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