NEWTON’S LAWS OF MOTION
Motion of objects are caused by unbalanced forces that act on the objects. Therefore, forces are related to motions of objects. This was discovered by Sir Isaac Newton ( 1642 – 1727 ). He stated three laws of motion which we shall look into in the treatment.
Newton’s First Law Of Motion:
Newton’s first law of motion states that an object will continue in its state of rest or motion in a straight line unless it is acted upon by an external force.
Inertia:
Inertia is the unwillingness of an object in motion to stop moving or an object at rest to start moving.
Inertia simply means unwillingness. Every object is unwilling to do other wise apart from what its is doing at the moment. Inertia is inherent in every object and is a property of very matter.
Mass:
Mass is the quantity of matter that is contained in an object. The more the quantity of matter that is contained in an object, the larger the mass of the object. Mass is measured in kilogram ( kg ). It is a scalar quantity.
Inertial Mass:
Inertial mass is a property of a matter which represent a resistance that the resistance f the object to any type of force.
Weight:
Weight of an object is the gravitational pull on the object. Or It can be defined as the force that is acting on the object due to the earth gravitational pull. Weight is measured in Newton. It is a vector quantity.
Relationship Between Inertia And Mass:
Mass of an object is the quantity of matter in the object. Mass of an object is a measure of the inertia of the object. The mass of an object determines the amount of inertia / resistance that the object will offer to a force that act on it and try to change its state. The more the mass of an object, the more the inertia of that object will be, the more the resistance and the larger the force required to change the state of rest of the object or uniform motion .
Relationship Between Mass And Weight:
The relationship between mass and weight is the formula that connect mass and weight. This formula is used in calculations, to either calculate the weight of a given mass of object or to calculate the mass of a given weight.
The formula is : weight = mass * acceleration due to gravity
Weight = m*g
Worked Examples:
1. Calculate the weight of a mass 2.6gram.
Solution:
Data given in the question:
Mass = 225g = 225 /1000 = 0.225 kg.
Formula: weight = mass * acceleration due to gravity
Substitution: weight = 0.225 * 10. ➡ weight = 2.25N
2. What is the mass of bject whose weight is 35N? ( g = 10 m/s² )
Solution:
Data given in he question:
Weight = 35 N, g = 10 m/s²
Formula: weight = mass * acceleration due to gravity
Substitution: 35 = mass * 10
Make mass the subject: mass = 35 / 10. ➡ mass = 3.5 kg
Significance Of Newton’s First Law Of Motion:
The significance of Newton’s first law of motion is that, Newton’s first law of motion make us to understand that inertia ( unwillingness ) is present in every object.
Short Coming Of Newton’s First Law Of Motion:
Though Newton’s first law of motion make us to know that inertia is inherent in every object and also explain what force does on an object. Newton’s first law of motion does not suggest how force should be measured.
Newton’s Second Law Of Motion:
Newton’s second law of motion states that the rate of change of momentum is directly proportional to the force that caused the change and take place in the direction of the force.
Relationship Between Force and Mass:
The formula that connect force and mass is as stated below:
Force = mass of object * acceleration of the object
F = m * a
Worked Examples:
1. Calculate the magnitude of the force that act on 12kg mass such that its acceleration is 1.75 m/s².
Solution:
Data given in the question:
Mass = 12kg, acceleration = 1.75 m/s², force = ?
Formula: force = mass * acceleration
Substitution: force = 12 * 1.75
Force = 21Newtons
2. Given that 35 N force act on 2.6 kg mass. If the acceleration of the object is, determine the acceleration of the and hence calculate the time if the velocity of the object is 0.25 m/s.
Solution:
Data given in the question:
Force = 35 N, =2.6 kg and velocity = 0.25 m/s
Formula: Force = mass * acceleration
Substitution: 35 = 2.6 * acceleration
Make acceleration the subject: acceleration = 35 / 2.6
Acceleration = 13.46 m/s²
3. An unbalance force of 25 N act on a 8.0 kg mass. What does it gives to the mass?
Solution:
Data given in the question:
Mass = 8.0 kg, force = 25 N, acceleration = ?
Formula: force = mass * acceleration
Substitution: 25 = 8.0 * acceleration
Make acceleration the subject: acceleration = 25 / 8.0
Acceleration = 3.13m/s²
Formula Of Newton’s Second Law Of Motion:
From the above statement of Newton’s second law of motion, the formula Newton’s second law of motion is obtained as follow:
Mathematically,
Force is directly proportional to rate of change in momentum
Force ∞ rate of change in momentum
Force ∞ change in momentum / time
Force ∞ (MV – MU) / time
Factorize m: Force ∞ m ( v - u ) / t
Recall that acceleration a = (final velocity v – initial velocity u) / time t
a = (v – u) / t
Substitute for a in the above equation: force = mass * acceleration
Force ∞ mass = acceleration
Replay the proportional symbol with k: force = k* m * a.
If constant k is 1, then the formula will become,
Force = ma ( or acceleration = force / mass or mass = force / acceleration )
Worked Examples:
1. N object of mass 950g is pulled along a tarmac by a horizontal force of 20 N. A frictional force of 6 N opposed the motion . what is e acceleration of the stone?
Solution:
Data given in the question:
Mass of stone = 950 g = 950 / 1000 = 0.95 kg, force = 20 N ,
frictional force = 6 N,
Note:
Since friction force is present to oppose the motion of the object, the force that moves the object or the effective force will be the difference between the give force and the frictional force. Therefore,
Effective force = force – friction force
Effective force = 20 – 6 = 14 N
Formula: force = mass * acceleration
Substitution: 14 = 0.95 * acceleration
Make acceleration the subject: acceleration = 14 / 0.95
Acceleration = 14.74 m/s²
2. A ball of mass 3 kg that is moving t a velocity of 26m/s is suddenly hit by a force of 6 N for 0.05 second. Find the new velocity of the object.
Solution:
Data given in the question:
Mass = 3 kg, velocity =26 m/s, force = 6N, time = 0.05 second
Note:
Velocity of 26 m/s is given in the question and we asked to calculate the new velocity of he object. That velocity is the initial velocity u, of the object while the new velocity that we are asked to calculate is the final velocity v, of the object.
Formula: force = ( MV – MU ) / t
Factorize m: Force = m ( v – u ) / t
Substitution: 6 = 3 ( v – 26 ) / 0.05
Cross multiply: 6 * 0.05 = 3 ( v – 26 )
0.3 = 3 ( v – 26 )
. 0.3 / 3 = v – 26
0.10 = v – 26
Make v the subject: v = 0.10 + 26
V = 26.10 m/s
3. A ball of mass 0.5 kg dropped from a height of 6 m onto a floor bounces back to a height of 2.5 m . calculate the change of momentum.
If the ball is in contact with the floor for 0.3 second, what is the force exerted on the ball?
Solution:
Data given in the question:
Mass of ball = 0.5kg, height = 6m, renounced height = 2.5m,
time = 0.3 second,
Formula: change in momentum = MV – MU
Note:
The change in momentum that the question want us to calculate, is the difference between the momentum of the object as it hits the ground and the momentum as it bounced back to the stated height. So you need the final V, on hitting the floor and the denounced initial velocity, U.
To calculate change in momentum, you must calculate:
I. The final velocity v of the object as it hits the ground
II. The bouncing velocity u
Therefore, final velocity of the object just before it hits the ground:
V² = U² + 2*g*S
Before the ball was dropped, its initial velocity was zero ( U = 0 m/s )
S = 6 m , g = 10 m/s²
Substitution: V² = 0² + 2 * 10 * 6
V² = 120. V = √120 = 10.95 m/s
You have to calculate the velocity with which bounced back:
At the renounced height, final velocity v = 0 m/s. Then,
Using the formula: V² = U² + 2* g*S
Substitution: 0² = U² - 2 * 10 * 2.5
Make U² the subject: U² = 2 * 10 * 2.5
U² = 50. U = √50 . U = 7.07 m/s
Change in momentum = MV – MU
Note that Since the ball bounced back, its bouncing momentum is opposite to the momentum on hitting the ground. Therefore,
Change in momentum = MV – ( - MU )
Change in momentum = MV + MU
Substitution: change in momentum = 0.5 * 10.95 – 0.5 * 7.07
Change in momentum = 5.48 + 3.54 = 9.02 m/s
To calculate force, we use the formula,
Force = change in momentum / time
Substitution: force = 9.02 / 0.3. force = 30.06 Newtons.
4. 4 kg mass fall freely from rest through a height of 30 m and ome to rest when it penetrated 0.7 m through sand. Calculate : ( I ) the velocity with which the body hit e ground. ( ii ) the force exerted by the sand in stopping the object ( iii ) the time the object took in falling.
Solution:
Data given in the question:
Mass = 4 kg, height = 30 m, penetration = 0.7 m, u = 0 m/s ( before the object falls, its initial velocity u = 0m/s )
I. Velocity with which the object hit the ground:
Using the equation of motion:
V² = U² + 2 * g * S
Substitution: V² = 0² + 2 * 10 * 30
V² = 2 * 10 * 30 . V² = 600. V = √600. V = 24.49 m/s
II. Force exerted by the sand:
Force = mass * acceleration
Note:
Acceleration is not given in the question. We have to calculate acceleration before we can calculate the force.
24.49 m/s which was the final velocity as the object hit the floor will become the initial velocity of the object as the object penetrates the sand, while the final velocity after the object penetrate the sand by 0.7 m is 0 because the sand stopped the object.
V = 0 m/s
Therefore,
We use equation of motion: V² = U² + 2 * a * S
Substitution: 0 = 24.49² + 2 * a * 0.7
Make a the subject: - 2 * a * 0.7 = 24.94:
-1.4* a = 24.94
A = 24.94 / ( - 1.4 ) . a = - 17.81 m/s²
We can now calculate force.
Force = mass * acceleration
Substitution: force = 4.0 * 17.81. Force = 71.24 Newtons
Force:
Force is an agent which changes or try to change the state of an object.
The state of an object refers to what the object is doing at the said time. The state of an object might bet that the object is stationary or the object is moving in a straight line.
Unit of Force:
The s.I. unit of force is Newton and the symbol that is used to represent newton is N.
Definition Of Newton:
Newton can be defined as the unit of force which make a mass of 1 kg to accelerates 1 m/s².
Impulse I:
Impulse is the product of the force that act on an object and the time which he object act on the object.
