.

After the particles have been moving for 7s, the string breaks. The particle Q remains under the actionof the force of magnitude 40N.

(d) Show that P continues to move for a further 3 seconds.
(e) Calculate he speed of Q at the instant when P com to rest.

I know I know how to these questions, I might be being silly and doing careless mistakes.

Any help is useful, Thanks.

c)......... then draw another diagram for P but not with the T pulling it farward then show decelleration using the same method as before (F=ma) then apply the new deceleration to suvat with
a=thing u just got
u=value of speed at 7s
v=0
t=t
v=u+at

d)again much the same use speed of P at 7s as will be same as Q, then again draw diagram of Q but without T will probably accelerate do F=ma again and then suvat the acceleration as u know initial speed, acceleration and time

and that is that

a) first treat as 1 paritcle as tention in string will cancel.
draw a good diagram
apply newtons 2nd law (f=ma) horizontally and end up with an equation with 2 unknowns the frictional force (F) and acceleration
to find friction resolve vertically (f=ma) and then using the value u get for R use F=(mu)R and mu as 2/7
put new value for F into first equation.

b) do same as before but just use particle p or q ill use p.
draw diagram of forces acting on P
get equation ie T-F=ma
sub in all values

c) find speed of P at seven seconds use suvat assume constant acceleration and start at 0ms^-1

d) after 7 secs speed of the two particles is 7*1.2 = 8.4 m/s

when string breaks there is only friction acting against P so it decellerates.
friction force = 4*9.8*2/7 = 11.2 N

deceleration d = 11.2/4 = 2.8 m/s² (Fr=md)

v=u+at
0 = 8.4 - 2.8t
t = 8.4/2.8
t = 3 s
=====

e)
Friction against Q is 6*9.9*2/7 = 16.8 N

T-Fr = ma
40 - 16.8 = 6a
a = 23.2/6
a = 3.867 m/s²

v = u + at
v = 8.4 + 3.867*3
v = 8.4 + 11.6
v = 20 m/s
========

You're not setting one direction to be positive and the other direction to be negative, I remember getting stuck in the same place.

Principle of Conservation Of Linear Momentum:

mpup+mquq=mpvp+mqvq

Remember the direction of p is reversed after the collision. Since we are dealing with velocities, which are vectors, the velocity of p therefore becomes negative. Hence:

2u=(-2(1/3))v+4v

Multiply out by 3:

6u=-2v+12v

=>6u=10v

=>v=((6/10)um(s^-1))=((3/5)um(s^-1))

But that is the velocity of q after the collision. It says that the velocity of p is ((1/3)v). Therefore:

vp=((1/3)v)=(((1/3)(3/5))u)=((3/15)v)=((1/5)v) Q. E. D.

Newton.