IIT JEE , AIEEE Forum - Physics , Mechanics - Find the time elapsed before the smaller block separates from the bigger block.

adithya7

A small block of mass m kept at the left end of the larger block of mass M and length l. The system can slide on a horizontal road. The system is started towards right with an initial velocity v. The frictional coefficient between the road and the bigger block is

and that between the blocks is

/2. Find the time elapsed before the smaller block separates from the bigger block.

ganesha1991

is it 2v/

g ....correct me if i'm wrong

hash_include

ganesha1991, how can the answer be independent of 'l'??

anyway, i think the answer is $\frac {2v +/- \sqrt {4v^2 - 4 \mu gl}}{\mu g}$

EDIT: its 2 v (plus or minus) root of....

varun.tinkle

pls expln this sum

madman

frictional force acting on the larger block =(mu) (m+M)g
acceleration=(mu)g
in the frame of reference of the big block there will be a pseudo force on the small block towards the right
F(p) = m(mu)g
frictional force between the blocks = m(mu)g/2
so net force on the small block = m(mu)g/2 towards the right
acceleration = (mu)g/2
let the time taken be t
then

l=1/2mug/2t^2
t=2(l/(mu)g)^1/2

BALGANESH

from biki

Let the retardation of the mass m be a1 and that of M be a2
For the mass m to cross the mass M and fall through a distance l, a1 > a2
Let t s is reqd. for falling....So for falling,
distance travelled by m(in t s) = distance travelled by M(in t s) + l.....{since m is on M}
=> vt - 1/2(a1)t^2 = vt - 1/2(a2)t^2 + l
=> - 1/2(a1)t^2 = - 1/2(a2)t^2 + l ________(1)

Now let us consider F.B.D. of the bodies.
for m..... The only force acting on the body m is the frictional force between m and M
this is equal to (u/2)mg
Now as m moves towards right...the frictional force (u/2)mg must have caused its rightward motion.
So retardation of m = a1 = [(u/2)mg]/m = (u/2)g

Now frictional force is after all a force. So being a member of the force family, it must satisfy Newton's 3rd law of motion.
So a reaction of the friction (u/2)mg (on m rightward) acts on M (towards left)
So force reqd. to cause its motion = u(m+M)g.
So the rightward motion of M can be written as......
...M(a2) = u(m+M)g - (u/2)mg
So a2 = [ug(2M + m)] / 2M ______(by simplification)

Putting the values of a1 and a2 in equ^n (1)

- (u/4)g.t^2 = -[ug(2M+m)/4M].t^2 + l

[ug(2M+m)/4M].t^2 - (u/4)g.t^2 = l

t^2 x [{ug(2M+m) - uMg] / 4M}] = l

t^2 x {ug(2M+m) - uMg} = 4Ml

t^2 x (2uMg +umg - uMg) = 4Ml

t^2 x (uMg + umg) = 4Ml

t^2 = 4Ml/[ug(M+m)]

Now t = <sq. root> [4Ml/{ug(M+m)}]

iitkgp_bipin

Perfectly done by balganesh.

Ask & Discuss Questions with Community & Experts