IIT JEE , AIEEE Forum - Physics , Mechanics

akshax

Refer the figure..

The smaller block has a mass of 5kg and the wedge a mass of 30kg.

The angle made by the wedge with the smooth horizontal floor is 30⁰.

The horizontal length of the wedge is 0.5m

Friction is negligible on all surfaces

Neglect the size of smaller block

Determine the distance the wedge moves when the smaller block moves to the bottom of the plane....

ramyani

Let v be the velocity of the small block along the incline at the instant it touches the ground. Let V is the velocity of the triangular block w.r.t the ground. ( opp to dirn of v _x )

We resolve v along X and Y axis. ( X axis horizontal and Y axis downwards )

v _x = v cos

- V .......( 1)

v_y = v sin

from the pl of conservation of momentum,

MV = mv _x = m ( v cos

- V ) ....... (2 )

using conservation of energy,

mgh = MV² / 2 + mv² / 2

mgh = MV² / 2 + m ( v _x ² + v_y ² ) / 2

mgh = MV² / 2 + ( 1/ 2 ) [ ( v cos

- V )² + v² sin²

mgh = MV² / 2 + mV² / 2 + mv² / 2 - ( m/ 2) ( 2 V v cos

) .....(3)

eliminating v from 2 amd 3 , we get

2 m²ghcos²

V = -------------------------------------

(M+m) (M+m sin ²

)

now plug the values.

m= 5kg

M = 30kg.

= 30 degree

The horizontal length of the wedge is 0.5m ( slightly edited )

h = horizontal length * tan alpha ( ooof again edited _)

g = 10 m/s/s

edited : The above answer is wrong. I keep it to show how we sometimes do not read the question at all in our hurry. Or, earlier solution has a hangover on recent question.

The above is the soln of question of the following question :

Q.A block of mass m is placed on a triangular block of mass M. Find the velocity of the triangular block when the smaller block reaches the bottom.

ramyani

edited.

ramyani

edited.

anant_c

just do it by using the fact that the position of centre of mass duznt change horizontally

zane3004

yep i guess conservation of centre of mass is how its done...
find centre of mass of system along x-axis before block moves.... and after it moves taking the initial position of block as origin
and equate the two... assume the shift to be X cm

ramyani

that method using centre of mass concept can be done.

But this method also gives result.

look here http://www.goiit.com/posts/list/mechanics-centre-of-mass-verma-34903.htm#174573

i just want to chk the result.

edited : this method does not give result.

ramyani

thanks ......my friend !!!

*** Determine the distance the wedge moves when the smaller block moves to the bottom of the plane

thats the Q. ohhh !!!

zane3004

hey can u try it with the Centre of mass method........... i find it scary.......... jus need help with that....

ramyani

yes it shd be tried in that way. But i want 1 more day as my CM concept is unclear.
I m just starting CM frm hcv. actually i tried cm months ago, then shifted to something else.

ramyani

This is , i think, correct answer :

If the 30 kg block moves towards right by a distance X m, the 5kg block will move towards left by a distance

( 0.5- X )m.

taking the two blocks together as the system, there is no horizontal external force on it. the center of mass which was at rest initially, will remain at the same horizontal position. thus,

     5 ( 0.5 - X ) = 30 X

    2.5 - X = 30 X

           X = 2.5 / 35 = 0.07 m

therefore the 30 kg block moves towards right by a distance 0.07 m

zane3004

hey thanks but im sounding stupid but srry. i dint get this equation:

5(0.5 - X) = 30 X

i mean taking block as m1 and position as x1
taking wedge to be m2 and position x2

m1x1=m2x2 holds true ????

i thought it was (centre of mass of system)(m1+m2)= m1x1+m2x2 rite ?????

ramyani

I solved it frm a solved example no. 9 of hcv in page 149 , vol I

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