MOMENTUM

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MOMENTUM


Introduction

A runner illustrates the momentum in the image.

A runner runs fast means he increses his momentum, he slows down means decreases his momentum.

If we throw two stones of equal mass with different velocities, we observe that the effect of motion,they have, are different. Similarly, if two stones of different mass be thrown with a same velocity, the effect of motion, they possess ,are also different. Here, the property which brings difference in the effect of motion, in the language of science, it is termed as Momentum.

Here, we will learn momentum, its definition, mathematical and graphical representations, types, relation with energy , law of conservation of momentum, application and some problems with explanation.

What is momentum?

Magnitude of motion is called momentum. It is a fundamental concept in physics and it describes the motion of an object.It depends on both - mass and velocity of the object.
Momentum is measured by the product of mass and velocity of an object.


i.e momentum = mass x velocity


or, p = mv

Illustrative image of momentum.

Unit of momentum :-


As, p = m x v
therefore, SI unit of momentum = unit of m x unit of v
= kg x ms-1 = kgms-1

Momentum is a vector quantity

As, p = mv
, Here m is scalar but v is a vector. Hence , the product of scalar and a vector is a vector. So, momentum is a vector.
Direction of momentum is same as the direction of velocity.

Momentum is a useful quantity

Why is Momentum a Useful Quantity?

Momentum is a very useful physical quantity because it helps us describe the motion of objects more completely than velocity alone. It depends on both the mass and velocity of an object:

p = mv

An object with a large mass moving slowly may have the same momentum as a light object moving very fast. Therefore, momentum gives a better measure of the "quantity of motion" of a body.

Momentum is especially important because it obeys the Principle of Conservation of Momentum. In an isolated system, the total momentum remains constant before and after interactions such as collisions, explosions, recoil of guns, rocket propulsion, and nuclear reactions.

The concept of momentum helps scientists and engineers analyze and predict the motion of objects in many real-life situations, making it one of the most important quantities in physics.

Key Point: Momentum combines both mass and velocity into a single quantity and remains conserved in isolated systems, making it extremely useful for studying motion and interactions.

GRAPHICAL REPRESENTATION


(A) Momentum–Velocity Graph (Mass Constant)

Graph between momentum and velocity

The graph shows the relationship between momentum (p) and velocity (v) when the mass (m) of an object remains constant.

Momentum is given by the equation:

p = mv

Explanation of the Graph

  • The horizontal axis (X-axis) represents velocity (v).
  • The vertical axis (Y-axis) represents momentum (p).
  • The straight line passing through the origin indicates that momentum is directly proportional to velocity.
  • The condition m = constant means the mass of the object does not change.
  • As velocity increases, momentum increases in the same proportion.
  • If the velocity doubles, the momentum also doubles.
  • If the velocity triples, the momentum also triples.
  • The line passes through the origin because when velocity is zero, momentum is also zero.

Slope of the Graph

The slope of the momentum–velocity graph is:

Slope = Δp / Δv = m

Therefore, the slope of the graph represents the mass of the object. A larger mass would produce a steeper line, while a smaller mass would produce a less steep line.

Obviously,

The graph confirms that for a constant mass, momentum is directly proportional to velocity.
Hence, the momentum–velocity graph is a straight line passing through the origin, and its slope represents the mass of the object.


(B) Momentum–Mass Graph (Velocity Constant)

Graph between momentum and mass.

The graph shows the relationship between momentum (p) and mass (m) when the velocity (v) of an object remains constant.

Momentum is given by the equation:

p = mv

Explanation of the Graph

  • The horizontal axis (X-axis) represents mass (m).
  • The vertical axis (Y-axis) represents momentum (p).
  • The straight line passing through the origin indicates that momentum is directly proportional to mass.
  • The condition v = constant means the velocity of the object remains unchanged.
  • As mass increases, momentum increases in the same proportion.
  • If the mass doubles, the momentum also doubles.
  • If the mass triples, the momentum also triples.
  • The line passes through the origin because when mass is zero, momentum is also zero.

