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# Newton’s laws of motion, Conservation of Linear Momentum, Collisions [Physics Tutorial]

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We will treat Physics and focus on Newton’s laws of motion, Conservation of Linear Momentum, Collisions.

How well do you know Newton’s laws of motion, Conservation of Linear Momentum, and Collisions? Whether or not you know those topics, this page is for you!

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## Newton Laws of Motion, Conservation of Linear Momentum, Collisions

### Newton’s First Law of Motion

Newton’s First Law of Motion – Idea of Inertia

States that object continues in its states of rest or uniform motion in a straight line unless an external force acts on it.

It is a common experience that a body at rest will remain at rest. For example, a book at rest on top of a table will remain there unless something pushes it or pulls at it. Also as was first pointed out by Galileo, the motion of a body once started, would continue along a straight line path unless some forces cause it to change.

This means that a moving object, if left to its self, will move in the same direction forever. That is once a body has been set in motion it is no longer necessary to on it to maintain it in motion. But our everyday experience does not seem to agree with this. We know that any sliding object slows to a stop.

Why? Because friction or gravity or other external forces slow moving objects to a stop. But in the absence of any unbalanced force, an object will move forever with the same velocity. It would neither slow down nor deflect.

This tendency of bodies to remain in their state of rest or of uniform linear motion in the absence of applied forces is known as Inertia. Newton’s first law shows that inertia is inherent in a body at rest or the one moving in a constant velocity. Inertia is a property of matter. Mass is a measure of inertia, the more massive an object is, the more inertia it has.

Newton’s first law explains what force does, but does not suggest how force should be measured.

### Newton’s 2nd Law of Motion

The second law of motion states that the rate of change of momentum is proportional to the impressed force and takes place in the direction of that force.

Newton’s second law enables us to define an absolute unit of force which remains constant under all conditions. In symbols the law states that

F a change in momentum / time taken for the change
F a mv – mu/t
Where m, u, v, t are mass, initial velocity, final velocity and time respectively of motion of the body acted upon by a force F and the product of m and v is called the momentum.

F a m (v – u/t)
F a ma
Where v – u/t = acceleration, a.
Thus, F = kma, where k is a constant.
The unit of force is chosen so that k = 1. Hence we can write F = ma.

In the SI unit F is in Newton, m is in kilogram, while the acceleration a, is in meter per square second (ms-2). The Newton is the unit of force which gives a mass of 1 kg an acceleration of 1ms-2.

Thus, F = m (v – u/t) or Ft = mv – mu

The product F.t is the impulse, I of the force.

I = Ft = change in momentum.

Unit of I is Newton-second (Ns). This is also a unit of momentum.

We see that Newton’s second law gives a measure of force as the product of mass and the acceleration of a body. Hence acceleration, a, is given by a = F/m

This second law of motion also gives an operational definition of force as the rate of change of momentum with time.

Example 1: An unbalanced force of 20 N acts on a 4 kg mass. What acceleration does it give?

Solution:
F = ma
20 N = 4a
a = 5 ms-2

Example 2: A body of mass 2 kg falls freely from rest through a height of 50 m and comes to rest having penetrated 5.0 cm of sand. Calculate (i) the velocity with which the body hits the sand (ii) the time taken in falling (iii) the average force exerted by the sand in bringing  the body to rest.

Solution
(i) v2 = u2 +2gs
Since u = 0, we have
v2 = 2gs =2 x 10 x 50 = 1000
v = Ö1000 = 31.62 ms-1

(ii) From v = u + gt
u = 0
t = v/g = 31.62/10
= 3.16 s

(iii) The initial velocity (u) when body hits the sand = 31.62 ms-1.
Final velocity after penetrating 5.0 cm of sand = 0
Using v2 = u2 + 2as, we have
0 = (31.62)2 + 2 x a x 5/100
0=(31.62)2+a/10
A crossing sign of equality
-a/10=(31.62)2
Cross multiplying
A= -10(31.62)2
a = -9998 ms-2
F = ma
= 2 x 9998 N or 19.996 x 103 N

### Newton’s Third Law of Motion

States that action and reaction are equal and opposite. Or to every action there is an equal and opposite reaction.

