72 - Newton's laws of motion and friction 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 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.
If a particle (strong enough to survive in space) is released in space from earth. Where it will go? What about its motion? Will it revolve around the sun or remain stationary?
it depends on the speed with which it is thrown
a large open tank has two holes in the wall. one is a square hole of side L at a depth Y from the top and the other is a circular hole of radius R At a depth 4Y from the top . when the tank is completely filled with H2O , the quantities of H2O flowing / sec from both holes are same . then R IS equal to??
ans L / [√(2π)]
speed of water from any hole = √2gh
so volume of water per sec. = A√2gh
for square hole volume/sec = L2√2gY
for circular hole volume/sec = πR2√2g4Y
compare and get answer