Formula Of Impulse:
Impulse = force * time
Also, remember that from Newton’s second law of motion,
Impulse = change in momentum
Change in momentum = momentum itself. Then,
Change in momentum = momentum = mass * velocity
Change in momentum = MV. Then,
Impulse = mass * velocity
Relationship between Impulse and Change in Momentum:
Relationship between two or more quantities is the formula that connect the quantities together.
Therefore, from Newton’s second law of motion,
Force = m ( v – u ) / t
Cross multiply: force * time = m ( v – u )
Force * time = mass ( final velocity v –initial velocity u )
F * t = m ( v – u )
Therefore,
Impulse = change in momentum
Newton’s Third Law Of Motion:
Newton’s third law of motion states that to every action, there is equal and opposite reaction. Action = - Reaction.
Explanation Of Newton’s Third Law Of Motion:
Newton’s third law of motion says that if object A hits another object B with a force Fa, then object B in turn also hit object A with the same amount of force as Fa .
force Fa of object A = force Fb of object B but in opposite direction.
Therefore,
Fa = - Fb
6.3.0 Conservation Of Linear Momentum Principle Of Conservation:
Principle of conservation of linear momentum States that if two or more bodies in a closed or isolated system collide, their total momentum before collision is equal to their total momentum after collision.
In a closed system,
Momentum before collision = momentum after collision.
(Mass * velocity)before collision = (Mass * velocity) after collision
Note that the principle of the conservation law of linear momentum emanates from Newton’s second and third law of motion.
A Closed / Isolated System:
A closed or isolated system means a system on which no external force act. Or it is a system in which there is no interference by any external factor however.
Types Of Collision:
There are two types of collisions: elastic collision and inelastic collision.
1. Elastic collision:
Elastic collision is the type of collision whereby the colliding objects bounce back with the same amount of velocity or different amount of velocities. The colliding objects do not stick together. In elastic collision, both kinetic energy and momentum are conserved.
Examples Of Elastic Collision:
Collision of ball and a wall
Collision of gas molecules
2. Inelastic Collision:
Inelastic collision is the type of collision whereby the colliding objects stick together and move with a common velocity v. In inelastic collision, only momentum that is conserved. Kinetic energy is loss because the kinetic energy is converted into sound energy, heat energy.
Different Cases Of Collision:
I will lead you into different cases of collisions and their corresponding concepts of calculations both for kinetic energy and momentum.
1.When A Moving Object Collides With A Stationary Object, Stick Together and Move With Common Velocity, V:
The figure above represent a collision type whereby an object of M1 that is moving with initial U1 collides with a stationary object of mass M2. They stick together and move with a common velocity V.
Momentum:
Form the conservation law of momentum:
Momentum of the two objects before collision:
= momentum of M1 + momentum of M2
= M1*U1 + M2*U2
Momentum of the two objects after collision:
= momentum of M1 + momentum of M2
= M1*V + M2*V = V ( M1 + M2 )
Let us combine the momentum of the
Momentum of objects before collision = momentum of objects after collision
M1*U1 + M2*U2 = M1*V + M2*V
Factorize V: M1*U1 + M2*U2 = V ( M1 + M2 )
Remember that U2 = 0 m/s ( the is at rest / stationary )
Then we will get
M1*U1 + M2* ( 0 ) = v ( M + M2 )
M1*U1 + 0 = V ( M1 + M2 )
M1*U1 = V ( M1 + M2 )
Make V the subject: V = ( M1*U1 ) / ( M1 + M2 )
Kinetic Energy:
The common velocity V can also be calculated from the concept of the conservation law of kinetic energy.
Conservation law of energy state the total energy of an isolated / closed system is constant.
Therefore,
Kinetic energy of the two objects before collision = K.E of M1 + K.E of M2
= ½*(M1)*(U1)² + ½*(M2)*(U2)²
Also,
Kinetic energy of the two objects after collision = K.E of M1 + K.E of M2
= ½*M1*V² + ½*M2*V²
Total Kinetic before collision = Total kinetic energy after collision
½*(M1)*(U1)² + ½*(M2)*(U2)² = ½*M1*V² + ½*M2*V²
Since M2 is stationary, U2 = 0 m/s. Then
½*(M1)*(U1)² + ½*(M2)*(0)² = ½*M1*V² + ½*M2*V²
½*(M1)*(U1)² + 0 = ½*M1*V² + ½*M2*V²
½*(M1)*(U1)² = ½*M1*V² + ½*M2*V²
Factorize V and ½: ½*(M1)*(U1)² = ½* (M1 + M2)V²
½ in both sides cancels one another: (M1)*(U1)² = (M1+ M2)V²
Make V² the subject: V² = (M1)*(U1)² / ((M1) + (M2))
Make V the subject: V = √( ((M1)*(U1)²) / ((M1) + (M2)) )
Loss in Kinetic Energy:
Loss in kinetic energy can be calculated as follows:
• Calculate the kinetic energy of the two objects before collision
• Calculate the kinetic energy of the two objects after collision
• Then find the difference between the two kinetic energise to get the kinetic energy loss.
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
K.E of the two objects before collision = ½*(M1)*(U1)² + ½*(M2)*(0)²
= ½*(M1)*(U1)² + 0
. = ½*(M1)*(U1)²
Also,
K.E of the two objects after collision = ½*M1*V² + ½*M2*V²
Factorize V and ½ : ½*( M1 + M2 )V²
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
2.When A Moving Object Collides With A Stationary Object and They Move In The Same Direction With Different Velocities:
From the figure above, object of mass M1 moving with initial velocity U1 collides with a stationary object of mass M2 and the two move with different final , V1 and V2.
Momentum:
Form the conservation law of momentum:
Total Momentum of the two objects before collision:
= momentum of M1 + momentum of M2
= M1*U1 + M2*U2
Momentum of the two objects after collision:
= momentum of M1 + momentum of M2
= M1*V1 + M2*V2
Let us apply the conservation law of linear momentum:
Momentum of objects before collision = momentum of objects after collision
M1*U1 + M2*U2 = M1*V1 + M2*V2
Remember that U2 = 0 m/s ( the is at rest / stationary )
Then we will get
M1*U1 + M2* ( 0 ) = M1*V1 + M2*V2
M1*U1 + 0 = M1*V1 + M2*V2
M1*U1 = M1*V1 + M2*V2
At this point, you can make any of the variable the subject of formula:
Kinetic Energy:
Going by the concept of the conservation law of kinetic energy, the relationship between kinetic energy before collision and after collision can also be established.
Conservation law of energy state the total energy of an isolated / closed system is constant.
Therefore,
Kinetic energy of the two objects before collision = K.E of M1 + K.E of M2
= ½*(M1)*(U1)² + ½*(M2)*(U2)²
Also,
Kinetic energy of the two objects after collision = K.E of M1 + K.E of M2
= ½*M1*(V1)² + ½*M2*(V2)²
Total Kinetic before collision = Total kinetic energy after collision
½*(M1)*(U1)² + ½*(M2)*(U2)² = ½*M1*(V1²)² + ½*M2*(V2)²
Since M2 is stationary, U2 = 0 m/s. Then
½*(M1)*(U1)² + ½*(M2)*(0)² = ½*M1*(V1)² + ½*M2*(V2)²
½*(M1)*(U1)² + 0 = ½*M1*(V1)² + ½*M2*(V2)²
½*(M1)*(U1)² = ½*M1*(V1)² + ½*M2*(V2)²
Factorize : ½*(M1)*(U1)² = ½*(M1*(V1)² + M2*(V2)²)
½ in both sides cancels one another: (M1)*(U1)² = M1*(V1)²+ M2*(V2)²
At this point make any of the variable that you want to calculate, the .
For example, if you want to calculate V1, you have to make V1 the subject of the after you might have substituted for other variables.
Thus:
(M1)*(U1)² = M1*(V1)²+ M2*(V2)²
(M1)*(U1)² - M2*(V2)² = M1*(V1)²
Make (V1)² the subject: (V1)² = ( M1*(V1)² - M2*(V2)² ) / M1
Make V1 the subject: V1 = √ ((M1*(V1)² - M2*(V2)²) / M1 ).
In then same way, you make any variable that you want to calculate, subject of the formula.
How To Calculate Loss in Kinetic Energy:
Loss in kinetic energy can be calculated as follows:
• Calculate the kinetic energy of the two objects before collision
• Calculate the kinetic energy of the two objects after collision
• Then find the difference between the two kinetic energise to get the kinetic energy loss.
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
K.E of the two objects before collision = ½*(M1)*(U1)² + ½*(M2)*(0)²
= ½*(M1)*(U1)² + 0
. = ½*(M1)*(U1)²
Also,
K.E of the two objects after collision = ½*M1*(V1)² + ½*M2*(V2)²
Factorize ½ : ½*( M1*(V1)²+ M2*(V2)²)
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
Loss in kinetic energy = ½*M1*(U1)² - ½*( M1*(V1)²+ M2*(V2)²)
3.When Two Objects That Move In The Same Direction Collide, Stick Together And Move With A Common Velocity:
From the figure above, two objects moving in the same direction collide with each other, stick together and move with a common velocity, V.
From this case of collision, we can obtain important formula for momentum and kinetic energy, that are useful in calculation.
Momentum:
From the conservation law of linear momentum which states that, in a closed system,
Total momentum before collision = Total momentum after collision
Total Momentum of the two objects before collision:
= momentum of M1 + momentum of M2
= M1*U1 + M2*U2
Momentum of the two objects after collision:
= momentum of M1 + momentum of M2
= M1*V1 + M2*V2
We will substitute for the momenta in the conservation law of linear momentum.
Therefore,
M1*U1 + M2*U2 = M1*V1 + M2*V2
At this point, you make any of the variable that you want to calculate the subject of formula.
For example, if we want to calculate V2, we will make V2 subject of the formula thus:
M1*U1 + M2*U2 = M1*V1 + M2*V2
M2*U2 = M1*V1 + M2*V2 - M1*V1
Make U2 the subject: U2 = (M1*V1 + M2*V2 - M1*V1) / M2
Just like that.
Kinetic Energy:
Going by the concept of the conservation law of kinetic energy, the relationship between kinetic energy before collision and after collision can also be established.
Conservation law of energy state the total energy of an isolated / closed system is constant.