The law implies that when a body A exerts a force FA on a body B, the body B always exerts a force FB on the body A. The force FA and FB are equal in magnitude but opposite in direction and since force is a vector, we can write FA = -FB. FA is the action force, FB is the reaction force.

For example when you push down on your desk with your fingers, the desk pushes up on your finger with an equal force. If you hit a wall with your head, the force exerted by the head acts on the wall; at the same time the wall exerts an equal and opposite force on your head.

You will feel this reaction as pain in your head. In the above examples, either force may be considered the action or the other treated as the reaction.

### Conservation of Linear Momentum

Imagine standing on a giant skateboard that is at rest. What is the total momentum of you and the skateboard? It must be zero because it is at rest.

Now suppose you walk on the skateboard, what happens to the skateboard? When you walk in one direction, the skateboard moves in the other direction. This situation can be understood well, if you know the concept of Conservation of Linear Momentum.

Conservation of any quantity includes:

• The quantity to be constant
• The quantity does not change until and unless any other quantity for example force is applied on it.
• The density (sigma) of the quantity over the system is a constant term.

The conservation of momentum can be of two types:

• Conservation of linear momentum
• Conservation of angular momentum

### Principle of Conservation of Linear Momentum

The conservation of linear momentum is based on the principle of Newton’s first law of motion. It implies that for an isolated system, i.e., for a system with no external force, the momentum remains a constant quantity.

It also implies the Newton’s third law of motion, i.e., the law of reciprocal actions which states that the force acting between systems is opposite in sign and equal to each other.

Law of Conservation of Linear Momentum

All the bodies have the power of exerting a force on other bodies that are in their line of motion. The magnitude or quantity that measures this capacity of the bodies is called Linear Momentum.

The law of conservation of linear momentum states that the momentum will remain constant no matter what until and unless any external force comes into action.

This results into the fact that the center of mass of the system of objects will move with the same or constant velocity unless and until it is being acted upon by external force.

The Conservation of momentum is mathematically the result of the homogeneity of space, i.e., conservation of momentum implies that the physical laws are independent of the position.

Example 3: When a gun is fired if we assume that the initial position was at rest and hence the initial momentum to be zero the final momentum should also be zero according to the law of conservation of momentum.

Let us suppose the file cabinet is in the middle of the room, a room with a smooth floor, we give it a push in order to move it close to the wall and before we realize it slams into the wall. It is difficult to stop because it has linear momentum.

Conservation of Linear Momentum Equation

For two objects with initial masses of m1 and m2 and initial velocity of u1 and u2 with final velocities after collision to be v1 and v2, we can write the law as:

• m1u1+m2u2 = m1v1+m2v2

The momentum remains conserved, i.e., the momentum that is lost initially is equal to the momentum gained afterwards.

The vector sum of all the given momenta for a closed system with no external force acting on it remains constant:

• P1 + P2 + P3 + P4 + .. + Pi = K

Where ‘K’ is a constant.

Also Linear momentum of a system of particles for a system of particles with a mass of m1, m2, m3 and so on and a velocity v1, v2, v3 and so on the linear momentum can be expressed as:

• P = ∑mv = m1v1+m2v2+m3v3+m4v4+……….+mivi

For two objects with initial masses of m1 and m2 and initial velocity of u1 and u2 with final velocities after collision to be v1 and v2 we can write the law

• m1 u1 + m2 u2 = m1 v1 + m2 v2
• Also Δp1 = −Δp2
• P-The product of mass and the velocity of a particle
• P = m v

MAGNITUDE:

P = m v
or P2 = m2v2 = 2m(1/2mv2) = 2mK
Thus, P=Ö2Km
K = P2/ 2m

### Collisions

There are two principal types of collisions – the elastic and the inelastic collisions.

