302 - Mechanics part 1 Questions Answers
see you tube video of iitbrain by the name I E Irodov problem 1.3
Two Particles move in a uniform gravitational field with an acceleration g. At The Initial moment the particles were located at one point and moved with velocities v1= 3.0ms-1 and v2 = 4.0ms-1 horizontally in opp. Direction.Find The distance between the particles at the moment when their velocity vector become mutually Perpendicular?
see you tube video of iitbrain by the name I E Irodov problem 1.11
A Train Accelerates from rest at a constant rate α for sometime and then it retards to rest at a constant rate β. If te total distance covered by the particle is x, then what is the maximum velocity of the train ?
for first part of motion
v2 = 2αx1 (1)
similarly for second part of motion
v2= 2βx2 (2)
now by given conditions
x = x1 + x2
so x = v2/2α + v2/2β
so x = v2/2[1/α + 1/β]
now solve
Find the moment of inertia of a uniform square plate of mass M and edge a about one of its diagonals.
(1) Ma2 / 12
(2) 2/3 x Ma2
let the moment of inertia about a diagonal = I
then by perpendicular axis theorem 2I = 1/6 Ma2
now calculate I
let the angular velocities are ω1, ω2
then angular acceleration of one w. r. t. other = ω1 ω2
A taxi leaves the station X for Station Y every 10 min. Simulataneously, a taxi also leaves the station Y for station X Every 10min. The Taxis move at the same constant speed and go from X to Y or Viceversa in 2Hours. How Many taxis coming from the other side meet each taxi enroute from Y To X?
the answer depends on the condition that where taxi meets with the first taxi coming from opposite side.
If we consider that at the starting point taxi meets with the first taxi coming from other side, then after every 5 min it will meet with a taxi from opposite side. This will be due to the double relative speed of two taxi moving opposite. Total time length of the path = 2 hrs = 120 min
so no. of taxi = 120/5 + 1 = 25
if first taxi meet at the point a little ahead the starting point then no of taxi = 120/5 = 24
similarly other cases
A train accelerates from rest at a constant rate α for distance x1 and time t1.After that it retards to rest at constant rate β for distamce x2 and time t2.Which of the following relation is correct?
| A. x1/x2=α/β=t2/t1 | B. x1/x2=β/α=t2/t1 |
| C. x1/x2=α/β=t1/t2 | D. x1/x2=β/α=t1/t2 |
for first part x1 = αt12/2 and v = αt1
similarly for second part x2 = βt22/2 and v = βt2
on comparing v we get αt1 = βt2
so β/α = t1/t2 (1)
and x1/x2 = αt12/βt22 = β/α (2) by using (1)
A ball of mass 1 kg is projected with a velocity of 20√2 m/s from the origin of an xy coordinates axis system at an angle 45° with x-axis (horizontal). The angular momentum of the ball about the point of projection after 2s of projection is [take g=10 m/s2] (y-axis is taken as vertical)
initial vertical and horizontal velocities are 20 m/s and 20 m/s
after 2 sec. vertical and horizontal velocities are 0 m/s and 20 m/s
and y and x displacements are 20 m and 40 m
so about point of projection angular momentum after 2 sec.
= mvxy - mvyx = 1*20*20 - 1*0*40 = 400
A disc of mass 3kg rolls down an inclined plane of height 5 m. The translational kinetic energy of the disc on reaching the bottom of the inclined plane is
mgh = 1/2 mv2 + 1/2 Iω2
I for the disc about centre = 1/2 MR2
and for pure rolling v = Rω
calculate v then 1/2 mv2
A heavy solid sphere is thrown on a horizontal rough surface with initial velocity u without rolling. What will be its speed, when it starts pure rolling motion
for translational motion
0-μMg = Ma so a = -μg
so after time t velocity v = u-μgt (1)
for rotational motion
about centre, μMgr = Iα = 2Mr2α/5
so α = 5μg/2r
so ω = 0 + αt
so ω = 5μgt/2r (2)
for pure rolling
v = rω implies that u-μgt = r5μgt/2r
implies that u = 7μgt/2
so t = 2u/7μg