302 - Mechanics part 1 Questions Answers
Two particles each of mass m are moving in horizontal circle with same angular speed. If both string are of same length then the ratio of tension in string T1 / T2 is
1. 3/2 2. 3 3. 2 4. 1/3
it will be 1:1
A particle is moving on a circular path of radius 1 m with 2 m/s. If speed starts increasing at a rate of 2 m/s² , then acceleration of particle is
centripetal acceleration = v2/r = 2*2/1 = 4
and tangential acceleration = 2 [given]
so acceleration = [42+22]1/2 = √ 20
A particle is revolving in a circular path of radius 2 m with constant angular speed 4 rad/s. The angular acceleration of particle is
1. zero 2. 8π² 3. 16π²
since angular velocity is constant, so zero will be the answer
If a particle is moving on a circular path with constant speed, then the angle between the direction of acceleration and its position vector w.r.t. centre of circle will be
1. zero 2. π 3. π/2 4. 2π
position will be radically outward and acceleration will be radically inward, so angle = π
If a particle of mass 2 kg is moving in circular path with constant speed 20 m/s. The magnitude of change in velocity when particle travels from A to P will be
1. 20√2 m/s 2. 40 m/s
in this question you should define A and P first then we will give you an answer.
A particle of mass m strikes elastically on a wall with velocity v, at an angle 60° from the wall the magnitude of change in momentum of ball along the wall is
(1) zero (2) mv
since momentum of the ball before strike and after strike are same along the wall so change will be 0
sir i had asked a ques and its ans was also replied by you, but still i have a doubt that in 1st part you take g as -ve because motion is against gravity and it is clear to me but in 2nd why g is taken as -ve? here, the motion is not against gravity, so plz tell me the concept of g? wten it is -ve or +ve.
b) in second part we have to calculate the time of journey from top of 25 m high building to ground.
for this we should use
h = -ut + gt2/2 [in this case we are using upward direction as - ve and downward as + ve]
so 25 = -20t + 10t2/2
or 5t2 - 20t - 25 = 0
or t2- 4t - 5 = 0
or (t+1)(t-5) = 0
or t = 5 , -1
so t = 5 sec.
This is answer of the same problem but here i am using g +ve, I want to say only that it will always be your opinion to consider the sign of direction. You can consider any of the direction as positive and negative.
A smooth sphere of radius R is made to translate in a straight line with constnt acceralation a . A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. find the speed of the particle with respect to the sphere as a function of the angle θ it slide
use energy conservation to solve it
if not possible then send a reply we will give you complete answer.
a ball is thrown vertically upwars with a velocity 20m/s from the top of a multistorey building. the height of the point from where the ball is thrown is 25m from the ground.
a) how high will the ball rise?
b) how long will it be before the ball hits the ground? take g=10m/s square
a) At heighest poing velocity will be 0
so by formula v2 = u2 - 2gh (here - is because you are moving against gravity)
so h = 20*20/2*10 = 20 m
so ball will rise upto 25+20 m from the ground.
b) in second part we have to calculate the time of journey from top of 25 m high building to ground.
for this we should use
-h = ut - gt2/2 [in this case we are using upward direction as + ve and downward as - ve]
so -25 = 20t - 10t2/2
or 5t2 - 20t - 25 = 0
or t2- 4t - 5 = 0
or (t+1)(t-5) = 0
or t = 5 , -1
so t = 5 sec.
A particle starts from rest and moves with constant acceleration. The magnitude of its displacement is
1) more than its distance traversed
2) less '' '' '' ''
3) equal to '' '' ''
4) zero
ans given is (3) but I want to know why it vyl not be less than dist. travelled & why (4) optn is wrong as in case of uniform circular motion displacement is 0 but have const. acc.??
I think the particle is moving in a straight line