Can anyone help me with 9709 may/June/2017 paper 43 Question number 7 part (ii) I have found velocity and acceleration as v=1.68v=1.68 and a=2.83a=2.83 I don't understand how it will move further?
2 Answers
Use the kinematic expression
v^2-u^2=2asv2−u2=2as
(Assuming your calculations are correct, Inserting given values we get
0^2-1.68^2=2xx2.83s02−1.682=2×2.83s
=>s=-1.68^2/(2xx2.83)⇒s=−1.6822×2.83
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.
- Force of friction along the incline
F_f=muR=mumgcosthetaFf=μR=μmgcosθ
whereRR is normal reaction. Takingg=10\ ms^-2 F_f=0.2xx3xx10cos 30^@=3sqrt3\ N - Once system is released from rest,
For4\ kg mass
It moves down. Let its velocity as it touches ground bev_4 . If tension in the string isT , 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)
For3kg 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 . EliminatingT from (1) and (2)
40-4v^2=15+3v^2+3sqrt3
=>v=sqrt((25-3sqrt3)/7)=1.68\ ms^-1
After4\ kg mass touches groundT=0 , the net force equation for mass3\ kg becomes
"Force due to gravity along the incline"\ +" F_f=F_"net" –3xx10sin30 - 3sqrt3 = 3a_n
wherea_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 mass3\ 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 mass3\ kg=0.5+0.21=0.71\ m