Therefore,
Kinetic energy of the two objects before collision = K.E of M1 + K.E of M2
= ½*(M1)*(U1)² + ½*(M2)*(U2)²
Also,
Kinetic energy of the two objects after collision = K.E of M1 + K.E of M2
= ½*M1*V² + ½*M2*V²
Total Kinetic before collision = Total kinetic energy after collision
½*(M1)*(U1)² + ½*(M2)*(U2)² = ½*M1*V² + ½*M2*V²
Factorize : ½*((M1)*(U1)² + (M1)*(U2)²) = ½*(M1*(V1)² + M2*(V2)²)
½ in both sides cancels one another:
(M1)*(U1)² + (M1)*(U2)² = M1*(V1)²+ M2*(V2)²
At this point make any of the variable that you want to calculate, subject of the formula.
For example, if you want to calculate V2, you have to make V2 the subject of the formula after you might have substituted for other variables.
Thus:
((M1)*(U1)² + (M1)*(U2)² = M1*(V1)²+ M2*(V2)²
Make (V2)² the subject: M2*(V2)² = (M1)*(U1)² + (M1)*(U2)² - M1*(V1)²
Make V2 the subject: (V2)² = ((M1)*(U1)² + (M1)*(U2)² - M1*(V1)²) / M2
V2 = √(((M1)*(U1)² + (M1)*(U2)² - M1*(V1)²) / M2)
In then same way, you make any variabIe that you want to calculate, subject of the formula.
How To Calculate Loss in Kinetic Energy:
Loss in kinetic energy can be calculated as follows:
• Calculate the kinetic energy of the two objects before collision
• Calculate the kinetic energy of the two objects after collision
• Then find the difference between the two kinetic energise to get the kinetic energy loss.
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
K.E of the two objects before collision = ½*(M1)*(U1)² + ½*(M2)*(U2)²
Factorize ½: = ½*((M1)*(U1)² + (M2)*(U2)²)
Also,
K.E of the two objects after collision = ½*M1*(V1)² + ½*M2*(V2)²
Factorize ½ : = ½*( M1*(V1)²+ M2*(V2)²)
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
Loss in kinetic energy = ½*M1*(U1)² - ½*( M1*(V1)²+ M2*(V2)²)
4.When Two Objects Move In Opposite Directions Collide and Move With Different Velocity In The Same Direction:
From the figure by the right, it is observed that two objects move in opposite direction, collide with each other and move in the same direction with different velocities.
As usual, we can obtain important formulae from the conservation law of momentum and kinetic energy, that are very useful n solving problems related to momentum and kinetic energy.
Therefore,
For momentum:
From the conservation law of linear momentum which states that, in a closed system,
Total momentum before collision = Total momentum after collision
Total Momentum of the two objects before collision:
= momentum of M1 + ( - momentum of M2 )
= M1*U1 + (- M2*U2)
= M1*U1 - M2*U2
Momentum of the two objects after collision:
= momentum of M1 + momentum of M2
= M1*V1 + M2*V2
We will substitute for the momenta in the conservation law of linear momentum as follow:
Therefore,
M1*U1 - M2*U2 = M1*V1 + M2*V2
At this point, you make any of the variable that you want to calculate the subject of formula.
For example, if we want to calculate U2, we have to make U2 subject of the formula thus:
M1*U1 - M2*U2 = M1*V1 + M2*V2
- M2*U2 = M1*V1 + M2*V2 - M1*U1
U2 = ( M1*V1 + M2*V2 - M1*U1) / (- M2)
Make U2 the subject: U2 = (M1*V1 + M2*V2 - M1*V1) / M2
Just like that.
Kinetic Energy:
Going by the concept of the conservation law of kinetic energy, the relationship between kinetic energy before collision and after collision can also be established.
Conservation law of energy state the total energy of an isolated / closed system is constant.
Therefore,
Kinetic energy of the two objects before collision = K.E of M1 + ( - K.E of M2 )
= ½*(M1)*(U1)² + ( - ½*(M2)*(U2)² )
= ½*(M1)*(U1)² - ½*(M2)*(U2)²
Factorize ½: = ½ ((M1)*(U1)² - (M2)*(U2)²)
Also,
Kinetic energy of the two objects after collision = K.E of M1 + K.E of M2
= ½*M1*V1² + ½*M2*V2²
Total Kinetic before collision = Total kinetic energy after collision
½*(M1)*(U1)² - ½*(M2)*(U2)² = ½*M1*V1² + ½*M2*V2²
Factorize ½ : ½*((M1)*(U1)² - (M1)*(U2)²) = ½*(M1*(V1)² + M2*(V2)²)
½ in both sides cancels one another:
(M1)*(U1)² - (M1)*(U2)² = M1*(V1)²+ M2*(V2)²
At this point make any of the variable that you want to calculate, subject of the formula.
For example, if you want to calculate V2, you have to make V2 the subject of the formula after you might have substituted for other variables.
Thus:
((M1)*(U1)² - (M1)*(U2)² = M1*(V1)²+ M2*(V2)²
Make (V2)² the subject: M2*(V2)² = (M1)*(U1)² - (M1)*(U2)² - M1*(V1)²
Make V2 the subject: (V2)² = ((M1)*(U1)² - (M1)*(U2)² - M1*(V1)²) / M2
V2 = √(((M1)*(U1)² - (M1)*(U2)² - M1*(V1)²) / M2)
In then same way, you make any variable that you want to calculate, subject of the formula.
How To Calculate Loss in Kinetic Energy:
Loss in kinetic energy can be calculated as follows:
• Calculate the kinetic energy of the two objects before collision
• Calculate the kinetic energy of the two objects after collision
• Then find the difference between the two kinetic energise to get the kinetic energy loss.
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
K.E of the two objects before collision = ½*(M1)*(U1)² + ( - ½*(M2)*(U2)² )
= ½*(M1)*(U1)² - ½*(M2)*(U2)²
Factorize ½: = ½*((M1)*(U1)² - (M2)*(U2)²)
Also,
K.E of the two objects after collision = ½*M1*(V1)² + ½*M2*(V2)²
Factorize ½ : = ½*( M1*(V1)²+ M2*(V2)²)
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
Loss in kinetic energy =
= (½*(M1)*(U1)² - ½*(M2)*(U2)² ) – (½*( M1*(V1)²+ ½M2*(V2)²)
Factorize ½: = ½*( M1*(U1)² - M2*(U2)²) - ½*(M1*(V1)² - M2*(V2)²)
5.When Tow Objects Move In Opposite Directions,Collide, Stick Together And Move With A Common Velocity In The Same Direction:
From the figure above, it is observed that two objects move in opposite direction, collide with each other and move in the same direction with different velocities.
As usual, we can obtain important formulae m1or m2 or u1 or u2 or v from the conservation law of momentum and kinetic energy. Any of the variable can be calculated when others are given in the question, by simply making it the subject of the formulaAn
Therefore,
For momentum:
From the conservation law of linear momentum which states that, in a closed system,
Total momentum before collision = Total momentum after collision.
Total Momentum of the two objects before collision:
= momentum of M1 + ( - momentum of M2 )
= M1*U1 + (- M2*U2)
= M1*U1 - M2*U2
Momentum of the two objects after collision:
= momentum of M1 + momentum of M2
= M1*V + M2*V
Factorize V: = ( M1 + M2 ) V
We will substitute for the momenta in the conservation law of linear momentum as follow:
Therefore,
M1*U1 - M2*U2 = ( M1 + M2 )*V
At this point, you make any of the variable that you want to calculate the subject of formula.
For example, if we want to calculate U2, we have to make U2 subject of the formula thus:
M1*U1 - M2*U2 = ( M1 + M2 )*V
- M2*U2 = ( M1 + M2 )*V - M1*U1
U2 = ( M1 + M2 )*V - M1*U1) / (- M2)
Make U2 the subject: U2 = (M1 + M2 )*V - M1*U1) / ( - M2)
Just like that.
Kinetic Energy:
Going by the concept of the conservation law of kinetic energy, the relationship between kinetic energy before collision and after collision can also be established.
Conservation law of energy state the total energy of an isolated / closed system is constant.
Therefore,
Kinetic energy of the two objects before collision = K.E of M1 + ( - K.E of M2 )
= ½*(M1)*(U1)² + ( - ½*(M2)*(U2)² )
= ½*(M1)*(U1)² - ½*(M2)*(U2)²
Factorize ½: = ½ ((M1)*(U1)² - (M2)*(U2)²)
Also,
Kinetic energy of the two objects after collision = K.E of M1 + K.E of M2
= ½*M1*V² + ½*M2*V²
Factorize ½ and V: = ½*(M1 + M2)*V²
Total Kinetic before collision = Total kinetic energy after collision
½*((M1)*(U1)² - (M2)*(U2)²) = ½*(M1 + M2)*V²
½ cancels ½: (M1)*(U1)² - (M1)*(U2)² = (M1 + M2)*V²
At this point make any of the variable that you want to calculate, subject of the formula.
For example, if you want to calculate V, you have to make V the subject of the formula after you might have substituted for other variables.
Thus:
(M1*(U1)² - M2*(U2)² = (M1 + M2)*V²
Make V² the subject: V² = ((M1)*(U1)² - (M2)*(U2)²) / (M1 + M2)
Make V the subject: V = √(((M1)*(U1)² - (M2)*(U2)²) / (M1 + M2))
In then same way, you make any variable that you want to calculate, subject of the formula.
How To Calculate Loss in Kinetic Energy:
Loss in kinetic energy can be calculated as follows:
• Calculate the kinetic energy of the two objects before collision
• Calculate the kinetic energy of the two objects after collision
• Then find the difference between the two kinetic energise to get the kinetic energy loss.
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
K.E of the two objects before collision = ½*(M1)*(U1)² + ( - ½*(M2)*(U2)² )
= ½*(M1)*(U1)² - ½*(M2)*(U2)²
Factorize ½: = ½*((M1)*(U1)² - (M2)*(U2)²)
Also,
K.E of the two objects after collision = ½*M1*V² + ½*M2*V²
Factorize ½ and V²: = ½*( M1 + M2 )*V²
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
Loss in kinetic energy =
= ½*((M1)*(U1)² - (M2)*(U2)² ) – ½*( M1 + M2 )*V²
Factorize ½: = ½*(((M1)*(U1)² - M2*(U2)²) - (M1 + M2)*V²)
6.When Two Objects Travel In The Same Direction Collide And Move In Opposite Direction With Different Velocities:
Application Of Newton’s Law And Conservation Law Of Momentum:
We will look into some areas in which the conservation laws of momentum and Newton’s laws are applied.
