PHYSICS FORM 5- NEWTON’S LAWS OF MOTION-COLLISION
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PHYSICS FORM 5- NEWTON’S LAWS OF MOTION-COLLISION
Collision
A collision is the process in which two or more bodies suddenly smash into each other. An impact at the point of collision causes an impulse on each of the colliding bodies’ results into 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
Figure 4 .17
Impulse,
The impulse of a force is the product of force applied and time interval remain in action, that is
The unit of impulse is Newton-second (Ns). From the Newton’s second law of motion we have also seen that
And therefore
This means that the impulse is a kilogram-meter per second (
PHYSICS FORM 5- NEWTON’S LAWS OF MOTION-COLLISION
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
Since the action and reaction are taken at an equal interval of time
Where
Or
Collecting initial terms together and final terms together we have
Equation (4.27) summarizes principle of conservation of momentum.
Conservation of kinetic energy:
“Work is done when the force moves a body through a distance,
But from
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
Likewise the virtual done on;
–
Collecting the initial quantities together on one side and the final quantities together on the other side we get
Equation (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.
Figure 4.18
PHYSICS FORM 5- NEWTON’S LAWS OF MOTION-COLLISION
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
For perfectly elastic collision,
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
Figure 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
Where
(b) Motion along y-direction
If initially the bodies are not moving in y-direction then
Which is
Applying the definition of coefficient of restitution in equation (4.30), we have
Or
PHYSICS FORM 5- NEWTON’S LAWS OF MOTION-COLLISION
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
Figure 4 .20 shows the ballistic pendulum.
Since the collision is inelastic, only the momentum is conserved. If
From which
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
Substituting for v in equation (4.34) we get
Thus the initial velocity of the bullet is found to be
In fig 4.21, suppose the length of the wire is
Figure 4.21
The height
Or
The angle the wire makes with vertical is
PHYSICS FORM 5- NEWTON’S LAWS OF MOTION-COLLISION
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
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.
PHYSICS FORM 5- NEWTON’S LAWS OF MOTION-COLLISION
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
Figure. 4 .23
From the principle of conservation of linear momentum
(
Hence
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
If the large thrust is to be obtained, the rocket designer has to make the velocity
Rocket moving vertically upwards
Let us consider the rocket fired vertically upwards from the surface of the earth as shown in fig 4.24
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
PHYSICS FORM 5- NEWTON’S LAWS OF MOTION-COLLISION
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.
Figure. 4.25
Let
Where
The time rate of change in momentum of water is therefore
Where
i.e.
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
Reaction on a gun
Consider a gun mass
Figure. 4. 26
Initially the velocity of the gun and that of the bullet are zero
Therefore the initial momentum
When the bullet leaves the gun with final velocity
From the principle of conservation of linear momentum
Thus
PHYSICS FORM 5- NEWTON’S LAWS OF MOTION-COLLISION
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
In this case
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
Likewise the net force on the body along the vertical direction is zero
To obtain the net forces we have to find the x- and y-components of tensions
Where
For the y-direction the net force
Where
Figure. 4. 27
Solving for
PHYSICS FORM 5- NEWTON’S LAWS OF MOTION-COLLISION
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
(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
4 (a) From the second law of motion show the expression for the force.
(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
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.
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
(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
Determine their final velocities.
7(a) A body is hanging in equilibrium as shown in fig 4.29, find the tensions
Figure. 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