Slope of the Graph

The slope of the momentum–mass graph is:

Slope = Δp / Δm = v

Therefore, the slope of the graph represents the velocity of the object. A higher velocity produces a steeper line, while a lower velocity produces a less steep line.

Mathematical Interpretation

Since velocity is constant,

p ∝ m

Hence, momentum increases linearly with mass.

Obviously,

The graph confirms that for a constant velocity, momentum is directly proportional to mass.
Therefore, the momentum–mass graph is a straight line passing through the origin, and its slope represents the velocity of the object.


(C) Velocity–Mass Graph (Momentum Constant)

Graph between mass and velocity.

The graph shows the relationship between velocity (v) and mass (m) when the momentum (p) of an object remains constant.

Momentum is given by the equation:

p = mv

Since momentum is constant,

v = p/m

Explanation of the Graph

  • The horizontal axis (X-axis) represents mass (m).
  • The vertical axis (Y-axis) represents velocity (v).
  • The graph is a downward-curving rectangular hyperbola.
  • The condition p = constant means the momentum of the object does not change.
  • As mass increases, velocity decreases.
  • As mass decreases, velocity increases.
  • This shows that velocity is inversely proportional to mass when momentum remains constant.

Mathematical Interpretation

Since momentum is constant,

mv = constant

Therefore,

v ∝ 1/m

This inverse relationship produces the hyperbolic curve shown in the graph.

Examples

  • If the mass doubles, the velocity becomes half.
  • If the mass becomes three times larger, the velocity becomes one-third.
  • If the mass is reduced to half, the velocity doubles.

Real-Life Example

Consider a gun firing a bullet. The bullet has a very small mass, so it moves with a very high velocity. In contrast, the gun has a much larger mass, so its recoil velocity is comparatively very small. Both possess equal momentum in opposite directions.

Therefore,

The graph confirms that when momentum remains constant, velocity varies inversely with mass. Hence, the velocity–mass graph is a rectangular hyperbola, showing that heavier objects move more slowly and lighter objects move more quickly for the same momentum.

Principle of Conservation of momentum

According to this principle,

The total momentum of an isolated system remains constant.

or,

In the absence of external force, the total momentum of a system of particles remains conserved. The total linear momentum is the vector sum of the linear momenta of  all the particles of the system.

Here, we should notice that the momentum of an individual particle of a system may change, but the total momentum ,i.e., the sum of momenta of the constituting particles of a system remains constant.

Types of momentum

Momentum is the quantity of motion possessed by a body due to its mass and velocity. Depending on the nature of motion, momentum can be classified into two main types:

1. Linear Momentum

Linear momentum is the momentum possessed by an object moving in a straight line. It is the product of the mass and velocity of the object.

p = mv

Where:

  • p = Linear momentum
  • m = Mass of the body
  • v = Velocity of the body

Examples:

  • A moving car on a road.
  • A flying bullet.
  • A football kicked by a player.

2. Angular Momentum

Angular momentum is the momentum possessed by a body due to its rotational motion about an axis.

L = Iω

Where:

  • L = Angular momentum
  • I = Moment of inertia
  • ω = Angular velocity

Angular momentum remains conserved when no external torque acts on the system.

Examples:

  • The rotation of Earth about its axis.
  • A spinning top.
  • A skater spinning on ice.
Linear Momentum Angular Momentum
Associated with translational motion. Associated with rotational motion.
Formula: p = mv Formula: L = Iω
Vector quantity directed along velocity. Vector quantity directed along the axis of rotation.
Example: Moving car. Example: Rotating wheel.

Key Takeaway

Momentum is broadly classified into Linear Momentum and Angular Momentum. Linear momentum is associated with motion in a straight line, whereas angular momentum is associated with rotational motion. Both are conserved quantities and play a vital role in understanding the motion of objects.

Momentum and Kinetic energy

Linear Momentum, p = mv ............ (i)

Kinetic Energy, K = ½mv² ............ (ii)

K = ½ (m²v²/m)
K = (mv)²/(2m)

K = p²/(2m)
Therefore,
relationship between kinetic energy and momentum.

