Home PHYSICS NEWTON’S LAWS OF MOTION-COLLISION PHYSICS FORM 5 AND 6

NEWTON’S LAWS OF MOTION-COLLISION PHYSICS FORM 5 AND 6

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NEWTON’S LAWS OF MOTION-COLLISION PHYSICS FORM 5 AND 6 NEWTON’S LAWS OF MOTION - MOTION PHYSICS FORM 5 AND 6 PROJECTILE MOTION PHYSICS FORM 5 AND 6 CIRCULAR MOTION PHYSICS FORM 5 AND 6 ERRORS PHYSICS FORM 5 AND 6 MEASUREMENT PHYSICS FORM 5 AND 6

NEWTON’S LAWS OF MOTION-COLLISION PHYSICS FORM 5 AND 6

A collision is a process in which two or more bodies suddenly smash into each other.

An impact at the point of the collision causes an impulse on each of the colliding bodies’ resulting in a change in momentum.

There are mainly two types of collision.

a) Elastic collision.
b) Inelastic collision.

(a) Elastic collision

The collision whereby the colliding bodies take very short time to separate is known as elastic collision. In this kind of collision, both the momentum and kinetic energy are conserved. Fig 4.18 illustrates the collision of two spherical bodies with masses C:\thlb\cr\tz\NEWTON_files\image007.gif

and m2initially moving with velocities C:\thlb\cr\tz\NEWTON_files\image165.gif

and C:\thlb\cr\tz\NEWTON_files\image166.gif

respectively where C:\thlb\cr\tz\NEWTON_files\image167.gif

. At the point of collision, the rear body exerts a force C:\thlb\cr\tz\NEWTON_files\image168.gif

on the front body and at the same time the front body exerts an equal but opposite force -F2on the rear body. After collision the rear body slows down to, velocity C:\thlb\cr\tz\NEWTON_files\image159.gif

whereas the front body picks up the motion attaining velocity C:\thlb\cr\tz\NEWTON_files\image169.gif

C:\thlb\cr\tz\__i__images__i__\VELOC1.pngFigure 4 .17

Impulse

The impulse of a force is the product of force applied and time interval remain in action, that is

C:\thlb\cr\tz\NEWTON_files\image171.gifC:\thlb\cr\tz\NEWTON_files\image172.gifThe unit of impulse is Newton-second (Ns). From the Newton’s second law of motion we have also seen that C:\thlb\cr\tz\NEWTON_files\image173.gifAnd therefore

C:\thlb\cr\tz\NEWTON_files\image174.gif. This means the impulse is equal to the change in momentum

C:\thlb\cr\tz\NEWTON_files\image175.gifThis means that the impulse is a kilogram-meter per second ( C:\thlb\cr\tz\NEWTON_files\image011.gif)

Principle of conservation of linear momentum:

“In system of colliding particles, the total momentum before collisions is equal to the total momentum after collision so long as there is no interference to the system”

From Newton’s third law of motion, action and reaction are equal but opposite.

For example at the point of impact in fig 4.18

C:\thlb\cr\tz\NEWTON_files\image176.gif

Since the action and reaction are taken at an equal interval of time C:\thlb\cr\tz\NEWTON_files\image177.gif

to remain in action each body experiences the same impulse that is

C:\thlb\cr\tz\NEWTON_files\image178.gif

Where C:\thlb\cr\tz\NEWTON_files\image168.gif causes change in momentum on  C:\thlb\cr\tz\NEWTON_files\image008.gif and C:\thlb\cr\tz\NEWTON_files\image179.gifcauses change in momentum on  C:\thlb\cr\tz\NEWTON_files\image007.gif such that C:\thlb\cr\tz\NEWTON_files\image180.gif) and C:\thlb\cr\tz\NEWTON_files\image181.gif

C:\thlb\cr\tz\NEWTON_files\image182.gif

 Or

C:\thlb\cr\tz\NEWTON_files\image183.gif

Collecting initial terms together and final terms together we have

C:\thlb\cr\tz\NEWTON_files\image184.gifEquation (4.27) summarizes principle of conservation of momentum.

Conservation of kinetic energy:

“Work is done when the force moves a body through a distance, C:\thlb\cr\tz\NEWTON_files\image185.gif

In motion the work done is translated into a change kinetic energy as it can be shown from the second law of motion and third equation of linear motion”.