The principle of conservation of linear momentum is applied in the following which we will discuss below:
1. Recoil Of A Gun:
The principle of conservation of linear momentum is applied in the recoil of a gun when it is fired. When a gun is fired, the bullet is energized by a forward force ( action ) which moves it forward with a certain velocity. The bullet therefore has a forward momentum. As the bullet leaves the gun, it exert a backward force ( reaction ) on the gun. The gun then jerk backward with certain amount of velocity. In accordance with the conservation law of linear momentum, therefore the gun has a backward momentum which is equal in magnitude to the forward momentum of the fired bullet.
Let,
Mg be mass of gun
VG be velocity of gun
Mb be mass of bullet
Vg be velocity of bullet
Then,
momentum of gun = Mg*Vg
momentum of bullet = Mb*Vb
By principle of the law of conservation of linear momentum:
Forward momentum of Bullet = Backward momentum of gun
Therefore,
Mb*Vb = Mg*Vg
At this point, you make any of the variables that you want to calculate, subject of the formula.
For example, assuming we want to calculate Mg, we have to make Mg the subject of the formula.
Therefore,
Mb*Vb = Mg*Vg
Mb*Vb / Vg = Mg
Why is it that when a bullet is fired from a gun of mass m, the bullet travels with high velocity while the gun only jerk backward. It is becausethe gun is designed to have massive , the massiveness of the guns is experience.
Worked Examples:
*50 gram mass of bullet is fired from a gun of mass 25 kilogram with a velocity of 130 m/s. Calculate the recoil velocity of the rifle.
Solution:
Data given in the question:
Mass of gun = 2.5 kg = 2.5 * 1000 = 2500g, mass of bullet = 30g,
velocity of bullet = 130 m/s, velocity of gun = ?
formula= Mb*Vb = Mg*Vg
Substitution: 30 * 130 = 2500 * Vb
Make Vb subject of the formula: Vb = 30 * 130 / 2500. ➡ Vb = 1.56 m/s
*A gun of mass 2.4kg fired 0.23kg mass of . if the recoil velocity of the gun is 21.4 m/s, calculate the velocity of the bullet.
Solution:
Data given in the question:
Mass of gun = 2.4kg, mass of bullet = 0.23kg, velocity of gun = 21.4 m/s.
Formula: Mb*Vb = Mg*Vg
Substitution: 0.23*Vb = 2.4*21.4
Make Vb the subject: VB = 2.4*21.4 / 0.23. ➡ VB = 223.30 m/s
*A jet burns at the rate of 250 g/s. If all the gas is ejected in one direction at the rate of 500m/s, what is the maximum weight the can have so that it can move vertically upward?
Solution:
Data given in the question:
Mass of bass per second = 250 g/s =250 / 1000 = 0.25kg/s, velocity = 500 m/s
Formula: momentum = mass * velocity
Substitution: momentum = 0.25 * 500 = 125kgm/s
From Newton’s third law of motion,
Force = weight = 125 N
2. Jet and Rocket Propulsion:
The principle of conservation of linear momentum is also utilized in the propulsion of jet, aircraft and rockets.
Principle of Utilization Of Principle Of Conservation of Linear Momentum And Newton’s Third Law Of Motion In Jets, Aircraft and Rockets
Gases are contained in a combustion chamber of an aircraft, rocket and jets. When the gas is burnt in the chamber, it expands and exert pressure within the chamber. The hot gas is therefore expel through a nozzle at a very high speed from the rocket. The ejected gas has mass and velocity and therefore has momentum. The momentum of the ejected gas produce equal and opposite momentum on the rocket which propel the rocket forward.
3. Why Walking Is Possible:
Man walk by pushing his foot against the ground. According to the principle of conservation law of linear momentum and Newton’s third law of motion, as the man push the ground with his foot ( action force ), the ground in turn exerts an equal and opposite force ( reaction force) on the man. This force move the man forward.
Weight Of A Object Inside A Lift
When a man stand in a lift or elevator, two forces act on the man. The forces are:
I. The true weight of the man which act vertically downward (w = m*g)
II. The reaction force R of the floor of the lift on the man, which act upward.
We are going to look into different cases when the man is on the lift
Case 1: when the lift is stationary or moving with a constant velocity:
When the lift is stationary, the downward force due to the weight of the man is equal to the upward reaction force of the floor of the lift. The lift is stationary or moving with constant velocity. Its acceleration is zero. ( a = 0 m/s² )
Therefore,
Weight of man = Reaction of floor = m*g
It means that there is no unbalance force acting on the man because the man’s weight is cancelled by the upward reaction force of the floor of he lift. If the man stands on a spring scale, the scale will record the man’s true weight.
Case2: when the lift accelerates upward with acceleration a:
If the lift accelerates upward with acceleration a, there is unbalance force acting on the man. This means that the upward reaction force of the floor of the lift on the man is greater than the downward force due to the weight of the man on the floor of the lift.
From Newton’s second law of motion,
Force F = m*a
Since the lift move upward, it means that floor Reaction is greater than man’s weight ( I.e R > W )
Effective force that move the lift is the difference between Man’s weight and floor reaction.
Effective force F = R – W
Therefore R – W = m*a
R – m*g = m*a
R = m*a + m*g
Factorize m: R = m ( a + g )
R = m ( a + g ) is also regarded as the apparent weight of the man when the lift is accelerating upward. The scale will record a value that is greater than the mans true weight. ( apparent weight ).
That is apparent weight W = R = m(a + g)
At this point, you can calculate R, m or a. This we shall see in the worked example below.
Case 3: If the lift is accelerating downwards with an acceleration, a:
If the lift accelerates downwards with acceleration a, it means that there is unbalancing force that is acting on the man. It means that the downward force due to the man’s weight is greater than the upward reaction force of the lift floor. ( W > R )
Man’s weight > Reaction force of lift floor
From Newton’s second law of motion,
Force F = m*a
Since the lift accelerates downwards, it means that the man’s weight is greater than the reaction force of the lift floor. ( W > R )
The effective force is the difference between the mans weight and the reaction force of the lift.
Effective force F = man’s weight – reaction force of lift floor
F = W – R
If we substitute for F in F = m*a, we will get
W – R = m*a
M*g – R = m*a
M*g – m*a = R
Factorize m: m ( g – a ) = R. ➡ R = m ( g – a )
At this point, you can calculate m, R or a by making it the subject of the formula. This we shall see in the worked example below.
R = m ( g – a ) can also be called apparent weight of the man.
W' = R = m { g – a ).
The man will fill lighter in weight. The scale will record a value the is smaller than the true weight of the man.
Case 4: when the lift is falling freely:
If the lift accelerate downward with acceleration that is equal to acceleration due o gravity, we say the lift is falling freely. His situation happen when the rope / cable of the lift cut.
From Newton’s second law of motion,
Force F = m*a
W > R ( the lift is falling )
Effective force F = W – R
Then. W – R = m*a
M*g – R = m*a
M*g - m*a = R
Factorize m: m( g - a )
Recall that the lift is falling freely, then a = g . Therefore,
R = m ( g – g ) = ( 0 ) = 0 N
We can see that R = 0. The man appears to have no weight ( weightless ). The man and the floor of the lift are not exerting force on each other. The scale of the weighing instrument reads zero.
Weightlessness In Satellites:
People and object is a satellite that is orbiting close to the earth experience weightlessness.
Exercises:
1. A ball of mass 6 kg moving with a velocity of 10 m/s collides with another ball of equal mass at rest. If the two balls off after the impact, calculate their common velocity.
2. A ball of mass 6 kg moving with a velocity of 10 m/s collides with a 2.0 kg mass ball moving in the opposite direction with a velocity of 5 m/s . After the collision, the two balls coalesce and move in the same direction. Calculate their common velocity.
3. A constant force of 5 N act for 7 seconds on a mass of 7 kg initially at rest. Calculate final momentum .
4. A foot baller applied a force of 39N when taking a penalty kick for a period of 0.24 second. If the mass of the ball is 0.045 kg, calculate the speed with which the ball took off.
5. A body of mass 100 g moving with a velocity of 10 m/s collides with a wall. Of after the collision, it moves with a velocity of 3 m/s in the opposite direction, calculate the change in momentum.
6. Define impulse. How is the rate of change of momentum related to the force that act on the body?
7. Explain she the velocity of a recoiling gun is lesser than that of the bullet shot out of he gun.
8. A 30 g bullet moving at 250 m/s hits a bag of sand and come to rest in 0.011 second. What is the momentum of the bullet just before hitting the bag? Find the average force that stopped the bullet.
9. State Newton’s laws of motion. Derive from one of the relation between the momentum produce in a body and the force applied on the body.
10. State the conservation law of linear momentum. A 250 g riffle lay on a smooth table when its suddenly discharges , firing a bullet of 0.05 g with a speed of 450 m/s . calculate the recoil speed of the gun.
11. Distinguish between mass, weight, momentum and inertia. Describe an experiment to demonstrate that momentum is conserved in a collision.
12. An object of mass 1.3kg falls a distance of 5 m to a horizontal surface and rebounds to a vertical height of 3 m . calculate the change in momentum.
13. A man whose mass is 85 kg stand on a spring weighing machine inside a lift. What is the reading of the weighing machine when:
I) When the lift starts to ascend with an acceleration of 2.0 m/s²?
II) The lift is moving with a uniform velocity?
III) The lift is coming to rest with a retardation of 4.0 m/s²
14. Explain the principle underlying the propulsion of rocket.
15. A rocket of mass 5800kg containing a propellant gas of 2000kg is to be launched vertically. If the fuel is consumed at a steady rate of 70kg/s, calculate the least velocity of the exhaust gases if the rocket and content will just lift off the launching pad immediately after firing.
16. Distinguish between impulse and momentum. How are they related? Calculate the momentum of : (i) The moon of mass 7x10^²²kg travelling with a velocity of 1.25 m/s. (ii) a tanker of mass 7x10^⁴ kg travelling at a speed of 6 m/s. (iii) an aircraft of mass 3.5 x 10^³ kg travelling at a speed of 400 m/s
17. Distinguish between elastic collision and inelastic collision. (ii) inertial mass and weight. From Newton’s law of motion show that F = ma. ( relationship between mass, force and acceleration ).
18. A 15 kg monkey hang from a cord suspended from the ceiling of a lift. The cord can withstand a tension of 200N and brakes as the lift accelerates. What was the lift minimum acceleration ( in magnitude and direction) g = 10.0 m/s².
