IIT JEE , AIEEE Forum - Physics , Mechanics

s.mirchi

a block slides down a smooth inclined plane to the ground when released at the top, in time t secs. another block is dropped vertically from the same point, in the absence of the incline plane and reaches the ground int/2 sec.then the angle on inclination with vertical is.....?????

aditya_arora04

what is int/2 ?

shubham.123

I THINK IT IS T\2 SEC.

s.mirchi

sorry ppl..........its t/2 secs

shubham.123

is the ans is 30....

coolank2

Let the angle of incline be

, the vertical height h and the hypotnuse l.

Thus, (h/l) = sin

For motion 1, l = (1/2)(g.sin

)(t)

For motion 2, h = (1/2)(g)(t/2)

Dividing 2 by 1,

(h/l) = (1 / (2.sin

) )

i.e. sin

= (1 / (2.sin

) )

Thus,

= sin^-1( 1/

2 )

atavistic

Use conservation of energy.

Velocity of both the bodies will be the same at the ground.
mgh=mv^2/2

Gives v = (2gh)^1/2

using v=u+at

we get t=v/a

Means t/2 = [(2gh)^1/2]/g

And t = [(2gh]^1/2]/gsin@

Comparing

[(2gh)^1/2]gsin@ = 2[(2gh)^1/2]g

IMplies 1/sin@ = 2

sin@ = 1/2

@=30 degree

atavistic

Coolank is wrong as he has used the formula wrongly.ITs s = ut+1/2at^2 and he has just used ut +1/2at.

And u have given him points.lol

jasmeetawal

using s=ut+1/2at²

h=g/2*(t/2)²

h=5/4t²

t²=4/5h

in case of inclined plane distance travelled=h/sin#

h/sin# =gsin#/2t²

t²=4/5h

h/sin# =5*4/5h*sin#

sin²# =1/4

sin# =1/2

# = 30⁰

hope it is clear

priyesh

u = 0 for both

& by energy conservation v is same for both

therefore using v = u + at for both & equating

we get

g sin

t = g(t/2)

=> sin

= 1/2

=>

= 30 degrees or pi/6 radians

spideyunlimited

Simple dude.
use S = ut + at^2/2

so the 2 eqns will be
s = 1/2*g t^2 /4
and
s/ cos@ = 1/2 * gcos@ t^2

dividing the 2 equations

cos@ = 1/4cos@
cos^2 @ = 1/4
cos@ = 1/2
@ = 60 degrees ( with the vertical)

spideyunlimited

Priyesh ur method is wrong! first of all ur angle ( the angle ur using) in g sin@ is the angle with horizontal whereas we have to find with vertical.

the place where ur going wrong is that velocity is conserved all right!
but that is the net velocity conserved whereas uve taken velocity conserved in the direction of g sin@ which is wrong!

priyesh

hey gaurav the angle with horizontal is 30 degrees & with vertical it is 60 degrees

& by the way u have written

cos@ = 1/2
@ = 30 degrees ( with the vertical)

arey if cos@ = 1/2

then => @ = 60 degrees.

& as far as my method is concerned , it's correct because the net acceleration of the block will be gsin(theta) & thus the NET velocity will be u + gsin(theta) * t

also i've taken velocity in the direction of gsin(theta) because net velocity will be along the incline.

hope you understood

cheers!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

priyesh

arey atleast write that u've edited your post now

dinesh.the.disciple

ans :the angle of inclination is 30 .

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