302 - Mechanics part 1 Questions Answers
If a particle is moving on a circular path with constant speed, then the angle between the direction of acceleration and its position vector wrt centre of circle will be
it should be 180 degree
Two blocks of masses 2 kg and 4 kg are hanging with the help of massless string passing over an ideal pulley inside an elevator. The elevator is moving upward with an acceleration g/2. The tension in the string connected between the blocks will be
equation 1
4g+4g/2-T = 4a
equation 2
T-2g-2g/2 = 2a
now solve
let tangential acc is a
then velocity after covering half circle
v2 = 2a(πr)
so centripetal acc = v2/r = 2aπr/r = 2aπ
so angle = tan-1[2aπ/a] = tan-12π
here notice that velocity will be in the direction of tangential acc.
For what angle should a body be projected with velocity 24m/s just to pass over the obstacle 16 m high at horizontal distance of 32 m? Take g=10 m/s^2
use the eq.
y = xtanθ-(gx2/2u2cos2θ)
A rocket is fired vertically from the ground. It moves upward with a constant acceleration 10 m/s2 for 30s after which the fuel is consumed. After what time from the instant of firing the rocket will attain the maximum height?
total height covered in upward journey = at2/2 = 10*30*30/2 = 4500 = 4.5 km
after consuming all the fuel, it will fall with gravity and g = 10 m/s2 so it will move the same distance again upto stop
so max H = 9 km.
a bullet is fired vertically upwards with an initial velocity of 50 m/s. It covers a distance h1 during the first second and a distance h2 during the last 3 seconds of its upward motion. If g = 10 m/s2, h1 and h2 will be related as
remember that gravity reduces velocity by 10 m/s on upward journey.
Also understand the diagram given
a car is going eastwards with a velocity of 8m/s. To the passengers in the car , a train appears to be moving northwards with a velocity 15m/s. What is the actual velocity of the train?
vobserver = 8i
vtrain,observer = 15j
so by formula
vtrain,observer = vtrain - vobserver
The displacement x of a particle varies with time t as x2= 1+t2 . What is its acceleration?
on diffrentiation we will get
2xdx/dt = 2t
so xv = t (1)
on differentiating again
xdv/dt + vdx/dt = 1
or xa + v2 = 1 (2)
now put v from (1) to (2) we will get answer
a particle moving with constant acceleration on a straight line ultimately comes to rest. What is the angle between its initial velocity and acceleration
180 degree because it is retarded particle
a particle moves along a straight line. Its position (x) at any instant is given by x= 32t - 8t3/3, where x is in metre and t in second. Find the acceleration of the particle at the instant particle comes to rest
differentiate the given function and find v, again differentiate find a
compare v to 0 for rest, get t and put in a