IIT JEE , AIEEE Forum - Physics , Mechanics

akshay107

PLS SOLVE HCV Q.NO. 61 PG. 164.

Q. figure shows a small body of mass m placed over a larger mass M whose surface is horizontal near the smaller mass and gradually curves to become vertical. the smaller mas is pushed on the larger one at a speed v and system is left to itself. Asume that all the surfaces are frictionless. (a) find the speed of the larger block when the smaller block is sliding on the vertical part. (b) find the speed of smaller mass when it breaks off the larger mass at height h (c) find the maximum height (from the ground) that the smaller mass ascends. (d) show that the smaller mass will again land on the bigger one. Find the distance travelled by the bigger block during the time when the smaller block was in its flight under gravity.

akshay107

somebody pls reply..................................................................

akshay107

atleast give solution of part (b)

akku

considering the small body and large body as a system, the only forces acting on the system-gravitational force in the vertical dirn
in the horizontal dirn momentum remains conserved
as the block remains in contact with the wedge both move with the same horizontal velocity(under effect of normal reaction -if u consider each body separately)
mv=(M+m)V
(a)V=mv/(M+m)
at ht h when the small block (its the small body u mentioned in the 1st line of the q)breaks of it has 2 velocities Vx=V and Vy
let u=vel of smalll block at ht h=^{[ 2]}

(Vx²+Vy²)
conserving energy
work done by gravity=KEf-KEi
-mgh=1/2*MV² +1/2mu^2-1/2 *m*v²
on solving
(b)u=^{[ 2]}

( v²(m+M+Mm)/(m+M)² -2gh)

.......

akku

for the maxm ht reached by small body
u=^{[ 2]}

( v²(m+M+Mm)/(m+M)² -2gh)=^{[2 ]}

Vx²+Vy²
Vx²=m²v²/(m+M)²
so Vy²=(M²+mM)v²/(m+M)²-2gh=Mv²/(M+m)-2gh
max ht reached frm the pt where the small body leaves the larger mass
ht=Vy²/2g= Mv²/2g(m+M)-h
frm the ground
H=h+ht
(c)H=Mv²/2g(m+M)

as both the small block and large mass have the same horizontal velocity the cover equal distances in the horizontal dirn and so the small mass lands again on the large mass
the distance travelled by the bigger block during the time when the smaller block was in its flight under gravity=VxT

where T=time of flight of the small mass=2Vy/g=2^{[ 2]}

(Mv²/(M+m)-2gh) /g
distance =l=2mv/(M+m) *^{[ 2]}

(Mv²/(M+m)-2gh) /g

(d)l=2mv*^{[ ]}

(Mv²-2(M+m)gh)/g(M+m)^3/2

hope it helps u
:):):):):)

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