C:\thlb\cr\tz\NEWTON_files\image186.gif  andC:\thlb\cr\tz\NEWTON_files\image187.gif

But from

C:\thlb\cr\tz\NEWTON_files\image188.gif

C:\thlb\cr\tz\NEWTON_files\image189.gif

C:\thlb\cr\tz\NEWTON_files\image190.gif

In the case of collision we talk in terms virtual distance and therefore virtual work done the forces of action and reaction.
The virtual work done on C:\thlb\cr\tz\NEWTON_files\image191.gifby C:\thlb\cr\tz\NEWTON_files\image168.gif is given by C:\thlb\cr\tz\NEWTON_files\image192.gif

Likewise the virtual done on;
C:\thlb\cr\tz\NEWTON_files\image007.gif

by C:\thlb\cr\tz\NEWTON_files\image193.gif is C:\thlb\cr\tz\NEWTON_files\image194.gif

and hence

C:\thlb\cr\tz\NEWTON_files\image195.gif

– C:\thlb\cr\tz\NEWTON_files\image196.gif

C:\thlb\cr\tz\NEWTON_files\image197.gif

Collecting the initial quantities together on one side and the final quantities together on the other side we get

C:\thlb\cr\tz\NEWTON_files\image198.gifEquation (4.29) is the summary of the conservation of kinetic energy in a system of colliding particle provided the collision is perfectly elastic.

(b) Inelastic collision

There are certain instances whereby the colliding bodies delay in separating after collision has taken place and at times they remain stuck together.

Delaying to separate or sticking together after collision is due to in elasticity and hence elastic collision. In this case it only the momentum which is conserved but not kinetic energy.

The deformation that takes place while the bodies are exerting onto each other in the process of colliding, results into transformation of energy from mechanical into heat and sound the two forms of energy which are recoverable.

Once the energy has changed into heat we say that it has degenerated, it is lost to the surroundings.

When the two bodies stick together after impact they can only move with a common velocity and if they do not move after collision then the momentum is .said to -have been destroyed.

C:\thlb\cr\tz\__i__images__i__\inelastic.png

Figure 4.18

Coefficient of restitution

One of the measures of elasticity of the body is the ratio of the different in velocity after and before the collision.

Before colliding, the space between the particles decreases as the rear body overtakes that in front but after collision the space between them widens as the front particle run away from the rear one.

The difference in velocity before collision is called velocity of approach and that after collision is called velocity of separation. The ratio of velocity of separation to velocity of approach is known as coefficient of restitution.

Let C:\thlb\cr\tz\NEWTON_files\image200.gifcoefficient of restitution, ( C:\thlb\cr\tz\NEWTON_files\image201.gifvelocity of separation, C:\thlb\cr\tz\NEWTON_files\image202.gifvelocity of approach.

C:\thlb\cr\tz\NEWTON_files\image203.gif

For perfectly elastic collision, C:\thlb\cr\tz\NEWTON_files\image204.gif in this case perfectly inelastic collision C:\thlb\cr\tz\NEWTON_files\image205.gif. But collision result into explosion, C:\thlb\cr\tz\NEWTON_files\image206.gifotherwise in normal circumstances, 0 C:\thlb\cr\tz\NEWTON_files\image207.gif the coefficient of restitution cannot be 1 due to the fact that it does not matter how hard the colliding bodies are they always undergo deformation at the moment of impact and hence take longer to recover to their original shape while separating. We only assume C:\thlb\cr\tz\NEWTON_files\image204.gif to make calculation simpler.

Oblique collision

In the previous discussion on collision we dealt with direct impingement of one body onto the other along the line joining their common center. However there are situations in which bodies collide at an angle. This is known as oblique collision. Fig 4.19 illustrates the oblique collision of two bodies of mass C:\thlb\cr\tz\NEWTON_files\image208.gif

and C:\thlb\cr\tz\NEWTON_files\image209.gifinitially moving at velocities u1 and urespectively in the x-direction. After collision the body in front moves along the direction making an-angle C:\thlb\cr\tz\NEWTON_files\image210.gifwith the initial direction whereas the rear body goes in the direction making an angle C:\thlb\cr\tz\NEWTON_files\image211.gifwith initial direction. Given the initial conditions of the colliding bodies, the final velocities and directions after collision can be found 

C:\thlb\cr\tz\__i__images__i__\COLLSION.pngFigure 4.19

Applying the principle of conservation of linear momentum in equation (4.27) we can come up with more,equations for solving problems on oblique considering the motion in x – and y – directions