19. A tractor of mass 5 x10³ kg s used to tow a car f mass 2.5 x10³ kg. The tractor moves with a speed of 3.0 m/s just before the towing rope become taut . calculate the (I) the speed of the tractor immediately the rope becames taut. (ii) loss kinetic energy of the system just after the car has started moving. (iii) impulse in the rope when it jerk the car into motion.
Motion of objects are caused by unbalanced forces that act on the objects. Therefore, forces are related to motions of objects. This was discovered by Sir Isaac Newton ( 1642 – 1727 ). He stated three laws of motion which we shall look into in the treatment.
Newton’s First Law Of Motion:
Newton’s first law of motion states that an object will continue in its state of rest or motion in a straight line unless it is acted upon by an external force.
Inertia:
Inertia is the unwillingness of an object in motion to stop moving or an object at rest to start moving.
Inertia simply means unwillingness. Every object is unwilling to do other wise apart from what its is doing at the moment. Inertia is inherent in every object and is a property of very matter.
Mass:
Mass is the quantity of matter that is contained in an object. The more the quantity of matter that is contained in an object, the larger the mass of the object. Mass is measured in kilogram ( kg ). It is a scalar quantity.
Inertial Mass:
Inertial mass is a property of a matter which represent a resistance that the resistance f the object to any type of force.
Weight:
Weight of an object is the gravitational pull on the object. Or It can be defined as the force that is acting on the object due to the earth gravitational pull. Weight is measured in Newton. It is a vector quantity.
Relationship Between Inertia And Mass:
Mass of an object is the quantity of matter in the object. Mass of an object is a measure of the inertia of the object. The mass of an object determines the amount of inertia / resistance that the object will offer to a force that act on it and try to change its state. The more the mass of an object, the more the inertia of that object will be, the more the resistance and the larger the force required to change the state of rest of the object or uniform motion .
Relationship Between Mass And Weight:
The relationship between mass and weight is the formula that connect mass and weight. This formula is used in calculations, to either calculate the weight of a given mass of object or to calculate the mass of a given weight.
The formula is : weight = mass * acceleration due to gravity
Weight = m*g
Worked Examples:
1. Calculate the weight of a mass 2.6gram.
Solution:
Data given in the question:
Mass = 225g = 225 /1000 = 0.225 kg.
Formula: weight = mass * acceleration due to gravity
Substitution: weight = 0.225 * 10. ➡ weight = 2.25N
2. What is the mass of bject whose weight is 35N? ( g = 10 m/s² )
Solution:
Data given in he question:
Weight = 35 N, g = 10 m/s²
Formula: weight = mass * acceleration due to gravity
Substitution: 35 = mass * 10
Make mass the subject: mass = 35 / 10. ➡ mass = 3.5 kg
Significance Of Newton’s First Law Of Motion:
The significance of Newton’s first law of motion is that, Newton’s first law of motion make us to understand that inertia ( unwillingness ) is present in every object.
Short Coming Of Newton’s First Law Of Motion:
Though Newton’s first law of motion make us to know that inertia is inherent in every object and also explain what force does on an object. Newton’s first law of motion does not suggest how force should be measured.
Newton’s Second Law Of Motion:
Newton’s second law of motion states that the rate of change of momentum is directly proportional to the force that caused the change and take place in the direction of the force.
Relationship Between Force and Mass:
The formula that connect force and mass is as stated below:
Force = mass of object * acceleration of the object
F = m * a
Worked Examples:
1. Calculate the magnitude of the force that act on 12kg mass such that its acceleration is 1.75 m/s².
Solution:
Data given in the question:
Mass = 12kg, acceleration = 1.75 m/s², force = ?
Formula: force = mass * acceleration
Substitution: force = 12 * 1.75
Force = 21Newtons
2. Given that 35 N force act on 2.6 kg mass. If the acceleration of the object is, determine the acceleration of the and hence calculate the time if the velocity of the object is 0.25 m/s.
Solution:
Data given in the question:
Force = 35 N, =2.6 kg and velocity = 0.25 m/s
Formula: Force = mass * acceleration
Substitution: 35 = 2.6 * acceleration
Make acceleration the subject: acceleration = 35 / 2.6
Acceleration = 13.46 m/s²
3. An unbalance force of 25 N act on a 8.0 kg mass. What does it gives to the mass?
Solution:
Data given in the question:
Mass = 8.0 kg, force = 25 N, acceleration = ?
Formula: force = mass * acceleration
Substitution: 25 = 8.0 * acceleration
Make acceleration the subject: acceleration = 25 / 8.0
Acceleration = 3.13m/s²
Formula Of Newton’s Second Law Of Motion:
From the above statement of Newton’s second law of motion, the formula Newton’s second law of motion is obtained as follow:
Mathematically,
Force is directly proportional to rate of change in momentum
Force ∞ rate of change in momentum
Force ∞ change in momentum / time
Force ∞ (MV – MU) / time
Factorize m: Force ∞ m ( v - u ) / t
Recall that acceleration a = (final velocity v – initial velocity u) / time t
a = (v – u) / t
Substitute for a in the above equation: force = mass * acceleration
Force ∞ mass = acceleration
Replay the proportional symbol with k: force = k* m * a.
If constant k is 1, then the formula will become,
Force = ma ( or acceleration = force / mass or mass = force / acceleration )
Worked Examples:
1. N object of mass 950g is pulled along a tarmac by a horizontal force of 20 N. A frictional force of 6 N opposed the motion . what is e acceleration of the stone?
Solution:
Data given in the question:
Mass of stone = 950 g = 950 / 1000 = 0.95 kg, force = 20 N ,
frictional force = 6 N,
Note:
Since friction force is present to oppose the motion of the object, the force that moves the object or the effective force will be the difference between the give force and the frictional force. Therefore,
Effective force = force – friction force
Effective force = 20 – 6 = 14 N
Formula: force = mass * acceleration
Substitution: 14 = 0.95 * acceleration
Make acceleration the subject: acceleration = 14 / 0.95
Acceleration = 14.74 m/s²
2. A ball of mass 3 kg that is moving t a velocity of 26m/s is suddenly hit by a force of 6 N for 0.05 second. Find the new velocity of the object.
Solution:
Data given in the question:
Mass = 3 kg, velocity =26 m/s, force = 6N, time = 0.05 second
Note:
Velocity of 26 m/s is given in the question and we asked to calculate the new velocity of he object. That velocity is the initial velocity u, of the object while the new velocity that we are asked to calculate is the final velocity v, of the object.
Formula: force = ( MV – MU ) / t
Factorize m: Force = m ( v – u ) / t
Substitution: 6 = 3 ( v – 26 ) / 0.05
Cross multiply: 6 * 0.05 = 3 ( v – 26 )
0.3 = 3 ( v – 26 )
. 0.3 / 3 = v – 26
0.10 = v – 26
Make v the subject: v = 0.10 + 26
V = 26.10 m/s
3. A ball of mass 0.5 kg dropped from a height of 6 m onto a floor bounces back to a height of 2.5 m . calculate the change of momentum.
If the ball is in contact with the floor for 0.3 second, what is the force exerted on the ball?
Solution:
Data given in the question:
Mass of ball = 0.5kg, height = 6m, renounced height = 2.5m,
time = 0.3 second,
Formula: change in momentum = MV – MU
Note:
The change in momentum that the question want us to calculate, is the difference between the momentum of the object as it hits the ground and the momentum as it bounced back to the stated height. So you need the final V, on hitting the floor and the denounced initial velocity, U.
To calculate change in momentum, you must calculate:
I. The final velocity v of the object as it hits the ground
II. The bouncing velocity u
Therefore, final velocity of the object just before it hits the ground:
V² = U² + 2*g*S
Before the ball was dropped, its initial velocity was zero ( U = 0 m/s )
S = 6 m , g = 10 m/s²
Substitution: V² = 0² + 2 * 10 * 6
V² = 120. V = √120 = 10.95 m/s
You have to calculate the velocity with which bounced back:
At the renounced height, final velocity v = 0 m/s. Then,
Using the formula: V² = U² + 2* g*S
Substitution: 0² = U² - 2 * 10 * 2.5
Make U² the subject: U² = 2 * 10 * 2.5
U² = 50. U = √50 . U = 7.07 m/s
Change in momentum = MV – MU
Note that Since the ball bounced back, its bouncing momentum is opposite to the momentum on hitting the ground. Therefore,
Change in momentum = MV – ( - MU )
Change in momentum = MV + MU
Substitution: change in momentum = 0.5 * 10.95 – 0.5 * 7.07
Change in momentum = 5.48 + 3.54 = 9.02 m/s
To calculate force, we use the formula,
Force = change in momentum / time
Substitution: force = 9.02 / 0.3. force = 30.06 Newtons.
4. 4 kg mass fall freely from rest through a height of 30 m and ome to rest when it penetrated 0.7 m through sand. Calculate : ( I ) the velocity with which the body hit e ground. ( ii ) the force exerted by the sand in stopping the object ( iii ) the time the object took in falling.
Solution:
Data given in the question:
Mass = 4 kg, height = 30 m, penetration = 0.7 m, u = 0 m/s ( before the object falls, its initial velocity u = 0m/s )
I. Velocity with which the object hit the ground:
Using the equation of motion:
V² = U² + 2 * g * S
Substitution: V² = 0² + 2 * 10 * 30
V² = 2 * 10 * 30 . V² = 600. V = √600. V = 24.49 m/s
II. Force exerted by the sand:
Force = mass * acceleration
Note:
Acceleration is not given in the question. We have to calculate acceleration before we can calculate the force.
24.49 m/s which was the final velocity as the object hit the floor will become the initial velocity of the object as the object penetrates the sand, while the final velocity after the object penetrate the sand by 0.7 m is 0 because the sand stopped the object.
V = 0 m/s
Therefore,
We use equation of motion: V² = U² + 2 * a * S
Substitution: 0 = 24.49² + 2 * a * 0.7
Make a the subject: - 2 * a * 0.7 = 24.94:
-1.4* a = 24.94
A = 24.94 / ( - 1.4 ) . a = - 17.81 m/s²
We can now calculate force.
Force = mass * acceleration
Substitution: force = 4.0 * 17.81. Force = 71.24 Newtons
Force:
Force is an agent which changes or try to change the state of an object.
The state of an object refers to what the object is doing at the said time. The state of an object might bet that the object is stationary or the object is moving in a straight line.
Unit of Force:
The s.I. unit of force is Newton and the symbol that is used to represent newton is N.
Definition Of Newton:
Newton can be defined as the unit of force which make a mass of 1 kg to accelerates 1 m/s².
