Question

Can anyone help me with 9709 may/June/2017 paper 43 Question number 7 part (ii) I have found velocity and acceleration as `v=1.68` and `a=2.83` I don't understand how it will move further?

http://www.teachifyme.com/wp-content/uploads/2016/05/9709_s17_qp_43.pdf

Answer 1

Use the kinematic expression

`v^2-u^2=2as`

(Assuming your calculations are correct, Inserting given values we get

`0^2-1.68^2=2xx2.83s`
`=>s=-1.68^2/(2xx2.83)`

We get a negative number. It appears your calculation of final acceleration is not correct. It should be a negative number. Perhaps next step missing.)

You need to post your work.

Answer 2

1. Force of friction along the incline `F_f=muR=mumgcostheta`
   where `R` is normal reaction. Taking `g=10\ ms^-2`

   `F_f=0.2xx3xx10cos 30^@=3sqrt3\ N`

2. Once system is released from rest,
   For `4\ kg` mass
   It moves down. Let its velocity as it touches ground be `v_4`. If tension in the string is `T`, keeping in view Law of consrvation of energy, energy equation is
   `"Loss of PE"\ -"Work done against "T="Gain in KE"`
   `=>4xx10xx0.5 -0.5T=1/2xx4xxv_4^2`
   `=>40 -T=4v_4^2`......(1)
   For `3kg` mass. Let its velocity at that time be `=v_3`
   `"Work done against "T="Gain in PE"+"Gain in KE"+"Work done against "F_f`
   (`T` is on both sides of smooth pulley.)
   `0.5T=3xx10xx0.5sin 30^@+1/2xx3xxv_3^2+3sqrt3xx0.5`
   `=>T=15+3v_3^2+3sqrt3` ....(2)
   As both masses are connected with same light inextensible string their velocities must always have equal magnitude. Let this be `=v`. Eliminating `T` from (1) and (2)
   `40-4v^2=15+3v^2+3sqrt3`
   `=>v=sqrt((25-3sqrt3)/7)=1.68\ ms^-1`
   After `4\ kg` mass touches ground `T=0`, the net force equation for mass `3\ kg` becomes
   `"Force due to gravity along the incline"\ +"` `F_f=F_"net"`

   `–3xx10sin30 - 3sqrt3 = 3a_n`
   where `a_n` is new acceleration and `-ve` sign is with respect to direction of motion.
   `=> a_n=-5-sqrt3=-6.73\ ms^-2`
   Using kinematic expression and imposing the given condition that mass `3\ kg` comes to rest momentarily
   `v^2-u^2=2as`
   We get
   `0^2-1.68^2=2(-6.73)s`
   `=>s=1.68^2/(2xx6.73)`
   `=>s=0.21\ m`
   Total distance traveled by mass `3\ kg=0.5+0.21=0.71\ m`