(a)Motion along x-direction

C:\thlb\cr\tz\NEWTON_files\image213.gif Where
  C:\thlb\cr\tz\NEWTON_files\image214.gifC:\thlb\cr\tz\NEWTON_files\image215.gifC:\thlb\cr\tz\NEWTON_files\image216.gifC:\thlb\cr\tz\NEWTON_files\image217.gif

(b) Motion along y-direction

C:\thlb\cr\tz\NEWTON_files\image218.gifIf initially the bodies are not moving in y-direction then
C:\thlb\cr\tz\NEWTON_files\image219.gif  and C:\thlb\cr\tz\NEWTON_files\image220.gif

C:\thlb\cr\tz\NEWTON_files\image221.gif

Which is

C:\thlb\cr\tz\NEWTON_files\image222.gif

Applying the definition of coefficient of restitution in equation (4.30), we have

C:\thlb\cr\tz\NEWTON_files\image223.gif

Or

C:\thlb\cr\tz\NEWTON_files\image224.gif………………………………………(4.33)

The ballistic balance

Ballistic balances are used in determining velocities of bullets as well as light comparison of masses. To do this a wooden block of mass M is suspended from light wires so that it hangs vertically. A bullet of mass m is fired horizontally towards a stationary block.

If the bullet is embedded inside the block, the two swings together as a single mass this is inelastic collision. The block will swing until the wires make an angle θ with the vertical as in figure 4.20

C:\thlb\cr\tz\__i__images__i__\ballistic.png

Figure 4 .20 shows the ballistic pendulum.

Since the collision is inelastic, only the momentum is conserved. If C:\thlb\cr\tz\NEWTON_files\image226.gif

and C:\thlb\cr\tz\NEWTON_files\image165.gif

are the mass and initial velocity of the bullet and M, C:\thlb\cr\tz\NEWTON_files\image227.gif

the mass and initial velocity of the block, then by principle of conservation of linear momentum.

C:\thlb\cr\tz\__i__images__i__\nit11.pngFrom which

C:\thlb\cr\tz\NEWTON_files\image229.gif

After impact the kinetic energy of the system at the beginning of the swing is transformed into gravitational potential energy at the end of the swing and therefore

C:\thlb\cr\tz\NEWTON_files\image230.gif

C:\thlb\cr\tz\NEWTON_files\image231.gif

C:\thlb\cr\tz\NEWTON_files\image232.gif

Substituting for v in equation (4.34) we get
C:\thlb\cr\tz\NEWTON_files\image233.gif

Thus the initial velocity of the bullet is found to be
C:\thlb\cr\tz\__i__images__i__\nit13.png

In fig 4.21, suppose the length of the wire is C:\thlb\cr\tz\NEWTON_files\image235.gif

before the block swings. After swinging, the center of gravity of the block rises by a distance C:\thlb\cr\tz\NEWTON_files\image236.gif

reducing the vertical distance to C:\thlb\cr\tz\NEWTON_files\image237.gif

. By forming the triangle of displacements the values of C:\thlb\cr\tz\NEWTON_files\image236.gif

and C:\thlb\cr\tz\NEWTON_files\image211.gif

can easily be found as shown in fig 4.21

C:\thlb\cr\tz\__i__images__i__\decve.png

Figure 4.21

C:\thlb\cr\tz\NEWTON_files\image239.gif

The height C:\thlb\cr\tz\NEWTON_files\image236.gifis

C:\thlb\cr\tz\NEWTON_files\image240.gif

……………………………………………….(4.37)

Or

C:\thlb\cr\tz\NEWTON_files\image241.gif

The angle the wire makes with vertical is

C:\thlb\cr\tz\NEWTON_files\image242.gif

………………………………………(4.38)

Reaction from a jet engine

The operation of a jet engine depends on the third law of motion where the escaping mass of hot gases exerts force on jet enabling it to move forward. Air  is first sucked in through the front side then compressed

The oxygen contained in this air intake is used in burning the fuel producing gases which when expelled  at a very high speed through the rear action forces are created and hence forward thrust . Fig 4.22 illustrate the principle of a jet in which the mass of air Ma is taken in at the rate of