Impulse I:
Impulse is the product of the force that act on an object and the time which he object act on the object.
Formula Of Impulse:
Impulse = force * time
Also, remember that from Newton’s second law of motion,
Impulse = change in momentum
Change in momentum = momentum itself. Then,
Change in momentum = momentum = mass * velocity
Change in momentum = MV. Then,
Impulse = mass * velocity
Relationship between Impulse and Change in Momentum:
Relationship between two or more quantities is the formula that connect the quantities together.
Therefore, from Newton’s second law of motion,
Force = m ( v – u ) / t
Cross multiply: force * time = m ( v – u )
Force * time = mass ( final velocity v –initial velocity u )
F * t = m ( v – u )
Therefore,
Impulse = change in momentum
Newton’s Third Law Of Motion:
Newton’s third law of motion states that to every action, there is equal and opposite reaction. Action = - Reaction.
Explanation Of Newton’s Third Law Of Motion:
Newton’s third law of motion says that if object A hits another object B with a force Fa, then object B in turn also hit object A with the same amount of force as Fa .
force Fa of object A = force Fb of object B but in opposite direction.
Therefore,
Fa = - Fb
6.3.0 Conservation Of Linear Momentum Principle Of Conservation:
Principle of conservation of linear momentum States that if two or more bodies in a closed or isolated system collide, their total momentum before collision is equal to their total momentum after collision.
In a closed system,
Momentum before collision = momentum after collision.
(Mass * velocity)before collision = (Mass * velocity) after collision
Note that the principle of the conservation law of linear momentum emanates from Newton’s second and third law of motion.
A Closed / Isolated System:
A closed or isolated system means a system on which no external force act. Or it is a system in which there is no interference by any external factor however.
Types Of Collision:
There are two types of collisions: elastic collision and inelastic collision.
1. Elastic collision:
Elastic collision is the type of collision whereby the colliding objects bounce back with the same amount of velocity or different amount of velocities. The colliding objects do not stick together. In elastic collision, both kinetic energy and momentum are conserved.
Examples Of Elastic Collision:
Collision of ball and a wall
Collision of gas molecules
2. Inelastic Collision:
Inelastic collision is the type of collision whereby the colliding objects stick together and move with a common velocity v. In inelastic collision, only momentum that is conserved. Kinetic energy is loss because the kinetic energy is converted into sound energy, heat energy.
Different Cases Of Collision:
I will lead you into different cases of collisions and their corresponding concepts of calculations both for kinetic energy and momentum.
1.When A Moving Object Collides With A Stationary Object, Stick Together and Move With Common Velocity, V:
Momentum:
Form the conservation law of momentum:
Momentum of the two objects before collision:
= momentum of M1 + momentum of M2
= M1*U1 + M2*U2
Momentum of the two objects after collision:
= momentum of M1 + momentum of M2
= M1*V + M2*V = V ( M1 + M2 )
Let us combine the momentum of the
Momentum of objects before collision = momentum of objects after collision
M1*U1 + M2*U2 = M1*V + M2*V
Factorize V: M1*U1 + M2*U2 = V ( M1 + M2 )
Remember that U2 = 0 m/s ( the is at rest / stationary )
Then we will get
M1*U1 + M2* ( 0 ) = v ( M + M2 )
M1*U1 + 0 = V ( M1 + M2 )
M1*U1 = V ( M1 + M2 )
Make V the subject: V = ( M1*U1 ) / ( M1 + M2 )
Kinetic Energy:
The common velocity V can also be calculated from the concept of the conservation law of kinetic energy.
Conservation law of energy state the total energy of an isolated / closed system is constant.
Therefore,
Kinetic energy of the two objects before collision = K.E of M1 + K.E of M2
= ½*(M1)*(U1)² + ½*(M2)*(U2)²
Also,
Kinetic energy of the two objects after collision = K.E of M1 + K.E of M2
= ½*M1*V² + ½*M2*V²
Total Kinetic before collision = Total kinetic energy after collision
½*(M1)*(U1)² + ½*(M2)*(U2)² = ½*M1*V² + ½*M2*V²
Since M2 is stationary, U2 = 0 m/s. Then
½*(M1)*(U1)² + ½*(M2)*(0)² = ½*M1*V² + ½*M2*V²
½*(M1)*(U1)² + 0 = ½*M1*V² + ½*M2*V²
½*(M1)*(U1)² = ½*M1*V² + ½*M2*V²
Factorize V and ½: ½*(M1)*(U1)² = ½* (M1 + M2)V²
½ in both sides cancels one another: (M1)*(U1)² = (M1+ M2)V²
Make V² the subject: V² = (M1)*(U1)² / ((M1) + (M2))
Make V the subject: V = √( ((M1)*(U1)²) / ((M1) + (M2)) )
Loss in Kinetic Energy:
Loss in kinetic energy can be calculated as follows:
• Calculate the kinetic energy of the two objects before collision
• Calculate the kinetic energy of the two objects after collision
• Then find the difference between the two kinetic energise to get the kinetic energy loss.
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
K.E of the two objects before collision = ½*(M1)*(U1)² + ½*(M2)*(0)²
= ½*(M1)*(U1)² + 0
. = ½*(M1)*(U1)²
Also,
K.E of the two objects after collision = ½*M1*V² + ½*M2*V²
Factorize V and ½ : ½*( M1 + M2 )V²
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
2.When A Moving Object Collides With A Stationary Object and They Move In The Same Direction With Different Velocities:
Momentum:
Form the conservation law of momentum:
Total Momentum of the two objects before collision:
= momentum of M1 + momentum of M2
= M1*U1 + M2*U2
Momentum of the two objects after collision:
= momentum of M1 + momentum of M2
= M1*V1 + M2*V2
Let us apply the conservation law of linear momentum:
Momentum of objects before collision = momentum of objects after collision
M1*U1 + M2*U2 = M1*V1 + M2*V2
Remember that U2 = 0 m/s ( the is at rest / stationary )
Then we will get
M1*U1 + M2* ( 0 ) = M1*V1 + M2*V2
M1*U1 + 0 = M1*V1 + M2*V2
M1*U1 = M1*V1 + M2*V2
At this point, you can make any of the variable the subject of formula:
Kinetic Energy:
Going by the concept of the conservation law of kinetic energy, the relationship between kinetic energy before collision and after collision can also be established.
Conservation law of energy state the total energy of an isolated / closed system is constant.
Therefore,
Kinetic energy of the two objects before collision = K.E of M1 + K.E of M2
= ½*(M1)*(U1)² + ½*(M2)*(U2)²
Also,
Kinetic energy of the two objects after collision = K.E of M1 + K.E of M2
= ½*M1*(V1)² + ½*M2*(V2)²
Total Kinetic before collision = Total kinetic energy after collision
½*(M1)*(U1)² + ½*(M2)*(U2)² = ½*M1*(V1²)² + ½*M2*(V2)²
Since M2 is stationary, U2 = 0 m/s. Then
½*(M1)*(U1)² + ½*(M2)*(0)² = ½*M1*(V1)² + ½*M2*(V2)²
½*(M1)*(U1)² + 0 = ½*M1*(V1)² + ½*M2*(V2)²
½*(M1)*(U1)² = ½*M1*(V1)² + ½*M2*(V2)²
Factorize : ½*(M1)*(U1)² = ½*(M1*(V1)² + M2*(V2)²)
½ in both sides cancels one another: (M1)*(U1)² = M1*(V1)²+ M2*(V2)²
At this point make any of the variable that you want to calculate, the .
For example, if you want to calculate V1, you have to make V1 the subject of the after you might have substituted for other variables.
Thus:
(M1)*(U1)² = M1*(V1)²+ M2*(V2)²
(M1)*(U1)² - M2*(V2)² = M1*(V1)²
Make (V1)² the subject: (V1)² = ( M1*(V1)² - M2*(V2)² ) / M1
Make V1 the subject: V1 = √ ((M1*(V1)² - M2*(V2)²) / M1 ).
In then same way, you make any variable that you want to calculate, subject of the formula.
How To Calculate Loss in Kinetic Energy:
Loss in kinetic energy can be calculated as follows:
• Calculate the kinetic energy of the two objects before collision
• Calculate the kinetic energy of the two objects after collision
• Then find the difference between the two kinetic energise to get the kinetic energy loss.
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
K.E of the two objects before collision = ½*(M1)*(U1)² + ½*(M2)*(0)²
= ½*(M1)*(U1)² + 0
. = ½*(M1)*(U1)²
Also,
K.E of the two objects after collision = ½*M1*(V1)² + ½*M2*(V2)²
Factorize ½ : ½*( M1*(V1)²+ M2*(V2)²)
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
Loss in kinetic energy = ½*M1*(U1)² - ½*( M1*(V1)²+ M2*(V2)²)
3.When Two Objects That Move In The Same Direction Collide, Stick Together And Move With A Common Velocity:
From the figure above, two objects moving in the same direction collide with each other, stick together and move with a common velocity, V.
From this case of collision, we can obtain important formula for momentum and kinetic energy, that are useful in calculation.
Momentum:
From the conservation law of linear momentum which states that, in a closed system,
Total momentum before collision = Total momentum after collision
Total Momentum of the two objects before collision:
= momentum of M1 + momentum of M2
= M1*U1 + M2*U2
Momentum of the two objects after collision:
= momentum of M1 + momentum of M2
= M1*V1 + M2*V2
We will substitute for the momenta in the conservation law of linear momentum.
Therefore,
M1*U1 + M2*U2 = M1*V1 + M2*V2
At this point, you make any of the variable that you want to calculate the subject of formula.
For example, if we want to calculate V2, we will make V2 subject of the formula thus:
M1*U1 + M2*U2 = M1*V1 + M2*V2
M2*U2 = M1*V1 + M2*V2 - M1*V1
Make U2 the subject: U2 = (M1*V1 + M2*V2 - M1*V1) / M2
Just like that.
Kinetic Energy:
Going by the concept of the conservation law of kinetic energy, the relationship between kinetic energy before collision and after collision can also be established.
Conservation law of energy state the total energy of an isolated / closed system is constant.
Therefore,
Kinetic energy of the two objects before collision = K.E of M1 + K.E of M2
= ½*(M1)*(U1)² + ½*(M2)*(U2)²
Also,
Kinetic energy of the two objects after collision = K.E of M1 + K.E of M2
= ½*M1*V² + ½*M2*V²
Total Kinetic before collision = Total kinetic energy after collision
½*(M1)*(U1)² + ½*(M2)*(U2)² = ½*M1*V² + ½*M2*V²
Factorize : ½*((M1)*(U1)² + (M1)*(U2)²) = ½*(M1*(V1)² + M2*(V2)²)
½ in both sides cancels one another:
(M1)*(U1)² + (M1)*(U2)² = M1*(V1)²+ M2*(V2)²
At this point make any of the variable that you want to calculate, subject of the formula.
