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?

2 Answers
May 16, 2018

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.

May 16, 2018
  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