IIT JEE , AIEEE Forum - Physics , Mechanics

ashok_p_ap

22.03.2008

Physics

1. A cubical block of mass m &edge ?a? slides down a rough inclined plane of inclination 0 with a uniform velocity. Find the torque of normal force acting on the block about its centre.

Ans. mga sin0

------------

2

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2. All surfaces frictionless. What is the velocity of smaller block when it reaches the bottom?

Ans. [2m²ghcos²?]^1/2

---------------------------

[(M+m)(M+msin²?)]^1/2

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3. All surfaces frictionless.

(1)Speed of smaller block when it breaks off at a height h.

(2)Max. height from ground the smaller block will ascend.

(3)Show that the smaller block will land again on the larger mass.

(4)Distance traversed by the bigger block during the time the smaller block was in flight under gravity.

Ans. (1) (M²+Mm+m²) v²

{ -------------------- ? 2gh}^1/2

(M+m)²

(2) Mv²(4) 2mv [Mv²-2(M+m)gh]^1/2

------------ -------------------------------

2g(M+m) g(M+m)^3/2

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4. A small block of superdense material has a mass M_e/2. It is situated at a height ?h? above the earth?s surface.(h<<R_e).

Find its speed when the height reduces to ?h?/2.

(M_e=6x 10²⁴ kg). Ans.(2gh/3)^1/2

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5. What is radial equilibrium?

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6.How to know whether cart is moving on an incline(with uniform velocity) or on a horizontal plane (accelerated)?

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ganesha1991

1)
torque = F*r
F = mgsin@
r = a/2
torque = magsin@/2

ganesha1991

6)
u can suspend a pendulum and if it starts moving then the cart is moving on an incline(with uniform velocity) or on a horizontal plane (accelerated)

ganesha1991

4) by conservation of energy
mg [ 1-2h/R ] h = 1/2mv^2 + mg[1-2h/2R]h/2
since h< therefore 1-2h/R will be almost = 1
and 1-h/R = 1-1/2 = 1/2

therefore
gh = v^2/2 + gh/4
v^2 = 2gh -gh/2
= 3gh/2

v = root 3gh/2

anchitsaini

2)
i have solved this earlier on goiit , can't find the link

taking a and A as accelerations of small and big block respectively

in the horizontal direction --

ma _x= (m + M )A

also a _x = g sin @ cos @

thus

A =m g sin @ cos @ / ( m + M )

a = g sin @ - A cos @

which on some solving gives

a = g sin @ [ {M + m sin ²@ } / { m + M} ]

using

s = at² / 2

and putting s = h/ sin @ and a as accn of small block

t =

2h / a sin @

now i think the velocity of big block is asked --

for big block --

v = A t

= [ m g sin @ cos @ / ( m + M ) ] * [

2h / a sin @]

on putting value of a

= [ [2m²ghcos²@]^1/2 ] / [(M+m)(M+msin²@)]^1/2

anchitsaini

the basic concept seems to be --

since all surfaces are frictionless momentum can be conserved in horizontal direction only and we would use it when the block of mass m rises on the vertical part of mass M

conserving momentum --

mv = (m+M) v1 ----------1

now conserving energy at the point of break off and point when the block of mass m is on the horizontal portion of M --

mv² / 2 = Mv1 ² / 2 + mv2² / 2 + mgh

where v2 is velocity of m during break off

putting value of v1 we get --

v2 =

[(M²+Mm+m²) v² / (M+m)² - 2gh ]

also

v2 ² = v2 _x ² + v2 _y² = v1 ²+ v2 _y ²

hence

v2 _y ² = v2 ² - v1 ²

. = Mv² / ( M + m) - 2gh

total height =

h + v2 _y ²/ 2g = Mv² / 2( M + m) g

the smaller block will fall on the big block because at the time of break off both have same components of horizontal velocity and there is no acceleration in horizontal direction

part iv of this qn is also easy now
just use the basic formulae and u'll get the answer easily

ramyani

Q.2

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

)

http://www.goiit.com/posts/listByUser/225/14124.htm

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