For example, if you want to calculate V2, you have to make V2 the subject of the formula after you might have substituted for other variables.
Thus:
((M1)*(U1)² + (M1)*(U2)² = M1*(V1)²+ M2*(V2)²
Make (V2)² the subject: M2*(V2)² = (M1)*(U1)² + (M1)*(U2)² - M1*(V1)²
Make V2 the subject: (V2)² = ((M1)*(U1)² + (M1)*(U2)² - M1*(V1)²) / M2
V2 = √(((M1)*(U1)² + (M1)*(U2)² - M1*(V1)²) / M2)
In then same way, you make any variabIe that you want to calculate, subject of the formula.
How To Calculate Loss in Kinetic Energy:
Loss in kinetic energy can be calculated as follows:
• Calculate the kinetic energy of the two objects before collision
• Calculate the kinetic energy of the two objects after collision
• Then find the difference between the two kinetic energise to get the kinetic energy loss.
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
K.E of the two objects before collision = ½*(M1)*(U1)² + ½*(M2)*(U2)²
Factorize ½: = ½*((M1)*(U1)² + (M2)*(U2)²)
Also,
K.E of the two objects after collision = ½*M1*(V1)² + ½*M2*(V2)²
Factorize ½ : = ½*( M1*(V1)²+ M2*(V2)²)
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
Loss in kinetic energy = ½*M1*(U1)² - ½*( M1*(V1)²+ M2*(V2)²)
4.When Two Objects Move In Opposite Directions Collide and Move With Different Velocity In The Same Direction:
As usual, we can obtain important formulae from the conservation law of momentum and kinetic energy, that are very useful n solving problems related to momentum and kinetic energy.
Therefore,
For momentum:
From the conservation law of linear momentum which states that, in a closed system,
Total momentum before collision = Total momentum after collision
Total Momentum of the two objects before collision:
= momentum of M1 + ( - momentum of M2 )
= M1*U1 + (- M2*U2)
= M1*U1 - M2*U2
Momentum of the two objects after collision:
= momentum of M1 + momentum of M2
= M1*V1 + M2*V2
We will substitute for the momenta in the conservation law of linear momentum as follow:
Therefore,
M1*U1 - M2*U2 = M1*V1 + M2*V2
At this point, you make any of the variable that you want to calculate the subject of formula.
For example, if we want to calculate U2, we have to make U2 subject of the formula thus:
M1*U1 - M2*U2 = M1*V1 + M2*V2
- M2*U2 = M1*V1 + M2*V2 - M1*U1
U2 = ( M1*V1 + M2*V2 - M1*U1) / (- M2)
Make U2 the subject: U2 = (M1*V1 + M2*V2 - M1*V1) / M2
Just like that.
Kinetic Energy:
Going by the concept of the conservation law of kinetic energy, the relationship between kinetic energy before collision and after collision can also be established.
Conservation law of energy state the total energy of an isolated / closed system is constant.
Therefore,
Kinetic energy of the two objects before collision = K.E of M1 + ( - K.E of M2 )
= ½*(M1)*(U1)² + ( - ½*(M2)*(U2)² )
= ½*(M1)*(U1)² - ½*(M2)*(U2)²
Factorize ½: = ½ ((M1)*(U1)² - (M2)*(U2)²)
Also,
Kinetic energy of the two objects after collision = K.E of M1 + K.E of M2
= ½*M1*V1² + ½*M2*V2²
Total Kinetic before collision = Total kinetic energy after collision
½*(M1)*(U1)² - ½*(M2)*(U2)² = ½*M1*V1² + ½*M2*V2²
Factorize ½ : ½*((M1)*(U1)² - (M1)*(U2)²) = ½*(M1*(V1)² + M2*(V2)²)
½ in both sides cancels one another:
(M1)*(U1)² - (M1)*(U2)² = M1*(V1)²+ M2*(V2)²
At this point make any of the variable that you want to calculate, subject of the formula.
For example, if you want to calculate V2, you have to make V2 the subject of the formula after you might have substituted for other variables.
Thus:
((M1)*(U1)² - (M1)*(U2)² = M1*(V1)²+ M2*(V2)²
Make (V2)² the subject: M2*(V2)² = (M1)*(U1)² - (M1)*(U2)² - M1*(V1)²
Make V2 the subject: (V2)² = ((M1)*(U1)² - (M1)*(U2)² - M1*(V1)²) / M2
V2 = √(((M1)*(U1)² - (M1)*(U2)² - M1*(V1)²) / M2)
In then same way, you make any variable that you want to calculate, subject of the formula.
How To Calculate Loss in Kinetic Energy:
Loss in kinetic energy can be calculated as follows:
• Calculate the kinetic energy of the two objects before collision
• Calculate the kinetic energy of the two objects after collision
• Then find the difference between the two kinetic energise to get the kinetic energy loss.
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
K.E of the two objects before collision = ½*(M1)*(U1)² + ( - ½*(M2)*(U2)² )
= ½*(M1)*(U1)² - ½*(M2)*(U2)²
Factorize ½: = ½*((M1)*(U1)² - (M2)*(U2)²)
Also,
K.E of the two objects after collision = ½*M1*(V1)² + ½*M2*(V2)²
Factorize ½ : = ½*( M1*(V1)²+ M2*(V2)²)
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
Loss in kinetic energy =
= (½*(M1)*(U1)² - ½*(M2)*(U2)² ) – (½*( M1*(V1)²+ ½M2*(V2)²)
Factorize ½: = ½*( M1*(U1)² - M2*(U2)²) - ½*(M1*(V1)² - M2*(V2)²)
5.When Tow Objects Move In Opposite Directions,Collide, Stick Together And Move With A Common Velocity In The Same Direction:
As usual, we can obtain important formulae m1or m2 or u1 or u2 or v from the conservation law of momentum and kinetic energy. Any of the variable can be calculated when others are given in the question, by simply making it the subject of the formulaAn
Therefore,
For momentum:
From the conservation law of linear momentum which states that, in a closed system,
Total momentum before collision = Total momentum after collision.
Total Momentum of the two objects before collision:
= momentum of M1 + ( - momentum of M2 )
= M1*U1 + (- M2*U2)
= M1*U1 - M2*U2
Momentum of the two objects after collision:
= momentum of M1 + momentum of M2
= M1*V + M2*V
Factorize V: = ( M1 + M2 ) V
We will substitute for the momenta in the conservation law of linear momentum as follow:
Therefore,
M1*U1 - M2*U2 = ( M1 + M2 )*V
At this point, you make any of the variable that you want to calculate the subject of formula.
For example, if we want to calculate U2, we have to make U2 subject of the formula thus:
M1*U1 - M2*U2 = ( M1 + M2 )*V
- M2*U2 = ( M1 + M2 )*V - M1*U1
U2 = ( M1 + M2 )*V - M1*U1) / (- M2)
Make U2 the subject: U2 = (M1 + M2 )*V - M1*U1) / ( - M2)
Just like that.
Kinetic Energy:
Going by the concept of the conservation law of kinetic energy, the relationship between kinetic energy before collision and after collision can also be established.
Conservation law of energy state the total energy of an isolated / closed system is constant.
Therefore,
Kinetic energy of the two objects before collision = K.E of M1 + ( - K.E of M2 )
= ½*(M1)*(U1)² + ( - ½*(M2)*(U2)² )
= ½*(M1)*(U1)² - ½*(M2)*(U2)²
Factorize ½: = ½ ((M1)*(U1)² - (M2)*(U2)²)
Also,
Kinetic energy of the two objects after collision = K.E of M1 + K.E of M2
= ½*M1*V² + ½*M2*V²
Factorize ½ and V: = ½*(M1 + M2)*V²
Total Kinetic before collision = Total kinetic energy after collision
½*((M1)*(U1)² - (M2)*(U2)²) = ½*(M1 + M2)*V²
½ cancels ½: (M1)*(U1)² - (M1)*(U2)² = (M1 + M2)*V²
At this point make any of the variable that you want to calculate, subject of the formula.
For example, if you want to calculate V, you have to make V the subject of the formula after you might have substituted for other variables.
Thus:
(M1*(U1)² - M2*(U2)² = (M1 + M2)*V²
Make V² the subject: V² = ((M1)*(U1)² - (M2)*(U2)²) / (M1 + M2)
Make V the subject: V = √(((M1)*(U1)² - (M2)*(U2)²) / (M1 + M2))
In then same way, you make any variable that you want to calculate, subject of the formula.
How To Calculate Loss in Kinetic Energy:
Loss in kinetic energy can be calculated as follows:
• Calculate the kinetic energy of the two objects before collision
• Calculate the kinetic energy of the two objects after collision
• Then find the difference between the two kinetic energise to get the kinetic energy loss.
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
K.E of the two objects before collision = ½*(M1)*(U1)² + ( - ½*(M2)*(U2)² )
= ½*(M1)*(U1)² - ½*(M2)*(U2)²
Factorize ½: = ½*((M1)*(U1)² - (M2)*(U2)²)
Also,
K.E of the two objects after collision = ½*M1*V² + ½*M2*V²
Factorize ½ and V²: = ½*( M1 + M2 )*V²
Therefore,
Kinetic energy loss = K.E before collision – K.E after collision
Loss in kinetic energy =
= ½*((M1)*(U1)² - (M2)*(U2)² ) – ½*( M1 + M2 )*V²
Factorize ½: = ½*(((M1)*(U1)² - M2*(U2)²) - (M1 + M2)*V²)
6.When Two Objects Travel In The Same Direction Collide And Move In Opposite Direction With Different Velocities:
Application Of Newton’s Law And Conservation Law Of Momentum:
We will look into some areas in which the conservation laws of momentum and Newton’s laws are applied.
The principle of conservation of linear momentum is applied in the following which we will discuss below:
1. Recoil Of A Gun:
The principle of conservation of linear momentum is applied in the recoil of a gun when it is fired. When a gun is fired, the bullet is energized by a forward force ( action ) which moves it forward with a certain velocity. The bullet therefore has a forward momentum. As the bullet leaves the gun, it exert a backward force ( reaction ) on the gun. The gun then jerk backward with certain amount of velocity. In accordance with the conservation law of linear momentum, therefore the gun has a backward momentum which is equal in magnitude to the forward momentum of the fired bullet.
