303 - Mechanics part 1 Questions Answers
A bus is begins to move with an acceleration of 1 m/s2. A boy who is 48 m behind the bus starts running at 10 m/s towards the bus. After what time bus will cross the boy?
according to the given condition
1/2 t2+ 48 = 10t
solutions of this eq. = 12, 8
so boy will cross the bus after 8 sec and then bus will cross the boy after 12 sec from start.
The acceleration of a particle, varies with time according to the relation a=-mw2sinwt.The displacement of this particle at a time t will be
1) m sin(wt)
2) mwcos(wt)
3) mwsin(wt)
4) -1/2(mw2sinwt)t2
a = dv/dt = -mw2sinwt
so ∫dv = ∫-mw2sinwtdt
so v = mwcoswt + C
at t=0, v=0 so 0 = mw+C or C = -mw
so v = -mw(1-coswt)
similarly put v = ds/dt and integrate again
A motor boat going down stream ,overtakes a floating object in water .One hour later ,the motor boat turned back .If speed of motor boat in still water is constant and flow velocity is also assumed constant,motor boat will again pass the floating object after time 't ' given by
1)Greater than 2 hrs
2)Equal to 2 hrs
3)less than 2 hrs
4)less than 1 hr
see video in you tube under the name iitbrain
video i e irodov 1.1
time will be 1 hour
A paper weight is dropped from the roof of a 35 storey building each storey being 3m high. It passes the ceiling of the 20th storey at x m/s [g=10 m/s2]. Find x.
total height = (35-20+1)*3 = 48 m
now solve
A lift with open top is moving up with acceleration f. A body is thrown up with speed u relative to lift and it comes back to lift in time t. Then speed u is ...
{g = acceleration due to gravity}
relative acc. = g+f
and relative velocity = u
now solve
A projectile is thrown with speed 40 ms-1 at angle θ from horizontal. It is found that projectile as at same height at 1s and 3s. What is the angle of projection?
It is found that projectile as at same height at 1s and 3s, so half time of flight = 2 sec
now solve
A particle is thrown with a velocity of u m/s. It passes A and B at time t1 = 1s and t2 = 3s. The value of u is (g=10 m/s2).
question is not complete
When a force F acts on a particle of mass m, the acceleration of particle becomes a. Now if two forces of magnitude 3F and 4F acts on the particle (both are at an angle of 90° from mass m) simultaneously , then the acceleration of the particle is ..............
resultant force = 5F
so acceleration = 5a
A man moves in an open field such that after moving 10m on a straight line, he makes a
sharp turn of 60° to his left. Find the displacement after 8 such turns
draw a hexagon
just after 8 such turn displacement = 10√3
a particle is moving eastwards with a speed of 6m/s. After 6s, the particle is found to be moving with same speed in a direction 60° north of east. The magnitude of average acceleration in this interval of time is ..............
initial velocity = 6i
final velocity = 6cos60i+6sin60j
now solve