Numerical Problems Based on K = p²/2m and p = √(2mK)


Numerical 1

Problem:
A body has a momentum of 20 kg·m/s and a mass of 5 kg. Find its kinetic energy.

Given:
p = 20 kg·m/s
m = 5 kg

Formula:

K = p² / 2m

Solution:

K = (20)² / (2 × 5)

K = 400 / 10

K = 40 J

Answer: Kinetic Energy = 40 J


Numerical 2

Problem:
A particle of mass 4 kg possesses kinetic energy 72 J. Calculate its momentum.

Given:
m = 4 kg
K = 72 J

Formula:

p = √(2mK)

Solution:

p = √(2 × 4 × 72)

p = √576

p = 24 kg·m/s

Answer: Momentum = 24 kg·m/s


Numerical 3

Problem:
A truck has a momentum of 6000 kg·m/s and a mass of 1500 kg. Find its kinetic energy.

Given:
p = 6000 kg·m/s
m = 1500 kg

Formula:

K = p² / 2m

Solution:

K = (6000)² / (2 × 1500)

K = 36000000 / 3000

K = 12000 J

Answer: Kinetic Energy = 12000 J


Numerical 4

Problem:
A ball of mass 2 kg has kinetic energy 100 J. Find its momentum.

Given:
m = 2 kg
K = 100 J

Formula:

p = √(2mK)

Solution:

p = √(2 × 2 × 100)

p = √400

p = 20 kg·m/s

Answer: Momentum = 20 kg·m/s


Key Formulae:

K = p² / 2m

p = √(2mK)

Practical applications based on the law of conservation of momentum

1. Recoil of a gun:

Image shows the firing of a bullet and how the gun gets recoi

Recoil Velocity of a Gun and the Principle of Conservation of Momentum

When a gun is fired, the bullet moves forward at a very high speed. At the same time, the gun moves backward. This backward motion of the gun is called recoil, and the velocity with which the gun moves backward is known as recoil velocity.

The phenomenon of recoil can be explained using the Principle of Conservation of Linear Momentum, which states:

If no external force acts on a system, the total linear momentum of the system remains constant.

Understanding the Situation

Consider the gun and bullet as a single system. Before firing, both the gun and bullet are at rest.

Before firing:

Velocity of gun = 0

Velocity of bullet = 0

Total momentum = 0

Since the initial momentum of the system is zero, the total momentum after firing must also remain zero.

After Firing

When the trigger is pressed, the bullet moves forward with a high velocity. To conserve momentum, the gun acquires an equal amount of momentum in the opposite direction.

Let:

  • M = mass of gun
  • Vg = recoil velocity of gun
  • m = mass of bullet
  • Vb = velocity of bullet

Derivation of Recoil Velocity

According to the conservation of momentum:

Initial Momentum = Final Momentum

Since the gun-bullet system is initially at rest:

0 = Momentum of Bullet + Momentum of Gun
0 = mVb - MVg

(The negative sign indicates that the gun moves in the direction opposite to the bullet.)

Rearranging:

MVg = mVb
Vg = (mVb) / M
Expression for Recoil Velocity:
Vg = (mVb) / M

What Does the Formula Tell Us?

  • If the bullet mass increases, recoil velocity increases.
  • If the bullet speed increases, recoil velocity increases.
  • If the mass of the gun increases, recoil velocity decreases.
  • Heavy guns therefore produce less noticeable recoil than light guns.

Numerical Example

Given:

  • Mass of bullet (m) = 0.02 kg
  • Velocity of bullet (Vb) = 400 m/s
  • Mass of gun (M) = 4 kg

Using:

Vg = (mVb) / M
Vg = (0.02 × 400) / 4
Vg = 8 / 4
Vg = 2 m/s

Therefore, the gun recoils backward with a velocity of 2 m/s.

Why Recoil Occurs

The bullet gains forward momentum due to the explosive force generated inside the gun barrel. Since momentum must remain conserved, the gun gains an equal amount of momentum in the opposite direction. This backward motion is the recoil of the gun.