### Application to collisions

By itself, the law of conservation of momentum is not enough to determine the motion of particles after a collision. Another property of the motion, kinetic energy, must be known. This is not necessarily conserved. If it is conserved, the collision is called an elastic collision; if not, it is an inelastic collision.

### Elastic collisions

An elastic collision is one in which no kinetic energy is lost. Perfectly elastic “collisions” can occur when the objects do not touch each other, as for example in atomic or nuclear scattering where electric repulsion keeps them apart.

A slingshot maneuver of a satellite around a planet can also be viewed as a perfectly elastic collision from a distance.

A collision between two pool balls is a good example of an almost totally elastic collision, due to their high rigidity; but when bodies come in contact there is always some dissipation.

A head-on elastic collision between two bodies can be represented by velocities in one dimension, along a line passing through the bodies. If the velocities are u1 and u2 before the collision and v1 and v2 after, the equations expressing conservation of momentum and kinetic energy are:

m1 u1 + m2 u2 = m1 v1 + m2 v2

½ m1u21 + ½ m2u22 = ½ m1 v21 + ½ m2v22

A change of reference frame can often simplify the analysis of a collision. For example, suppose there are two bodies of equal mass m, one stationary and one approaching the other at a speed v (as in the figure).

The center of mass is moving at speed v/2 and both bodies are moving towards it at speed v/2. Because of the symmetry, after the collision both must be moving away from the center of mass at the same speed. Adding the speed of the center of mass to both, we find that the body that was moving is now stopped and the other is moving away at speed v.

The bodies have exchanged their velocities. Regardless of the velocities of the bodies, a switch to the center of mass frame leads us to the same conclusion. Therefore, the final velocities are given by

v1 = u2
v2 = u1.

In general, when the initial velocities are known, the final velocities are given by

v1 =  (m1 – m2/m1 + m2) u1 + (2m2/m1 + m2) u2
v2 = (m2 – m1/m1 + m2) u2 + (2m1/m1 + m2) u1.

If one body has much greater mass than the other, its velocity will be little affected by a collision while the other body will experience a large change.

### Inelastic collisions

In an inelastic collision, some of the kinetic energy of the colliding bodies is converted into other forms of energy such as heat or sound.

Examples include:

• traffic collisions, in which the effect of lost kinetic energy can be seen in the damage to the vehicles;
• electrons losing some of their energy to atoms (as in the Franck–Hertz experiment); and
• particle accelerators in which the kinetic energy is converted into mass in the form of new particles.

In a perfectly inelastic collision (such as a bug hitting a windshield), both bodies have the same motion afterwards. If one body is motionless to begin with, the equation for conservation of momentum is

m1u1 = (m1 + m2) v,
so
v = (m1/m1 + m2)u1

In a frame of reference moving at the speed v), the objects are brought to rest by the collision and 100% of the kinetic energy is converted.

Application of Newton’s and Conservation of Momentum Laws

• Recoil of a gun
When a bullet is fired from a gun, the gun jerks backward or recoils. Before the gun is fired, the initial momentum of the gun-bullet moves forward with a certain velocity. It has a forward momentum equals to m1v1 where m1 is the mass of bullet and its velocity is v1. The momentum of the gun is given by m2v2 where m2 is the mass of the gun and v2 is its velocity. Since the momentum must be conserved we have,
m1v1 + m2v2 = 0; Hence m1v1 = – m2v2
The negative sign shows that the momentum of the gun is directed oppositely to that of the bullet. Therefore the gun jerks backward or recoils or recoils. Because the mass of the gun is much larger than that of the bullet (m2 >> m1) the velocity and kinetic energy are much smaller than those of the bullet (v2 < v1).
• We can also consider the system of a case of the operation of Newton’s third law. The propulsive force (action) acting on the bullet must be equal and opposite to the recoil force (reaction) acting on the gun.
• Jet and Rocket Propulsion
The principle of conservation of linear momentum is also utilized in the propulsion of jet aircraft and the rockets used for launching satellites. Gases are burnt within the combustion chambers of the engine. As jets of the hot gas are expelled downwards through the tail nozzle at very high speeds from the rockets or aircraft, an equal and opposite momentum is given to the rocket or aircraft which the moves. The principle underlying the propulsion of rocket can be illustrated in the laboratory using an inflated balloon. If the inflated balloon is pierced with a pin in one direction and released, the balloon will be seen to move in the opposite direction.
• Why Walking is Possible
A person walks by pushing with his or her foot against the ground. The ground exerts an equal and opposite force back on the person.
It is this force on the person due to the reaction of the ground that moves him or her forward. Thus a person walking is actually pushed forward by the reaction force of the ground on him and not by his own push on the ground.

