302 - Mechanics part 1 Questions Answers
Time of flight of a projectile thrown from ground is 8√5 s. It will rise up to a height of.........
T = 2usinθ/g
by formula calculate usinθ
then put in H formula
A resistor of 10kΩ having tolerance 10 % is connected in series with another resistor of 20 kΩ having tolerance 20 %. The tolerance of the combination will be approximately.
resultant = (10±1)+(20±4) = 30±5
so tolerance = 5*100/30 = 50/3 = 16.67 %
A projectile is projected with velocity u= ai+bj from ground if acceleration due ti gravity is g, then the change in velocity of the projectile in second of projection is
for t sec of projection v = u + at so v = ai+bj-gtj
now calculate v-u then mod
A shell at rest at large height explodes into two parts which move horizontally with speeds v1 and v2. After what time their instantaneous velocities are perpendicular?
please see iitbrain video for the answer in you tube
iitbrain.com in you tube
video name i e irodov problem 1.11
A particle moving in uniform circular motion undergoes an angular displacement α (less then 360°) from point P and Q. Then angle between average velocity from P to Q and velocity at Q is........
as clear from the figure that angle is a/2
A particle moves in the plane xy with velocity v = ai + bxj, where i and j are the unit vectors of the x and y axes, and a and b are constants. At the initial moment of time the particle was located at the point x = y = 0. Find:
(a) the equation of the particle's trajectory y(x);
(b) the curvature radius of trajectory as a function of x.
see the video in you tube by the name iitbrain.com
video i e irodov 1.35
A ball is thrown up with some velocity after time t second it is at height h from the ground which is given by h = at -1/2 bt2. The dh/dt is zero at t equal to
dh/dt = a - bt
so for dh/dt = 0
t = a/b
A ball is thrown up with velocity 20m/s relative to ground in open lift moving up with speed 10m/s at time of throwing and has an acceleration of 2 m/s upward, ball meets lift after
relative to lift
u of the ball = 20-10 = 10
a = 10-(-2) = 12 downward
and s = 0 because ball returns to initial point
now use s = ut + 1/2 a t2
Velocity of a particle is given by v= 3-t m/s. The distance travelled by particle in 1st four second is
∫04 |v|dt = ∫03 (3-t)dt + ∫34(t-3)dt
now solve
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.