Key Takeaway

Before firing, the total momentum of the gun-bullet system is zero. After firing, the forward momentum of the bullet is exactly equal and opposite to the backward momentum of the gun. Therefore, recoil velocity is a direct consequence of the Principle of Conservation of Momentum.

2. Working of Rocket and Jetplanes

In brief:

Initially, the rocket and its fuel are at rest. So, their total momentum is zero. For rocket propulsion, the fuel is exploded. The burnt gases are allowed to escape through a nozzle with a very high downward velocity. The gases carry a large momentum in the downward direction. To conserve momentum, the rocket also acqires an equal momentum in the upward direction and hence starts moving upwards.

Labelled diagram of a rocket.

In detail:

Working of a Rocket on the Basis of the Principle of Conservation of Momentum

A rocket moves forward by applying the Principle of Conservation of Momentum. This principle states that:

"When no external force acts on a system, the total momentum of the system remains constant."

How Does a Rocket Move?

As shown in the labelled diagrams, a rocket contains a fuel tank, an oxidizer tank, a combustion chamber, and an engine nozzle.

Inside the combustion chamber, fuel reacts with the oxidizer and produces a large amount of hot gases at very high speed. These gases are expelled backward through the nozzle.

Since the gases are pushed backward with high momentum, the rocket acquires an equal amount of momentum in the forward direction. As a result, the rocket moves upward or forward.

Application of Conservation of Momentum

Before ignition, both the rocket and the fuel are at rest. Therefore, the total momentum of the system is:

Total Momentum = 0

When the rocket engine starts, exhaust gases are ejected backward with momentum. To keep the total momentum constant, the rocket gains an equal momentum in the opposite direction.

Momentum of Rocket + Momentum of Exhaust Gases = 0

Thus,

mrvr = mgvg

where,

  • mr = mass of rocket
  • vr = velocity of rocket
  • mg = mass of exhaust gases
  • vg = velocity of exhaust gases

Role of Different Parts of the Rocket

  • Nose Cone: Reduces air resistance and protects the payload.
  • Payload: Carries satellites, instruments, or astronauts.
  • Fuel Tank: Stores the fuel required for combustion.
  • Oxidizer Tank: Supplies oxygen for burning fuel in space.
  • Combustion Chamber: Fuel and oxidizer react to produce high-temperature gases.
  • Engine Nozzle: Directs exhaust gases backward at very high speed.
  • Fins: Provide stability and directional control.

Why Can a Rocket Move in Space?

Unlike airplanes, rockets do not require air for propulsion. They carry both fuel and oxidizer with them. Therefore, even in the vacuum of space, rockets can expel gases backward and move forward according to the conservation of momentum.

Real-Life Observation

The backward movement of exhaust gases and the forward motion of the rocket are examples of equal and opposite momentum changes. The greater the speed of the expelled gases, the greater the thrust produced by the rocket.

Key Takeaway

A rocket works on the Principle of Conservation of Momentum. When high-speed gases are expelled backward through the nozzle, the rocket gains an equal amount of momentum in the forward direction, causing it to move upward or forward.

3. Astronaut in open space

In brief:

An astronaut in open space, who wants to return to the spaceship, throws some object in a direction opposite to the direction of motion of the spaceship. By doing so, he gains a momentum equal and opposite to that of the thrown object and so he moves towards the spaceship.

Astronaut in an open space and the principle of conservation of momentum.

In detail:

Astronaut in Space and the Principle of Conservation of Momentum

The image illustrates how an astronaut can move in space by using the Principle of Conservation of Momentum. In outer space, there is almost no air, so an astronaut cannot move by walking or pushing against the air. Instead, the astronaut uses a small thruster that ejects gas backward. As the gas moves backward, the astronaut moves forward with equal and opposite momentum.

Principle of Conservation of Momentum:
When no external force acts on a system, the total momentum of the system remains constant.

Step 1: Initial State

Initially, both the astronaut and the spacecraft are at rest. Therefore, the velocity of the astronaut is zero and the total momentum of the system is also zero.