C:\thlb\cr\tz\NEWTON_files\image243.gif

with relative velocity C:\thlb\cr\tz\NEWTON_files\image244.gif

and passes through the engine at the rate C:\thlb\cr\tz\NEWTON_files\image245.gif

with relative velocity of C:\thlb\cr\tz\NEWTON_files\image246.gif

.The mass of gases C:\thlb\cr\tz\NEWTON_files\image247.gif

produced at the rate of  C:\thlb\cr\tz\NEWTON_files\image248.gif

by combustion is ejected at a relative velocity C:\thlb\cr\tz\NEWTON_files\image246.gif

.The total rate of change of momentum of the system is therefore given as

C:\thlb\cr\tz\NEWTON_files\image249.gif

…………(4.39)

C:\thlb\cr\tz\__i__images__i__\newton.png

Figure  : 4 .22 Jet engine

From Newton’s second law of motion, the rate of change of momentum is equal to force.

Therefore equation (4.39) represents the forward thrust on the jet aircraft. Some jet aircraft have two identical engines and others have four. The total thrust is the product of number of engines and thrust of one engine.

Reaction from a rocket

Unlike the jet engine, the rocket carries all of its propellant materials including oxygen with it.

Imagine a rocket that is so far away from gravitational influence of the earth, then all of the exhaust hot gases will be available for the propelling and accelerating the rocket. Fig 4.24 is an illustration of a rocket of mass m carrying the fuel of mass C:\thlb\cr\tz\NEWTON_files\image251.gif

such that the total mass at time t is ( C:\thlb\cr\tz\NEWTON_files\image252.gif

) moving horizontally far away from the earth surface. As the fuel burns and gases formed expelled from the rocket at a velocity of .∨g after sometime C:\thlb\cr\tz\NEWTON_files\image177.gif

, the mass of the rocket becomes m but its velocity increases to ( C:\thlb\cr\tz\NEWTON_files\image253.gif

) as in fig.4.23(b) whilst the velocity of the ejected gases decreases from v to ( C:\thlb\cr\tz\NEWTON_files\image254.gifC:\thlb\cr\tz\__i__images__i__\jet_angine.png

Figure. 4 .23

From the principle of conservation of linear momentum

C:\thlb\cr\tz\NEWTON_files\image256.gif

C:\thlb\cr\tz\NEWTON_files\image257.gif

C:\thlb\cr\tz\NEWTON_files\image258.gif

Hence

C:\thlb\cr\tz\NEWTON_files\image259.gif

C:\thlb\cr\tz\NEWTON_files\image260.gif

Since the time rate of change in momentum is equal to thrust or force (F) on the rocket by the escaping mass of the gases the above relation can be written as   C:\thlb\cr\tz\NEWTON_files\image261.gifIf the large thrust is to be obtained, the rocket designer has to make the velocity C:\thlb\cr\tz\NEWTON_files\image262.gif

at which the hot gases are ejected and the rate at which the fuel is burnt C:\thlb\cr\tz\NEWTON_files\image263.gif

high as possible.

Rocket moving vertically upwards

Let us consider the rocket fired vertically upwards from the surface of the earth as shown in fig 4.24

C:\thlb\cr\tz\__i__images__i__\upwards.png

Figure. 4. 24

The thrust developed during combustion must be greater than the weight of the rocket if at all it is to accelerate vertically upwards C:\thlb\cr\tz\NEWTON_files\image265.gifwhich means that   C:\thlb\cr\tz\NEWTON_files\image266.gif

Reaction from the hose pipe

If a hose pipe connected to the running tap on the smooth horizontal surface, the free end that issues water seems to move backwards as the water flow out. This is yet another example of action and reaction forces.

Again if a jet of water from a horizontal hose pipe is directed at a vertical wall, it exerts an equal but opposite force on the water.

C:\thlb\cr\tz\__i__images__i__\reaction.png

Figure. 4.25

Let C:\thlb\cr\tz\NEWTON_files\image268.gif

be the initial velocity of water when leaving the pipe. On striking the wall its final velocity C:\thlb\cr\tz\NEWTON_files\image269.gifassuming that the water  does not rebound. If is the density of water, A is the cross-sectional area of the pipe, then the mass of water hitting the wall per second is given by

C:\thlb\cr\tz\NEWTON_files\image270.gif

Where

C:\thlb\cr\tz\NEWTON_files\image271.gif

C:\thlb\cr\tz\NEWTON_files\image272.gif

C:\thlb\cr\tz\NEWTON_files\image273.gif

The time rate of change in momentum of water is therefore

C:\thlb\cr\tz\NEWTON_files\image274.gif

Where C:\thlb\cr\tz\NEWTON_files\image275.gif

and C:\thlb\cr\tz\NEWTON_files\image269.gif

C:\thlb\cr\tz\NEWTON_files\image276.gif

i.e.

