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SEE THIS IS A SIMPLE SUM
THE FRICTION BETWEEN THE OWER BLOCK AND GROUND
15N
THE FRCICTION COFFIECIENT BEWTWEEN THE BLOCKS=5N
WHEN A FORCE 100N IS APLIED
FOR
LOWER BLOCK
100-15-5=10A
A=8
FOR UPPER BLOCK
5=5A
A=1
WHEN A FORCE 30 N
LOWER BOCK
30-15-5=10A
A=2
UPPER BLOCK
5=5A
A=1
WHEN A FORCE 20N
THE TOTAL FORCE ACTING OPP IS GREATER SO
THERE WILL BE NO RELATIVE MOTION BETWEEN THE BLOCKS
THIS DOESNT MEAN THE BLOCKS WONT MOVE THEY WILL MOVE WITH SAME ACC SO
20-15=15A
5=15A
A=1/3
PLS RATE ME IF U FIND ME USEFUL
!!!!!!CHEERS!!!!!!!!

THIS SUM IS LITTLE TRICKY
FIRST WE CALCULATE THE VELCOITY OF THE MASS 100 g AFTER FALLING =3MS
CONSERVING MOMENTUM WE GET THAT THE VELOCITY OF THE WHOLE SYSTEM IS 9/8
NOW LET THE SPRING COMPRESS BY A DISTANCE X
SO
1/2(M+m)(9/8)^2+MGX=1/2KX^2
SOLVE AND GET THE ANSWER
PLS RATE ME IF U FIN ME USEFUL
!!!CHEERES!!!!!!!

AND MIND U
U CANT USE ANY FORMULA

FIND THE SUM OF THE NUMBERS FROM 1 TO 100 THAT I 1+2+3......+99+100 IN 2 MINUTES

THIS QUESTION WAAS ASKED BY LSE TO MY FATHERS BOSS WHO IS A MD AND HE COULDNT SOLVE IT

TO MY FATHERS FRIEND COULDNT SOLVE IT AND TO MY DAD COULDNT SOLVE .

UNFORTUNATELY I COULD SOLVE IT LETS SEE WHO CAN DO IT UNDER TIME LIMIT

IT CON BE OUTIDE THE BODY
CONSIDER THE EXAMPLE OF A CIRCUAR RING
THE CENTRE OF GRAVITY WILL BE IN THE MIDDE

I THINK u=1/2
SEE MY WORKING
SINCE A IS INVERSELY PROP TO TIME
2WICE TIME =1/2 ACC
SO
MGSIN@=2(MGSIN@-uMGCOS@)
SOLVING IT WE GET u=1/2

SEE ABHI UR TOTALLY WRONG
THE WORKING IS THIS
THE SYSTEM WILL LOOSE V DUE TO GAIN IN POTENTIAL ENRGY
SO THE NEW VELCOITY
=
1.2MV2^2=1/2MV^2-MGL(1-SIN@)
SOLVING WE GET V2^2=20
THE REAL EQUATION IS
T-MGCOS@=MV^2/R
T=M(GCOS@+V^2/R)
SOLVING WE GET T AS 135 NEWTON

2ND PART OF THE EQUATION
FOR COMPLETING THE CIRCLE
AT THE HIGHEST POINT
T+MG=MV^2/R
T CAN TAJE ANY VALUES SO FOR SURETY
MG=MV^2/R
SOLVING IT WE GET V^2S 5
SO THE VELOCITY AT THE HIGHEST POINT =5
SO WE HV TO VERIFY IT BY THE WOK AND ENERGY METHOD
1/2MV^2=1.2MV1^2-2MGL
V^2=5
SO IT WILL COMPLETE THE CIRCEL
!PLS RATE ME IF U FIND ME USEFUL
!!!CHEERS!!!

THE MAXIMUM ELINGATION OF THE SPRING IS WHEN THE BLOCKS HV THE SAME VELCOITY
m1V=(m1+m2)V2
V2=M1V/M1+M2
SINCE NO EXT FORCE ACTS ON THE SYSTEM THE ENERGY WILL BE CONSERVED
SO LOSS IN K.E.=GAIN IN SPRING ENRGY
SOLVE AND GET THE ANSWER
PLS RATE ME IF U FIND ME USEFUL
!!!!CHEERS!!!!!!

SIMPLE TAKING THE FRAAME OF REFRENCE TO BE THE INCLINE WE HAVE TO OFFER A PSUEDO FORCE
MW^2R=MGSIN@
W^2R=GSN@
AND THEN U CAN SOLVE THE REST

I AM APLLYING THE CONSTRAINT EQUATION LIKE THIS
ACCORDING TO THE EQ
1) CHOOSE A SUITABE ORIGOIN
HERE I AM CHOOSING IT AS THE FIXED PULLEY
2)MARK THE DISTANCES OF THE MOVABLE POINTS FROM THE ORIGION
THE DISTANCE OF THE SIDE MASS m IS X1 AND THE MASS 2m IS X2 AND OF THE MOVABLE PULLEY IS X3 (ACTUALY THE FIGURE GIVEN IS LITTLE WRONG THE PULLEY IS UP)
SO
A1+A2+2A3=0
A3=ACC OF THE PULLEY=ACC OF THE MASS OBVIOUSLY
NOW LET THE TENSION IN THE MASS 2M BE 2T
SO
MG-2T=2MA
T-MG=2MA (SINCE ACC I TWICE THAT OF M)
T=2MG=4MA
SOLVE AND GET THE ANSWER
PLS RATE ME IF U FIND ME USEFUL
!!!!!!CHEERS!!!!!!!!
THEN THE TENSION IN THE MASS OF M (LOWER ONE)=2T
3)DERIVE AN EQUATION FROM THE FACT THAT THE STRING CONNECTING THE MASSES ARE EQUAL
SO HERE
X1+X12+2X3=L
DIFFERENTIATING IT TWICE WE GET
A1+A2+2A3=0 (SINCE L IS CONSTANT)

ANSWER
2

I HAVE SOLVED IT AND ITS NOT THAT A TOUGH A SUM
IST METHOD THE SIMPLEST ONE
CONSERVING ANGULAR MOMENTUM .LET US CONSERVE IT ABT THE SURFACE F THE ROD
mV1R=(m+m)V2R1 +IW HEREI=(ML^21/2+ML^2/4=ML^2/3)
BUT SINCE V IS PARALEL TO R m+mV2R=0
SO
MVIR=IW
AND IF WE APPLY THE CONSERVATION OF LINEAR MOMENTUM
MVV1=2MV2
WE GET V2
WE KNOW THAT THE VELCOITY OF ANY POINT OF AROTATIONG BODY CAN BE FOUND OUT FROM THE COM
SO
V=RW
R=V/W
PLS RATE ME IF U FNED ME USEFUL
!!!!!!!!!!CHEERS!!!!!!!!!!

NO W WILL NOT CHANGE
IN ELASTIC COLLISION THE REL VEL OF THE POINT OF CONTACT IS 0
IF U PICTURISETHE COLLISON U WILL UNDERSTAND

SEE THIS QUESTION HAS LOTS OF MATHS TO DO AND I CANT DO IT
SORRY
BUT UNDERSTAND THIS CONCEPT
FORCE APPLIED=RATE OF CHANGE OF MOMENTUM
=M(V-V/20)T
AND SO T CAN BE EASILLY CALCULATED