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
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.
- 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# - 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#