Example 4

• A rocket is burning fuel at the rate of 200g/s and ejecting all the gas in one direction at the rate of 400 m/s. What is the maximum weight of the rocket can have if it is going to move vertically upwards?

Solution

Mass of gas per second = 200/1000 kg/s
Velocity of expulsion = 400m/s
Momentum change per second = 200/1000 x 400kgm/s2
From Newton’s second law, we have
F = W = 200/1000 x 400 N
= 80 N.

### EXERCISES

Lets see how much you’ve learnt, the first person to answer all the questions correct gets N2,000 from Allschool. Please make sure you read the tutorial before and after attempting the questions. You the comment box below to tell us the answers:

1. A mass of 4 kg is moving at a speed of 10 m/s in a frictionless surface. It collides with a 3kg mass moving in the same direction at 5 m/s. What is the final velocity of the system after the collision?
A. 7.86 m/s B. 9.86 m/s C.  6.98 m/s  D. 3.21 m/s
2. There are two cars moving at a speed of 50 km/s and 70 km/s with mass of 100 kg and 60 kg respectively. Find the final speed of both after collision?
A. 60 km/s B. 57.5 km/s C. 68.5 km/s  D. 56 km/s
3. Which of these statements is in incorrect?
A. An elastic collision is one in which no kinetic energy is lost
B. Perfectly elastic “collisions” can occur when the objects do not touch each other
C.  Before the gun is fired, the initial momentum of the gun-bullet reduces with a certain velocity
D. A collision between two pool balls is a good example of an almost totally elastic collision, due to their high rigidity
4. A body will remain in its state of rest or uniform motion in a straight line unless an external force acts on it. Which of the Newton’s laws is this?
A. First Law B. Second Law C. Third Law  D. A reflection of both the first and the second laws.
5. Which is correct being the equation Conservation of Linear Momentum?
A. m1u1+m2u2 = m1v1+m2v2
B. m1u1 + m1v1 = m2u2 + m2v2
C. m1u1 – m2v2 ­=  m2u2 – m1v1
D. m1u1 + m2v2 ­=  m2u2 + m1v1
6. A cyclist of mass 30 kg exerts a force of 250 N to move his cycle. acceleration is 4 ms−2. force of friction between the road and tyres will be
A. 120 N B. 130 N C. 150N D. 115 N
7. Force that produces an acceleration of 1 ms−2 in a body of mass of 1 kg is called
A. slow newton
B. zero newton
C. one newton
D. two newton
8. Acceleration that is produced by a 15N force in a mass of 8 kg will be equal to
A. 1.5 ms−2
B. 1.87 ms−2
C. 2.35 ms−2
D. 2 ms−2
9. When a net force act on a body, it produces acceleration in the body in the direction of the net force which is directly proportional to the net force acting on the body and inversely proportional to its mass. This statement is called
A. newton’s 2nd law of motion
B. newton’s 1st law of motion
C. newton’s 3rd law of motion
D. law of momentum
10. Mass of a body (m) into acceleration (a) is equal to
(a) inertia
(b) displacement
(c) momentum
(d) force

Credit: This PHYSICS tutorial was prepared by Obiejemba Victor.
Victor is a pharmacist, physics tutor and an academic lover who loves teaching and reading.

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