C:\thlb\cr\tz\NEWTON_files\image277.gif

The negative sign means the force is the reaction of the wall on water. Thus the force exerted by the water on the wall is C:\thlb\cr\tz\NEWTON_files\image278.gif

Reaction on a gun

 Consider a gun mass C:\thlb\cr\tz\NEWTON_files\image279.gif

with bullet mass C:\thlb\cr\tz\NEWTON_files\image280.gif

in it initially at rest. Before firing the gun, their total momentum is zero as in fig. 4.27(a). At the point of firing there are equal opposite internal forces as in fig 4.27(b). As the bullet leaves the gun the total momentum of the system is still zero as in fig 4.27(c).

C:\thlb\cr\tz\__i__images__i__\gun.png

Figure. 4. 26

Initially the velocity of the gun and that of the bullet are zero

C:\thlb\cr\tz\NEWTON_files\image282.gif

Therefore the initial momentum

C:\thlb\cr\tz\NEWTON_files\image283.gif

C:\thlb\cr\tz\NEWTON_files\image284.gif

(0)

C:\thlb\cr\tz\NEWTON_files\image285.gif

When the bullet leaves the gun with final velocity C:\thlb\cr\tz\NEWTON_files\image286.gif

, the gun recoil with velocity of – C:\thlb\cr\tz\NEWTON_files\image262.gif

and the final total momentum

C:\thlb\cr\tz\NEWTON_files\image287.gif

From the principle of conservation of linear momentum

C:\thlb\cr\tz\NEWTON_files\image288.gif

Thus

C:\thlb\cr\tz\NEWTON_files\image289.gif

C:\thlb\cr\tz\NEWTON_files\image290.gif

Equilibrant forces

A body is said to be in static equilibrium if it does not move under the action of external forces. For example in fig 4.3, a block is in equilibrium since it neither moves up nor down under forces R and W.

These two forces are action and reaction which cancel each other out such that the net force on the body is zero. The net external force is an algebraic sum of all the forces acting on the body that is

C:\thlb\cr\tz\NEWTON_files\image291.gif

In this case

C:\thlb\cr\tz\NEWTON_files\image292.gif

C:\thlb\cr\tz\NEWTON_files\image293.gif

C:\thlb\cr\tz\NEWTON_files\image294.gif

The forces that keep the body in equilibrium are called equilibrant forces. These are the forces whose resultant is zero. Fig 4.28(a) shows a body in equilibrium
under the action of three forces, hanging vertically. The weight W of the body establishes the tensions C:\thlb\cr\tz\NEWTON_files\image295.gif

and C:\thlb\cr\tz\NEWTON_files\image296.gif

in the sections AB and BC of the string which make angles C:\thlb\cr\tz\NEWTON_files\image211.gif

and C:\thlb\cr\tz\NEWTON_files\image297.gifrespectively with the horizontal point B along the string experiences three forces as shown In fig 4.28(b). If this point is taken as an origin and two perpendicular axes drawn, the tensions C:\thlb\cr\tz\NEWTON_files\image295.gif

and C:\thlb\cr\tz\NEWTON_files\image298.gif

appear to make angles C:\thlb\cr\tz\NEWTON_files\image211.gif

and C:\thlb\cr\tz\NEWTON_files\image297.gif

with the x – axis. The body remains in equilibrium when no motion occurs either horizontally or vertically and for that reason the net force along the horizontal direction is zero

C:\thlb\cr\tz\NEWTON_files\image299.gif

Likewise the net force on the body along the vertical direction is zero

C:\thlb\cr\tz\NEWTON_files\image300.gif

To obtain the net forces we have to find the x- and y-components of tensions C:\thlb\cr\tz\NEWTON_files\image301.gif

and C:\thlb\cr\tz\NEWTON_files\image302.gif

as shown in fig 4.28 (C).  For the horizontal components of two tension gives

C:\thlb\cr\tz\NEWTON_files\image303.gif

Where

C:\thlb\cr\tz\NEWTON_files\image304.gif

C:\thlb\cr\tz\NEWTON_files\image305.gif

C:\thlb\cr\tz\NEWTON_files\image306.gif

For the y-direction the net force

C:\thlb\cr\tz\NEWTON_files\image307.gif

Where  C:\thlb\cr\tz\NEWTON_files\image308.gif

C:\thlb\cr\tz\__i__images__i__\nit17.pngC:\thlb\cr\tz\__i__images__i__\solving.png

