IIT JEE , AIEEE Forum - Physics , Mechanics - QUE IN MECH... PLZ HELP ME OUT..RATES ASSURED..

saignp

1.} A balloon of mass M is moving down with an accelaration a. The mass of the balloon that must be detached so that it moves up with same accelaration is _________

2.} A shell moving horizontally explodes into 2 fragments. One of hte fragments immediately after explosion attains a momentum "p" vertically up and other fragment attains a momentum $\sqrt{3}$ p in a direction inclined at arc{tan 1/ $\sqrt{2}$ } below the horizontal. The momentum of the shell before explosion is _________

3.} A stone is projected upwith velocity u. it crosses a point at a height h in its path after times t₁ &t₂ from the instant of projection. Then t₁ + t_2 ; t₁ t₂ = __________

4.} A ball is proj at an angle 45 with the horizontal just crosses a wall at a distance 'a' and falls at a distance 'b' beyond the wall. then wall's height is _________

5.} A stone of mass 1kg. tied to a string of lenght 10/3 m and rotated in a vertical plane. I f the ratio of max. tension to min. tension is 4, then the speed of the stone at the highest point of circle is _________

6.}A particle is moving along a straight line A,B,C. Its velocity at A and C are 5m/s &25m/s resp. If BC=2AB, then the ratio of time of travel from A to B, to that from B to C is __________

7.} A constant force acts for 5s on a body of mass 4 quintals which is at rest and then siezes to act. During the next 5s, the body travels 400m. Then the magnitude of force is ___________

varun.tinkle

1)
FIRST WE HV TO SET A FIXED POSITIVE AXIS AND A FIXED NEGATIVE AXIS
SUPPOSE ACC IN THE UPWARD DIRECTION IS POSITIVE THEN IN THE DOWNWARD DIRECTION IT WIL BE NEGATIVE
SUPPOSE A FORCE X ACTS ON THE BALLON WHICH IS RESPONSIBLE FOR THE UPWARD ACC OF THE BALLOON.
IN THE FIRST CASE
T-MG=Ma
IN THE SECOND CASE LET THE MASS BE M-m
SINCE IT SHOULD ACC IN THE DOWNWARDS DIRECTION THEREFORE
(M-m)G-T=-(M-m)a
THE ACC IS NEGATIVE SINCE IT IS MOVING IN THE DOWNWARDS DIRECTION
NOW EQUATE THE 2 EQUATIONS
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varun.tinkle

SEE AT THE TOPMOST POINT OF THE CIRCLE
THE EQUATION IS
MG +T=MV^2/L

AND A THE LOWERMOST POINT IS
T-MG=MV^2/L

OBVIOUSLY THE MAX TENSION IS AT THE LOWEST POINT WHICH IS EQUAL TO MG+MV^2/L
AND THE MIN TENSION IS AT THE HIGHEST POINT
WHICH IS EQUAL TO MV^2/L-MG
EQUATE THE EQUATIONS
AND FIND THE VARIABLES
THE SPPED AT THE TOPMOST POINT WILL BE FORMULATED BY THE EQUATION
MG +T=MV^2/L
REMEMBER THE RADIUS WILL BE L/2
WHERE L IS THE LENGTH OF THE SPRING
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varun.tinkle

6)
ET AB BEX THEN BC IS 2X AND THE TOTAL DIS IS 3X
IF THE BODY IS MOVING WITH A CONSTANT ACC.
THEN R=THE ACC CAN BE FOUND OUT FROM THE FORMUA
V^2=U^2+2AS
V=25 U=5
A=100/X
TIME TAKEN TO REACH POINT B WHICH IS AT A DISTANCE X FROM POINT A CAN BE GIVEN FROM THIS FORMULA
S=UT+1/2AT^2
FROM THE TIME SO FOUND OUT WE CAN CALCULATE THE VELOCITY AT POINT B FROM THE FORMULA
V=U+AT
THEN POINT C IS AT A DIS 2X FROM POINT B
THE VELOCITY AT POINT B IS ALREADY TAKEN OUT
SO
TIME CAN BE TAKEN OUT FROM THE FORMULA '
S=UT+1/2AT^2
AND THUS THE RATIO OF THE TIME TAKEN CAN BE EXTRACTED
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varun.tinkle

4)
I AM ASSUMING THAT THE AT THE HIGHEST POINT OF THE TRAJECTORY IT CROSSES THE WALL THEN
LET THE HIEGHT OF THE WALL BE X
THE HORIZONTAL COMPONENT OF THE VELOCITY IS UCOS 45
THEREFORE THE TIME TAKEN TO CROSS A DIST A IS U COS 45/A
SINCE AT THIS POINT IT REACHES THE HIGHEST POINT OF ITS JOURNEY SO THE TIME TRAVELLED=1/2OF THE TOTAL TIME OF FLIGHT WHICH IS EQUAL TO USIN 45/G
USIN 45/G=U COS 45/A
THE MAX HIEGHT REACHED IS A
SO U^2SIN^245/2G=X
THE RANGE IS B
SO U^2SIN90/G=B
SOLVE THE EQUATIONS AND U WILL GET THE VALUE OF X

oneyeartogo

Let theta be the angle at which lower fragment goes from the horizontal. (I don't know the meaning of arc(tan 1/(2)^2)

Let initial linear momentum = x

Conserving linear momentum in horizontal direction (we can do so as there is no external force in horizontal direction)

x = (3)^1/2 P cos(theta)

Thats the answer.

krishna.gopal

Saignp, first of all i would like to tell you that we answer only one question at a time in this forum.

For question number one there is no solution. The motion of baloon upwards and downwards does not depends on its mass only. It will depend on its volume as well (boyant force). So the answer depends on the densities of gas that is filled in the baloon as well as density of mass that you are detaching. Unless that information is given no one can solve this question

anchitsaini

anchitsaini

anchitsaini

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