Total Momentum = 0

Step 2: Gas is Expelled Backward

The astronaut activates a small thruster attached to the spacesuit. The thruster ejects gas backward at high speed. The expelled gas carries momentum in the backward direction.

Step 3: Equal and Opposite Momentum

According to the conservation of momentum, if the gas gains momentum in the backward direction, the astronaut must gain an equal amount of momentum in the forward direction.

Thus, the backward momentum of the gas produces an equal and opposite forward momentum on the astronaut.

Step 4: Astronaut Moves Forward

As a result of the reaction force produced by the expelled gas, the astronaut starts moving toward the spacecraft. The astronaut now possesses forward momentum.

Step 5: Motion Continues

Once the astronaut begins moving, the motion continues even after the gas stops being expelled because there is very little resistance in space. The astronaut keeps moving toward the spacecraft.

Step 6: Mathematical Explanation

Before the gas is expelled:

0 = mava + mgvg

Since the total momentum before the motion was zero, the total momentum after the motion must also remain zero.

mava = - mgvg

where,

  • ma = mass of astronaut
  • va = velocity of astronaut
  • mg = mass of expelled gas
  • vg = velocity of expelled gas

The negative sign indicates that the gas and the astronaut move in opposite directions.

Why This Happens in Space

Many people think that a rocket or astronaut needs air to push against. This is not true. The astronaut moves because momentum is conserved. The gas is pushed backward, and the astronaut gains an equal amount of forward momentum. Therefore, motion is possible even in the vacuum of space.

Key Takeaway

When the astronaut expels gas backward, the gas gains backward momentum and the astronaut gains equal forward momentum. This is a practical application of the Principle of Conservation of Momentum, which enables astronauts to maneuver in space.

4. While firing a bullet, the gun should be held tight to the shoulder

The recoiling gun can hurt the shoulder. If the gun is held tightly against the shoulder, then the body and the gun together will constitute one system. Total mass becomes large and the recoil velocity becomes small.

The gun is held tight against the shoulder, while firing, principle of conservation of momentum.

Why Should a Gun Be Held Tightly Against the Shoulder While Firing?

The image shows a gun being fired while it is firmly pressed against the shooter's shoulder. This practice is important because of the Principle of Conservation of Momentum. It not only improves stability and accuracy but also reduces the effect of recoil on the shooter.

Principle of Conservation of Momentum:
In the absence of external forces, the total momentum of a system remains constant.

What Happens When the Gun Is Fired?

Before firing, both the gun and the bullet are at rest. Therefore, the total momentum of the gun-bullet system is zero.

Initial Momentum = 0

When the trigger is pressed, the bullet moves forward with a very high velocity. As a result, the bullet gains a large forward momentum.

Momentum of Bullet = mbvb

According to the conservation of momentum, the gun must gain an equal amount of momentum in the opposite direction.

mbvb = Mgvg

This backward motion of the gun is called recoil.

Why Hold the Gun Tightly to the Shoulder?

If the gun is not pressed firmly against the shoulder, the gun alone experiences the recoil. Since only the gun's mass resists the backward motion, it recoils more sharply and can strike the shoulder forcefully.

However, when the gun is held tightly against the shoulder, the gun and the shooter's body effectively act as a single system.

The effective mass resisting recoil becomes:

Mass of Gun + Mass of Shooter

Since recoil velocity is inversely proportional to mass, increasing the effective mass reduces the recoil velocity.

Recoil Velocity ∝ 1 / Mass

Effect of a Larger Effective Mass

When the gun is tightly supported by the shoulder:

  • The recoil momentum is shared by both the gun and the shooter.
  • The backward velocity becomes much smaller.
  • The force felt by the shoulder is reduced.
  • The gun remains more stable.
  • The shooter can aim more accurately.

Momentum Explanation Using the Image

In the image, the bullet moves forward with a large momentum. To conserve momentum, the gun tends to move backward. Because the gun is firmly pressed against the shoulder, the recoil momentum is distributed over the combined mass of the gun and the shooter. Consequently, the backward motion becomes less noticeable and easier to control.

