A cylindrical solid of mass 10-2 kg and cross sectional area 10-4 m2 ismoving parallel to its axis (the x- axis) with a uniform speed of 103 m/s in the +ve direcn . At t=0 , its front face passes the plane x = 0 . The region to the rite of this plane is filled with stationary dust particles of uniform density 10-3 kg/m3. When dust particle collides with the face of cylinder , it sticks to its surface . Assuming that the dimensions of cylinder remain practically unchanged , and that the dust sticks only to the front face of the cylinder , find the x- coordinate of the front of the cylinder at t= 150 s.
11m' = 10-2 kg
A(cross sectional area) = 10 -4m2
v' = 103m/s
Ïdust = Ï = 10-3 kg/m3
m = m' + mass f the dust collected so far
= m' + AxÏ
The linear momentum at t =0 is
p'=m'v'
and momentum at t=t is
p = mv = (m' + AxÏ)v
From the law of conservation of momentum,
p' = p
m'v' = (m' + AxÏ)v
therefore, m'v' = (m' + AxÏ)dx/dt
or (m' + AxÏ)dx = (m'v')dt
Integrate (m' + AxÏ)dx from 0 to x and (m'v')dt from t = 0 to t = 150
Hence, m'x + AÏ(x2/2) = 150m'v'
Solving this quadratic equation and substituting the values of m', A, Ï, v' we get positive value of x as 105m
Therefore, the x- coordinate of the front of the cylinder at t= 150 s is 105m
11@ Akhil, type the question carefully, you have made many mistake in that
A cylindrical solid of mass 10-2 kg
and there are many more, these can confuse everone and even if someone gives the correct
answer using the values given by you ,they'll get the correct answer but you will find it wrong
because for you the values are different.
Tiit has made all the arrangement for typing all the symbols, you must use then without fail.
Be careful from next time.
Sorry if I hurt you.
23is this 10-2 = 10-2 ??
i m assuming m = 10-2kg, A = 10-4 m2, vo=103 m/s , Ï = 10-3kg/m3
hey i m not sure but wil giv it a try
suppose cylinder has moved by x
so mass accumulated on it = ÏAx
conserving momentum along x axis
mvo= ( m +ÏAx ) v = ( m +ÏAx )dxdt
mvot = mx + ÏAx22
10-2 103 (150) = 10-2x + 10-310-4x2/2
x2 + (2)(105)x - (3)108=0
Δ= (2)2(105)2 + 4(3)1010
= 4 (1010) + 12 (1010) = 16(1010)
x = -(2)(105) + 4 (10)52= 105 m
23edit : the eqn will be x2 + (2)(105)x - (3)1010=0
( corrected the constant term )
1wht does the red colour on some posts indicate? ans my 2 que plz! pljj!!
11@ Shubham,
The pink post (red color on some post) indicate that the answer given to the question are perfect, i.e, correct.
They are pinked by the moderators,i.e, Nishant Sir, Manish Sir or Lokesh Sir.
1Ok @ khyati thnx nd please answer the prev que...wil try to solve on my own :D..................this wue....... wht does it mean to say it passes the plane x=0 ????
11@ Shubham,
See there is cylindrical solid mass which is moving inside a plane in positive x direction, So the
meaning of passing through plane is that as the time is increasing it is moving inside the plane
and at t=0, it is at x=0.
Hope this clears your doubts. [1]
11@ Vivek, it's not necessary that pink is the favourite color of all gals [3]
1can nyone tell me whether we can copy da questn above n send to pendrive as i m facing a prob.in it.plz help
