Physics: questions and answers on momentum and gravity

questions  on momentum and gravity 1 t0 5

Momentum and gravity:

Give the direction and magnitude of the net force acting on the followings

(1)  drop of rain falling down with a constant speed

(2) A cork of mass 10g floating on water

(3)) A kite held stationary on the sky

(4) A can moving with a constant velocity of 30km/hr

(5) A high speed electron in a space far from all material object and free of

Magnetic and electric filed

Read also question and answer on simple machine

Answers for 1 to 5 question on momentum and gravity

(1)The net force is zero, this is because, the rain drop is falling with a constant speed , therefore it was not accelerating ( that is  F = ma=m x 0 =0)

(2)The net force is Zero, this is because the weight of the cock with is acting downward is balance by the up thrust force with is acting upward exerted by the water. Therefore not net force is acting on a floating cock

(3)The net force is zero, this is because the kite is stationary, no movement at all, hence as per first Newton’s law of motion, no net force is acting on the kite

(4)  The net force is zero, this because the car was moving on a rough road with a constant speed. Therefore acceleration is zero, so as per second Newton’s law of motion no net force is acting on the car.

(5) The net force is zero, this is because, the high speed electron is from all fields, such as magnetic and electric field, hence the net force on is zero.

questions on momentum and gravity 6 to 10

(6)What force will be required to accelerate an electron from rest to a velocity of 104m/s in 10 seconds ( mass of electron =9 x 10-31kg)

(7)Explain while the velocity of recoiling gun is lower than that of the ejected bullet

(8) Calculate the force require to impart an acceleration of 5ms-2 to a mass of  10kg

(9)  A bullet of mass 0.04kg is fired from a gun of 9kg the bullet moving with initial velocity of 200m/s. fined the initial backward velocity of the gun

(10)  Define the impulse of force. If  a force of 6.0N act on a body for 5 seconds what is the change in momentum

ANSWERS for 6 to 10  on momentum and gravity

(6)  Solution, using F = ma. And Acceleration a= v/t, where m = 9 x 10-31kg, v =104ms-1, t = 10 seconds

So,  a =10000/10 =103ms-2 therefore  F =ma = 103 x 9 x10-31

F = 9 x 10-31 +3, F =9.0 x 10-28N

 

(7) The velocity of the recoiling is lower than that of ejected bullet because, the mass of the gun is greater than the mass of the ejected bullet. But both the momentum of recoiling gun and that of the ejected gun are the same.

 

(8)  Solution , using F =ma , where m 10kg= and a =5ms-2

 

Then F = 10 x 5 = 50N

 

(9) .  Note that the momentum of gun is equal to the momentum of the bullet,

That is MgUg =MbUb, where Mg, Ug, Mb, and Ub, are the masses and the initial velocity of gun respectively.

But Mg =9kg, Ug = x , Mb = 0.04kg, Ub = 200m/s

Therefore MgUg =MbUb   =9  x Ug =0.04 x 200

9Ug = 8, Ug = 8/9 0.888m/s

Therefore the initial velocity of the gun = 0.89m/s

 

(10). The  impulse I, can be defined as change in momentum, which is given as

I =Ft =mv (change in momentum)

 

The change in momentum when a force 6.0N act on a body for 5 seconds, is give by

Ft = change in momentum = 6 x 5 = 30Ns

 

questions and answers on momentum and gravity 11 t0 15

(11).  A body of mass 10kg is pulled along horizontal floor by a horizontal force of 48N with acceleration of 3.0m/s2. Calculate the frictional force between the body and the floor.

(12). A ball of mass 200kg traveling with velocity of 100m/s, collide with another ball of mass 800kg , moving at 50m/s in the same direction  . If they stick together, what will be their common velocity?

(13).  A rocket expels gas at the rate of 0.4kg/s. if the average force of the gas is 120N; calculate the velocity of the gas. State the law used in your calculation.

(14). A jet engine takes in 20kg of air per a second at 150m. The air is compressed, heated and discharged at 500m/s. what force is developed in the engine.

(15). State the principle of conservation of momentum. A ball of 10g traveling with velocity of 10m/s collides with another ball of 60g with velocity of 5m/s in opposite direction. If they stick together, what will be their common velocity? And in what direction.

 

ANSWERS for 11 to 15 questions on momentum and gravity

(11). Using  48N = F + ma, where 48N = force applied, F = the frictional force, ma = the for force with which the body is moving on the floor and is = 3 x 10 = 30N

Therefore F = 48 -30 = 18N

So frictional force = 18N

(12). Using m1u1 +m2u2 = v(m1 +m2) ( law of conservation of momentum ) where m1, u1, and m2, u2 , are the masses  and initial velocity of first and second ball respectively. And v is their common velocity after collision

So m1u1 +m2u2 = v(m1 +m2) =  200 x 100 + 800 x 50 = v(200 + 800)

20000 + 40000 = 1000v

60000 = 1000v

v = 60m/s (the common velocity after collision)

(13). Using third Newton’s law of motion

Ft = mv

F/v = m/t, given from the question, m/t = 0.4kg/s and F = 120N

So 120/v = 0.4

V  = 120/0.4 = 300m/s

V = 300m/s (the velocity of the gas

(14). Using

F = ma

a = v/t,  so F = mv/t = vm/t, but from the question, m/t = 20kg/s and v = 500m/s

Therefore F = 20 x 500 = 10000

F = 10,000N

For the second question

(15).  Using m1u1 +m2u2 = v(m1 +m2) where m1 =50g, u1 =10m/s,  m2 = 60g and u2 =-5m/s (second ball velocity is with minus sign because it was moving in opposite direction to the first ball) . V= ?

So m1u1 +m2u2 = v(m1 +m2) = 50 x 10 + 60 x (-5)  = v (50 + 60)

500 + (-300)  = 110

500-300 = 110v

200 = 110v

V = 1.8m/s in the direction of the first ball.

questions on momentum and gravity 16

What is the change in momentum of body if a force of 6.0N acts on it for 5s

 

 Answer on momentum and gravity 6 to 10

Force = the rate of change of momentum, which is given as

force\quad =\quad \frac { change\quad in\quad momentum }{ time } \quad that\quad is\quad \\ \\ F\quad =\frac { MV-Mu }{ t } ,\quad and\quad f\quad =6.0N,\quad t=\quad 5s\\ \\  

then\quad \\ 6\quad =\frac { mv-mu }{ 5 } \\ mv-mu\quad =\quad 6\times 5\quad =30\\ \\ \\ 

The change in momentum = 30Ns

 

 

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