Forward Momentum of Bullet
=
Backward Momentum of Gun + Shooter

Real-Life Importance

  • Reduces shoulder injury due to recoil.
  • Improves shooting accuracy.
  • Provides better control of the firearm.
  • Prevents sudden backward jerks.
  • Makes repeated firing more comfortable.

Key Takeaway

A gun should always be held tightly against the shoulder because the backward recoil produced during firing is shared by the combined mass of the gun and the shooter. According to the Principle of Conservation of Momentum, increasing the effective mass reduces the recoil velocity, making firing safer, more stable, and more accurate.

Frequently Asked Questions (FAQs) on Conservation of Momentum

1. Why does a gun recoil on firing a bullet?

According to the Principle of Conservation of Momentum, the total momentum of the gun-bullet system must remain constant. Before firing, both the gun and bullet are at rest, so the total momentum is zero.

When the bullet is fired, it acquires a large forward momentum. To balance this momentum, the gun acquires an equal amount of momentum in the opposite direction. As a result, the gun moves backward. This backward motion is called recoil.

Forward Momentum of Bullet = Backward Momentum of Gun

2. Why is it advisable to hold a gun tight to one's shoulder when it is being fired?

When a gun is held tightly against the shoulder, the gun and the shooter effectively behave as a single body. This increases the effective mass resisting recoil.

Since recoil velocity is inversely proportional to mass, the larger combined mass of the gun and shooter produces a smaller recoil velocity.

As a result:

  • Recoil becomes less severe.
  • The gun is easier to control.
  • Accuracy improves.
  • The possibility of shoulder injury is reduced.
Greater Effective Mass → Smaller Recoil Velocity

3. A meteorite burns in the atmosphere before it reaches the Earth's surface. What happens to its momentum?

As the meteorite enters the atmosphere, it experiences strong air resistance and friction. These external forces oppose its motion and continuously reduce its velocity.

Since momentum is given by:

p = mv

A decrease in velocity causes a decrease in momentum. Moreover, some mass is also lost due to burning and vaporization, which further reduces the momentum of the meteorite.

Therefore, the momentum of the meteorite gradually decreases as it burns in the atmosphere.

4. Why does a heavy rifle not kick as strongly as a light rifle using the same cartridge?

Both rifles fire the same bullet with nearly the same forward momentum. According to the conservation of momentum, each rifle must acquire an equal backward momentum.

However, recoil velocity is given by:

Vrecoil = (mbullet × vbullet) / Mrifle

Since a heavy rifle has a larger mass, its recoil velocity is smaller. Therefore, it produces less kick and feels more comfortable to the shooter.

Heavier Rifle → Smaller Recoil Velocity → Less Kick

5. A body of mass 1 kg initially at rest explodes and breaks into three fragments of masses in the ratio 1 : 1 : 3. The two equal fragments fly off perpendicular to each other with a speed of 30 m/s each. What is the velocity of the heavier fragment?

Given:

  • Total mass = 1 kg
  • Mass ratio = 1 : 1 : 3
  • Total parts = 5

Mass of each smaller fragment:

m = 1/5 kg = 0.2 kg

Mass of heavier fragment:

M = 3/5 kg = 0.6 kg

Momentum of each smaller fragment:

p = mv = 0.2 × 30 = 6 kg·m/s

Since they move at right angles:

Resultant Momentum = √(6² + 6²)
= 6√2 kg·m/s

To conserve momentum, the heavier fragment must have equal momentum in the opposite direction.

0.6v = 6√2
v = (6√2)/0.6
v = 10√2 m/s
v ≈ 14.14 m/s

Answer: The heavier fragment moves with a velocity of approximately 14.14 m/s.

6. A 30 kg shell is flying at 48 m/s. When it explodes, one part of 18 kg stops, while the remaining part flies on. Find the velocity of the latter.