Figure. 4. 27

Solving for C:\thlb\cr\tz\NEWTON_files\image311.gif

in terms of C:\thlb\cr\tz\NEWTON_files\image312.gif

from equations (4.43) and (4.44 we get

C:\thlb\cr\tz\__i__images__i__\nit18.png

Exercise 

1. (a) State Newton’s laws of motion

(b) Give three examples in which Newton’s third law applies

(c) With the aid of labeled diagram explain what causes frictional force.

2   (a) A body of mass m rests on a rough inclined plane with angle of inclination C:\thlb\cr\tz\NEWTON_files\image314.gif

(i)   Explain why the body does not slide down the plane,

(ii)Draw the diagram indicating all the forces acting on the body and give each force its name, direction and magnitude.

(b) If the mass of the body in (a) is 5kg and the angle of inclination is 20° thenfind

(i)  The force that keeps the body in contact with the plane  (ii) The force that prevents the body to slide down the plane.

3(a) Differentiate between

(i)  A rough surface and a smooth surface

(ii) Static friction and kinetic friction

(iii) Coefficient of static friction and coefficient of dynamic friction .

(b) A body of mass 20kg is pulled by a horizontal force P. If it accelerates at 1.5 C:\thlb\cr\tz\NEWTON_files\image019.gif

and the coefficient of friction of the plane is 0.25, what is the magnitude of P? (Take  C:\thlb\cr\tz\NEWTON_files\image315.gif

4 (a) From the second law of motion show the expression for the force.

C:\thlb\cr\tz\NEWTON_files\image316.gif

, where C:\thlb\cr\tz\NEWTON_files\image317.gif

and C:\thlb\cr\tz\NEWTON_files\image100.gifare the mass and acceleration respectively

(b)Using the third law of motion, show that for the two colliding bodies of masses m{and m2moving along the common line at velocities u1 and u2 , before collision and at velocities v1and v2 respectively just after collision, the total momentum of the system is conserved and represented by relation

C:\thlb\cr\tz\NEWTON_files\image318.gif

. Assume elastic collision.

5 (a) Fig 4.28 shows two bodies of masses M and m connected by the light string,
if the body of mass m rests on a rough plane whose coefficient of friction is μ on releasing the system free, obtain an expression for

(i) The acceleration of the system.

(ii) The tension in the string.

C:\thlb\cr\tz\__i__images__i__\STRING.png

Figure. 4 .28

6   (a)   (i) What is the difference between the coefficient of restitution and the coefficient of friction?

(ii) Explain in details the implications of the following about the coefficient of restitution e

–         when e >1

–         when e =0

–         when  e <1

–         when C:\thlb\cr\tz\NEWTON_files\image320.gif

= l

(b) Two bodies A and B of masses 3kg and 2.5kg are moving towards each other along a common line with initial velocities 4 C:\thlb\cr\tz\NEWTON_files\image019.gif

and2.5 C:\thlb\cr\tz\NEWTON_files\image321.gif

respectively, after sometime they eventually collide elastically.

Determine their final velocities.

7(a) A body is hanging in equilibrium as shown in fig 4.29, find the tensions

C:\thlb\cr\tz\NEWTON_files\image295.gif

C:\thlb\cr\tz\NEWTON_files\image296.gif

and  C:\thlb\cr\tz\NEWTON_files\image322.gif

.Take C:\thlb\cr\tz\NEWTON_files\image323.gif

C:\thlb\cr\tz\__i__images__i__\tension.pngFigure. 4. 29

(b) A body of mass 2kg sits on a horizontal plane, and then the plane is accelerated vertically upwards at 4ms-2. Determine the magnitude of the reaction on the body by the plane.

8     The engine of a jet aircraft flying at 400 C:\thlb\cr\tz\NEWTON_files\image019.gif

takes in 1000m3 of air per second at an operating height where the density of air is 0.5 C:\thlb\cr\tz\NEWTON_files\image326.gif

, The air is used to burn the fuel at the rate of 50 C:\thlb\cr\tz\NEWTON_files\image327.gif

and the exhaust gases (Including incoming air) are ejected at 700 C:\thlb\cr\tz\NEWTON_files\image328.gifrelative to the aircraft . Determine the thrust.

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