302 - Mechanics part 1 Questions Answers
a body is projected vertically upward with speed 32 m/s. total time taken in its complete journey is
suppose it travelled h height thus from bottom to top it will covered h height in time t1 & at top its vel will be zero thus from eqn v= u-gt => 0= 32- 10t u can find t1= 3.2s , same time particle will take in downward journey so total time = 6.4 sec.
This answer is given by sarika and is correct answer, we made some modifications only
a body is p rojected vertically upwards with speed 20m/s .find the distance travelled by it during the last second of its upward journey(takeg=10m/s2
last sec. of upward journey means first sec. of free fall from the top so s=0(1) + 1/2(10)(1)2 = 5 meter, here g is taken 10m/s2
The velocity of a body moving along x- axis is given by v =√( 36+2x) where v is in m/s and x is in metre. Find the distance travelled by it in 2 nd second.
The answer is 7.5 m
dx/dt = √( 36+2x)
so dx/√( 36+2x) = dt
now integrate
limits of x are x1 to x2 and that of time are 1 to 2
A BALLOON WITH ITS CONTENTS WEIGHING 160N IS MOVING DOWN WITH AN ACCELERATION OF g/2 ms-2. THE MASS TO BE REMOVED FROM IT SO THAT THE BALLOOON MOVES UP WITH AN ACCELERATION g/3 ms-2 IS
according to the given conditions
for downward journey 160-R = (160/g)g/2 = 80 so R = 80
now after removing m kg the new equation
R - [160/g - m]g = [160/g - m]g/3
put the value of R and get m
एक नाव की शांत जल में चाल 5 km /h है ,तथा 1 km चोडी नदी को न्यूनतम पथ के अनुदिश 15 मिनट में पार करती है तो नदी के जल प्रवाह का वेग km/h होगा..?
let boat is moving in a direction making an angle θ with the perpendicular to the flow, then for minimum drift
5sinθ = r
and 5cosθ = 1/.25 = 4
so cosθ = 4/5
and sinθ = 3/5
so r = 3 km/hr
You can see video related to this question in you tube (iit brain) as the name irodov problems
Velocity of raindrops in still air is 4 m/s vertically downward. If wind starts blowing at the rate 3 m/s horizontally towards north, then find the direction along which a man standing on the ground would keep his umbrella ti protect himself from rain?
sir, i dont know the figure of this ques, so plz tell me this ques diagramatically.
angel of umbrella with vertical = tan-1[3/4]
एक लिफ्ट में एक सिक्का लिफ्ट के फिर्श से 2m की उचाई पर छोड़ा जाता है, लिफ्ट की उचाई 10m है | लिफ्ट 11m/s2के त्वरण से नीचे की ओर गति कर रही है | वह समय जिसके पश्चात सिक्का लिफ्ट से टकराएगा ?
the coin will fall with 10 m/s2 and acc. of lift is given as 11 m/s2. so coin will move upward with acc. 1 m/s2 relative to lift and initial velocity of coin = 0 relative to lift. so it will strike the roof of the lift
use s = ut + 1/2 at2
so 8 = 0 + 1/2 (1) t2
so t = 4
A block of mass 10 kg is released on rough incline plane. Block start decending with acceleration 2 m/s² . Kinetic friction force acting on block is ( take g= 10m/s² )
1. 10N 2. 30N
question is not correct
A bomb of mass M @ rest explodes into 2 fragments of masses m1 &m2 . The total energy released in the explosn is E if E1 &E2 represent the energies carried by m1 &m2. then which of the followg is correct
a) E1=(m2/M)E
B) E1= m1/m2 E
C) E1= m1/M E
D) E1= m2/m1 E
(1) is correct
the ratio of energy will be E1/E2 = m2/m1
A block of mass m takes t to slide down on a smooth inclined plane of angle of inclination and height h. If same block slids down on a rough inclined plane of same angle of inclination and some height and takes time n times of initial value, then coefficient friction between block and inclined plane is
the answer is [1- 1/n² ] tanθ
in first case acceleration is gsinθ
and in second case acceleration is gsinθ-μgcosθ
now use s = 1/2 a t2 in both the case and solve