Given:

  • Mass of shell = 30 kg
  • Initial velocity = 48 m/s
  • One fragment = 18 kg and comes to rest

Remaining mass:

30 − 18 = 12 kg

Initial momentum:

P = 30 × 48
P = 1440 kg·m/s

Using conservation of momentum:

1440 = (18 × 0) + (12 × v)
1440 = 12v
v = 1440/12
v = 120 m/s

Answer: The remaining fragment moves with a velocity of 120 m/s.

Quick Summary

  • Recoil occurs due to conservation of momentum.
  • Holding a gun tightly reduces recoil velocity.
  • Atmospheric friction decreases a meteorite's momentum.
  • Heavy rifles have smaller recoil velocities.
  • Explosion problems are solved using conservation of momentum.
  • Total momentum remains constant when no external force acts on the system.

Additional Numerical Problems on Conservation of Momentum

7. A hunter has a machine gun that can fire 50 g bullets with a velocity of 150 m/s. A 60 kg tiger springs at him with a velocity of 10 m/s. How many bullets must the hunter fire into the tiger to stop him in his track?

Given:

  • Mass of tiger, M = 60 kg
  • Velocity of tiger, V = 10 m/s
  • Mass of each bullet, m = 50 g = 0.05 kg
  • Velocity of bullet, v = 150 m/s

Momentum of tiger:

Ptiger = MV = 60 × 10 = 600 kg·m/s

Momentum of one bullet:

Pbullet = mv = 0.05 × 150 = 7.5 kg·m/s

Let the required number of bullets be n. To stop the tiger, total momentum of bullets must equal the momentum of the tiger.

n × 7.5 = 600
n = 600/7.5 = 80

Answer: 80 bullets must be fired to stop the tiger.

8. A 40 kg shell is flying at a speed of 72 km/h. It explodes into two pieces. One piece of mass 15 kg stops. Calculate the speed of the other piece.

Given:

  • Total mass = 40 kg
  • Initial speed = 72 km/h = 20 m/s
  • One fragment mass = 15 kg
  • This fragment comes to rest

Mass of second fragment:

40 − 15 = 25 kg

Initial momentum:

P = 40 × 20 = 800 kg·m/s

Applying conservation of momentum:

800 = (15 × 0) + 25v
v = 800/25
v = 32 m/s

Answer: The speed of the second fragment is 32 m/s.

9. A bomb at rest explodes into three fragments of equal masses. Two fragments fly off at right angles to each other with velocities 9 m/s and 12 m/s respectively. Calculate the speed of the third fragment.

Given:

  • Bomb initially at rest
  • Three equal fragments
  • Velocities of two fragments = 9 m/s and 12 m/s
  • Directions are perpendicular

Since all fragments have equal mass, we can use velocity vectors directly.

Resultant velocity of first two fragments:

v = √(9² + 12²)
v = √(81 + 144)
v = √225
v = 15 m/s

To keep total momentum zero, the third fragment must move with equal momentum in the opposite direction. Since masses are equal:

Speed of third fragment = 15 m/s

Answer: The third fragment moves with a speed of 15 m/s.

10. A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei, the products must move in opposite directions.

Initially, the nucleus is at rest. Therefore, its momentum is:

Pinitial = 0

Suppose after disintegration the two daughter nuclei have masses m1 and m2, and velocities v1 and v2.

According to the Principle of Conservation of Momentum:

0 = m₁v₁ + m₂v₂
m₁v₁ = −m₂v₂

The negative sign indicates that the two momenta are in opposite directions. Hence, the two nuclei must move away from each other along opposite directions so that the total momentum remains zero.

Since the original nucleus had zero momentum, the daughter nuclei must possess equal and opposite momenta.

Conclusion: The products of nuclear disintegration move in opposite directions.

Key Takeaway

  • Total momentum of an isolated system always remains constant.
  • Explosion problems are solved using conservation of momentum.
  • Momentum before interaction = Momentum after interaction.
  • When an object initially at rest breaks apart, the vector sum of all fragment momenta remains zero.
  • Equal and opposite momenta are a direct consequence of momentum conservation.

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