Let,
Mg be mass of gun
VG be velocity of gun
Mb be mass of bullet
Vg be velocity of bullet
Then,
momentum of gun = Mg*Vg
momentum of bullet = Mb*Vb
By principle of the law of conservation of linear momentum:
Forward momentum of Bullet = Backward momentum of gun
Therefore,
Mb*Vb = Mg*Vg
At this point, you make any of the variables that you want to calculate, subject of the formula.
For example, assuming we want to calculate Mg, we have to make Mg the subject of the formula.
Therefore,
Mb*Vb = Mg*Vg
Mb*Vb / Vg = Mg
Why is it that when a bullet is fired from a gun of mass m, the bullet travels with high velocity while the gun only jerk backward. It is becausethe gun is designed to have massive , the massiveness of the guns is experience.
Worked Examples:
*50 gram mass of bullet is fired from a gun of mass 25 kilogram with a velocity of 130 m/s. Calculate the recoil velocity of the rifle.
Solution:
Data given in the question:
Mass of gun = 2.5 kg = 2.5 * 1000 = 2500g, mass of bullet = 30g,
velocity of bullet = 130 m/s, velocity of gun = ?
formula= Mb*Vb = Mg*Vg
Substitution: 30 * 130 = 2500 * Vb
Make Vb subject of the formula: Vb = 30 * 130 / 2500. ➡ Vb = 1.56 m/s
*A gun of mass 2.4kg fired 0.23kg mass of . if the recoil velocity of the gun is 21.4 m/s, calculate the velocity of the bullet.
Solution:
Data given in the question:
Mass of gun = 2.4kg, mass of bullet = 0.23kg, velocity of gun = 21.4 m/s.
Formula: Mb*Vb = Mg*Vg
Substitution: 0.23*Vb = 2.4*21.4
Make Vb the subject: VB = 2.4*21.4 / 0.23. ➡ VB = 223.30 m/s
*A jet burns at the rate of 250 g/s. If all the gas is ejected in one direction at the rate of 500m/s, what is the maximum weight the can have so that it can move vertically upward?
Solution:
Data given in the question:
Mass of bass per second = 250 g/s =250 / 1000 = 0.25kg/s, velocity = 500 m/s
Formula: momentum = mass * velocity
Substitution: momentum = 0.25 * 500 = 125kgm/s
From Newton’s third law of motion,
Force = weight = 125 N
2. Jet and Rocket Propulsion:
The principle of conservation of linear momentum is also utilized in the propulsion of jet, aircraft and rockets.
Principle of Utilization Of Principle Of Conservation of Linear Momentum And Newton’s Third Law Of Motion In Jets, Aircraft and Rockets
Gases are contained in a combustion chamber of an aircraft, rocket and jets. When the gas is burnt in the chamber, it expands and exert pressure within the chamber. The hot gas is therefore expel through a nozzle at a very high speed from the rocket. The ejected gas has mass and velocity and therefore has momentum. The momentum of the ejected gas produce equal and opposite momentum on the rocket which propel the rocket forward.
3. Why Walking Is Possible:
Man walk by pushing his foot against the ground. According to the principle of conservation law of linear momentum and Newton’s third law of motion, as the man push the ground with his foot ( action force ), the ground in turn exerts an equal and opposite force ( reaction force) on the man. This force move the man forward.
Weight Of A Object Inside A Lift
When a man stand in a lift or elevator, two forces act on the man. The forces are:
I. The true weight of the man which act vertically downward (w = m*g)
II. The reaction force R of the floor of the lift on the man, which act upward.
We are going to look into different cases when the man is on the lift
Case 1: when the lift is stationary or moving with a constant velocity:
When the lift is stationary, the downward force due to the weight of the man is equal to the upward reaction force of the floor of the lift. The lift is stationary or moving with constant velocity. Its acceleration is zero. ( a = 0 m/s² )
Therefore,
Weight of man = Reaction of floor = m*g
It means that there is no unbalance force acting on the man because the man’s weight is cancelled by the upward reaction force of the floor of he lift. If the man stands on a spring scale, the scale will record the man’s true weight.
Case2: when the lift accelerates upward with acceleration a:
If the lift accelerates upward with acceleration a, there is unbalance force acting on the man. This means that the upward reaction force of the floor of the lift on the man is greater than the downward force due to the weight of the man on the floor of the lift.
From Newton’s second law of motion,
Force F = m*a
Since the lift move upward, it means that floor Reaction is greater than man’s weight ( I.e R > W )
Effective force that move the lift is the difference between Man’s weight and floor reaction.
Effective force F = R – W
Therefore R – W = m*a
R – m*g = m*a
R = m*a + m*g
Factorize m: R = m ( a + g )
R = m ( a + g ) is also regarded as the apparent weight of the man when the lift is accelerating upward. The scale will record a value that is greater than the mans true weight. ( apparent weight ).
That is apparent weight W = R = m(a + g)
At this point, you can calculate R, m or a. This we shall see in the worked example below.
Case 3: If the lift is accelerating downwards with an acceleration, a:
If the lift accelerates downwards with acceleration a, it means that there is unbalancing force that is acting on the man. It means that the downward force due to the man’s weight is greater than the upward reaction force of the lift floor. ( W > R )
Man’s weight > Reaction force of lift floor
From Newton’s second law of motion,
Force F = m*a
Since the lift accelerates downwards, it means that the man’s weight is greater than the reaction force of the lift floor. ( W > R )
The effective force is the difference between the mans weight and the reaction force of the lift.
Effective force F = man’s weight – reaction force of lift floor
F = W – R
If we substitute for F in F = m*a, we will get
W – R = m*a
M*g – R = m*a
M*g – m*a = R
Factorize m: m ( g – a ) = R. ➡ R = m ( g – a )
At this point, you can calculate m, R or a by making it the subject of the formula. This we shall see in the worked example below.
R = m ( g – a ) can also be called apparent weight of the man.
W' = R = m { g – a ).
The man will fill lighter in weight. The scale will record a value the is smaller than the true weight of the man.
Case 4: when the lift is falling freely:
If the lift accelerate downward with acceleration that is equal to acceleration due o gravity, we say the lift is falling freely. His situation happen when the rope / cable of the lift cut.
From Newton’s second law of motion,
Force F = m*a
W > R ( the lift is falling )
Effective force F = W – R
Then. W – R = m*a
M*g – R = m*a
M*g - m*a = R
Factorize m: m( g - a )
Recall that the lift is falling freely, then a = g . Therefore,
R = m ( g – g ) = ( 0 ) = 0 N
We can see that R = 0. The man appears to have no weight ( weightless ). The man and the floor of the lift are not exerting force on each other. The scale of the weighing instrument reads zero.
Weightlessness In Satellites:
People and object is a satellite that is orbiting close to the earth experience weightlessness.
Exercises:
1. A ball of mass 6 kg moving with a velocity of 10 m/s collides with another ball of equal mass at rest. If the two balls off after the impact, calculate their common velocity.
2. A ball of mass 6 kg moving with a velocity of 10 m/s collides with a 2.0 kg mass ball moving in the opposite direction with a velocity of 5 m/s . After the collision, the two balls coalesce and move in the same direction. Calculate their common velocity.
3. A constant force of 5 N act for 7 seconds on a mass of 7 kg initially at rest. Calculate final momentum .
4. A foot baller applied a force of 39N when taking a penalty kick for a period of 0.24 second. If the mass of the ball is 0.045 kg, calculate the speed with which the ball took off.
5. A body of mass 100 g moving with a velocity of 10 m/s collides with a wall. Of after the collision, it moves with a velocity of 3 m/s in the opposite direction, calculate the change in momentum.
6. Define impulse. How is the rate of change of momentum related to the force that act on the body?
7. Explain she the velocity of a recoiling gun is lesser than that of the bullet shot out of he gun.
8. A 30 g bullet moving at 250 m/s hits a bag of sand and come to rest in 0.011 second. What is the momentum of the bullet just before hitting the bag? Find the average force that stopped the bullet.
9. State Newton’s laws of motion. Derive from one of the relation between the momentum produce in a body and the force applied on the body.
10. State the conservation law of linear momentum. A 250 g riffle lay on a smooth table when its suddenly discharges , firing a bullet of 0.05 g with a speed of 450 m/s . calculate the recoil speed of the gun.
11. Distinguish between mass, weight, momentum and inertia. Describe an experiment to demonstrate that momentum is conserved in a collision.
12. An object of mass 1.3kg falls a distance of 5 m to a horizontal surface and rebounds to a vertical height of 3 m . calculate the change in momentum.
13. A man whose mass is 85 kg stand on a spring weighing machine inside a lift. What is the reading of the weighing machine when:
I) When the lift starts to ascend with an acceleration of 2.0 m/s²?
II) The lift is moving with a uniform velocity?
III) The lift is coming to rest with a retardation of 4.0 m/s²
14. Explain the principle underlying the propulsion of rocket.
15. A rocket of mass 5800kg containing a propellant gas of 2000kg is to be launched vertically. If the fuel is consumed at a steady rate of 70kg/s, calculate the least velocity of the exhaust gases if the rocket and content will just lift off the launching pad immediately after firing.
16. Distinguish between impulse and momentum. How are they related? Calculate the momentum of : (i) The moon of mass 7x10^²²kg travelling with a velocity of 1.25 m/s. (ii) a tanker of mass 7x10^⁴ kg travelling at a speed of 6 m/s. (iii) an aircraft of mass 3.5 x 10^³ kg travelling at a speed of 400 m/s
17. Distinguish between elastic collision and inelastic collision. (ii) inertial mass and weight. From Newton’s law of motion show that F = ma. ( relationship between mass, force and acceleration ).
18. A 15 kg monkey hang from a cord suspended from the ceiling of a lift. The cord can withstand a tension of 200N and brakes as the lift accelerates. What was the lift minimum acceleration ( in magnitude and direction) g = 10.0 m/s².
19. A tractor of mass 5 x10³ kg s used to tow a car f mass 2.5 x10³ kg. The tractor moves with a speed of 3.0 m/s just before the towing rope become taut . calculate the (I) the speed of the tractor immediately the rope becames taut. (ii) loss kinetic energy of the system just after the car has started moving. (iii) impulse in the rope when it jerk the car into motion.
The notes are well arranged and very easy to understand and explain
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